ML Aggarwal Class 9 Maths Solutions Chapter 04 Factorization

Exercise 4.1

Question 1(i). Factorise the following:
\( 8xy^3 + 12x^2y^2 \)
Answer: The H.C.F. of \( 8xy^3 \) and \( 12x^2y^2 \) is \( 4xy^2 \).

\( 8xy^3 + 12x^2y^2 = 4xy^2(2y + 3x) \)
In simple words: Find the biggest number and letters that divide both terms. That is the common factor. Write it outside, and put what is left inside the brackets.

Exam Tip: Always identify the H.C.F. of all coefficients and variables before factoring out - this ensures you pull out the largest possible common factor.

Question 1(ii). Factorise the following:
\( 15ax^3 - 9ax^2 \)
Answer: The H.C.F. of \( 15ax^3 \) and \( 9ax^2 \) is \( 3ax^2 \).

\( 15ax^3 - 9ax^2 = 3ax^2(5x - 3) \)
In simple words: The common factor is \( 3ax^2 \). When you divide each term by this, you get what remains in the parentheses.

Exam Tip: Check your answer by multiplying the common factor back through the brackets - you should get the original expression.

Question 2(i). Factorise the following:
\( 21py^2 - 56py \)
Answer: The H.C.F. of \( 21py^2 \) and \( 56py \) is \( 7py \).

\( 21py^2 - 56py = 7py(3y - 8) \)
In simple words: Both terms share a common factor of \( 7py \). Factor it out to simplify the expression.

Exam Tip: When subtracting factored terms, ensure the signs are correct inside the brackets - here, \( 21py^2 \) divided by \( 7py \) gives \( 3y \), and \( 56py \) divided by \( 7py \) gives \( 8 \).

Question 2(ii). Factorise the following:
\( 4x^3 - 6x^2 \)
Answer: The H.C.F. of \( 4x^3 \) and \( 6x^2 \) is \( 2x^2 \).

\( 4x^3 - 6x^2 = 2x^2(2x - 3) \)
In simple words: Pull out the common factor \( 2x^2 \) from both terms. What is left inside the brackets is \( 2x - 3 \).

Exam Tip: Always factor out the highest power of any repeated variable to ensure the H.C.F. is truly the greatest common factor.

Question 3(i). Factorise the following:
\( 2\pi r^2 - 4\pi r \)
Answer: The H.C.F. of \( 2\pi r^2 \) and \( 4\pi r \) is \( 2\pi r \).

\( 2\pi r^2 - 4\pi r = 2\pi r(r - 2) \)
In simple words: The common factor includes \( \pi \) along with the numbers and the letter. Take \( 2\pi r \) out front.

Exam Tip: Do not overlook constants like \( \pi \) when finding the H.C.F. - they are part of the common factor too.

Question 3(ii). Factorise the following:
\( 18m + 16n \)
Answer: The H.C.F. of \( 18m \) and \( 16n \) is \( 2 \).

\( 18m + 16n = 2(9m + 8n) \)
In simple words: Both 18 and 16 can be divided by 2. So 2 is the common factor. Inside the brackets, you write \( 9m + 8n \).

Exam Tip: When variables are different, focus on the numerical coefficients to find the H.C.F. of the entire expression.

Question 4(i). Factorise the following:
\( 25abc^2 - 15a^2b^2c \)
Answer: The H.C.F. of \( 25abc^2 \) and \( 15a^2b^2c \) is \( 5abc \).

\( 25abc^2 - 15a^2b^2c = 5abc(5c - 3ab) \)
In simple words: Look at each letter and number separately. For the letters, take the smallest power that appears in both terms. That gives you the common factor.

Exam Tip: When finding H.C.F. with multiple variables, take the lowest exponent of each variable that appears in all terms.

Question 4(ii). Factorise the following:
\( 28p^2q^2r - 42pq^2r^2 \)
Answer: The H.C.F. of \( 28p^2q^2r \) and \( 42pq^2r^2 \) is \( 14pq^2r \).

\( 28p^2q^2r - 42pq^2r^2 = 14pq^2r(2p - 3r) \)
In simple words: The common factor includes 14, and the letters \( p \), \( q \), and \( r \) each taken with their smallest power across both terms.

Exam Tip: Break down each coefficient into prime factors first - this helps you spot the H.C.F. of the numbers quickly.

Question 5(i). Factorise the following:
\( 8x^3 - 6x^2 + 10x \)
Answer: The H.C.F. of \( 8x^3 \), \( 6x^2 \), and \( 10x \) is \( 2x \).

\( 8x^3 - 6x^2 + 10x = 2x(4x^2 - 3x + 5) \)
In simple words: All three terms share the factor \( 2x \). When you divide each term by \( 2x \), the results go inside the brackets.

Exam Tip: With three or more terms, find the H.C.F. by checking what divides all of them evenly - do not stop after examining just two.

Question 5(ii). Factorise the following:
\( 14mn + 22m - 62p \)
Answer: The H.C.F. of \( 14mn \), \( 22m \), and \( 62p \) is \( 2 \).

\( 14mn + 22m - 62p = 2(7mn + 11m - 31p) \)
In simple words: Even though the terms have different variables, they all share 2 as a common factor. Pull it out and divide each term by 2.

Exam Tip: When variables differ across terms, rely on the numerical coefficients to identify the common factor.

Question 6(i). Factorise the following:
\( 18p^2q^2 - 24pq^2 + 30p^2q \)
Answer: The H.C.F. of \( 18p^2q^2 \), \( 24pq^2 \), and \( 30p^2q \) is \( 6pq \).

\( 18p^2q^2 - 24pq^2 + 30p^2q = 6pq(3pq - 4q + 5p) \)
In simple words: Take out \( 6pq \) as the common factor from all three terms, and simplify what remains in the brackets.

Exam Tip: Always verify by multiplying the H.C.F. back through each term in the brackets - the result should match the original expression.

Question 6(ii). Factorise the following:
\( 27a^3b^3 - 18a^2b^3 + 75a^3b^2 \)
Answer: The H.C.F. of \( 27a^3b^3 \), \( 18a^2b^3 \), and \( 75a^3b^2 \) is \( 3a^2b^2 \).

\( 27a^3b^3 - 18a^2b^3 + 75a^3b^2 = 3a^2b^2(9ab - 6b + 25a) \)
In simple words: The common factor is \( 3a^2b^2 \). Divide each term by this to get the expression inside the brackets.

Exam Tip: For variables with exponents, select the lowest exponent of each variable present in all three terms to form part of the H.C.F.

Question 7(i). Factorise the following:
\( 15a(2p - 3q) - 10b(2p - 3q) \)
Answer: The H.C.F. of \( 15a(2p - 3q) \) and \( 10b(2p - 3q) \) is \( 5(2p - 3q) \).

\( 15a(2p - 3q) - 10b(2p - 3q) = 5(2p - 3q)(3a - 2b) \)
In simple words: Both terms have \( (2p - 3q) \) as a shared part. Factor this out along with the common number 5.

Exam Tip: Look for common bracketed expressions - these can be treated as single units when extracting the H.C.F.

Question 7(ii). Factorise the following:
\( 3a(x^2 + y^2) + 6b(x^2 + y^2) \)
Answer: The H.C.F. of \( 3a(x^2 + y^2) \) and \( 6b(x^2 + y^2) \) is \( 3(x^2 + y^2) \).

\( 3a(x^2 + y^2) + 6b(x^2 + y^2) = 3(x^2 + y^2)(a + 2b) \)
In simple words: The bracket \( (x^2 + y^2) \) appears in both terms. Pull it out with the number 3 that divides all coefficients.

Exam Tip: When the same expression appears in multiple terms, it forms part of the common factor - extract it systematically.

Question 8(i). Factorise the following:
\( 6(x + 2y)^3 + 8(x + 2y)^2 \)
Answer: The H.C.F. of \( 6(x + 2y)^3 \) and \( 8(x + 2y)^2 \) is \( 2(x + 2y)^2 \).

\( 6(x + 2y)^3 + 8(x + 2y)^2 = 2(x + 2y)^2(3(x + 2y) + 4) = 2(x + 2y)^2(3x + 6y + 4) \)
In simple words: The expression \( (x + 2y) \) appears in both terms. Take out the lower power, which is \( (x + 2y)^2 \), along with the H.C.F. of the numbers.

Exam Tip: When the same bracketed term appears with different powers, always extract the lowest power as part of the common factor.

Question 8(ii). Factorise the following:
\( 14(a - 3b)^3 - 21p(a - 3b) \)
Answer: The H.C.F. of \( 14(a - 3b)^3 \) and \( 21p(a - 3b) \) is \( 7(a - 3b) \).

\( 14(a - 3b)^3 - 21p(a - 3b) = 7(a - 3b)[2(a - 3b)^2 - 3p] \)
In simple words: Both terms share \( (a - 3b) \) and are divisible by 7. Factor these out, and the remaining parts go into square brackets.

Exam Tip: After factoring out the common part, simplify what remains inside the brackets - ensure the power of the bracketed term is reduced by 1 in each case.

Question 9(i). Factorise the following:
\( 10a(2p + q)^3 - 15b(2p + q)^2 + 35(2p + q) \)
Answer: The H.C.F. of \( 10a(2p + q)^3 \), \( 15b(2p + q)^2 \), and \( 35(2p + q) \) is \( 5(2p + q) \).

\( 10a(2p + q)^3 - 15b(2p + q)^2 + 35(2p + q) = 5(2p + q)[2a(2p + q)^2 - 3b(2p + q) + 7] \)
In simple words: All three terms have \( (2p + q) \) in common, and 5 divides 10, 15, and 35. Extract these together.

Exam Tip: With three terms sharing a common bracketed expression, factor it out along with the H.C.F. of the numerical coefficients for a complete factorization.

Question 9(ii). Factorise the following:
\( x(x^2 + y^2 - z^2) + y(-x^2 - y^2 + z^2) - z(x^2 + y^2 - z^2) \)
Answer: The given expression can be written as:

\( x(x^2 + y^2 - z^2) + y(-1)(x^2 + y^2 - z^2) - z(x^2 + y^2 - z^2) \)

\( = x(x^2 + y^2 - z^2) - y(x^2 + y^2 - z^2) - z(x^2 + y^2 - z^2) \)

The H.C.F. of \( x(x^2 + y^2 - z^2) \), \( y(x^2 + y^2 - z^2) \), and \( z(x^2 + y^2 - z^2) \) is \( (x^2 + y^2 - z^2) \).

\( x(x^2 + y^2 - z^2) + y(-x^2 - y^2 + z^2) - z(x^2 + y^2 - z^2) = (x^2 + y^2 - z^2)(x - y - z) \)
In simple words: The three-term bracket \( (x^2 + y^2 - z^2) \) is the common part. Pull it out completely to get the final answer.

Exam Tip: When a term is preceded by a negative sign, rewrite it with a -1 multiplied through - this helps clarify the common factor hidden inside.

Exercise 4.2

Question 1(i). Factorise the following:
\( x^2 + xy - x - y \)
Answer:
\( x^2 + xy - x - y = x(x + y) - 1(x + y) = (x + y)(x - 1) \)

Therefore, \( x^2 + xy - x - y = (x - 1)(x + y) \)
In simple words: Group the first two terms and the last two terms separately. Each group has its own common factor. Then take out the bracket that both groups share.

Exam Tip: Group terms in pairs and look for a common bracketed expression across the groups - this is the key to factoring by grouping.

Question 1(ii). Factorise the following:
\( y^2 - yz - 5y + 5z \)
Answer:
\( y^2 - yz - 5y + 5z = y(y - z) - 5(y - z) = (y - z)(y - 5) \)

Therefore, \( y^2 - yz - 5y + 5z = (y - z)(y - 5) \)
In simple words: Split into two groups: the first two terms and the last two terms. Factor each group, and then extract the common bracket.

Exam Tip: Always verify by expanding: \( (y - z)(y - 5) \) should equal the original expression when multiplied out.

Question 2(i). Factorise the following:
\( 5xy + 7y - 5y^2 - 7x \)
Answer: Rearranging the terms:

\( 5xy + 7y - 5y^2 - 7x = 5xy - 7x + 7y - 5y^2 = x(5y - 7) - y(5y - 7) = (x - y)(5y - 7) \)

Therefore, \( 5xy + 7y - 5y^2 - 7x = (x - y)(5y - 7) \)
In simple words: Put similar terms next to each other first. Then group them in pairs and find what they share. Pull out that shared bracket.

Exam Tip: Rearranging terms before grouping often makes the common factors more obvious - do not feel stuck with the original order.

Question 2(ii). Factorise the following:
\( 5p^2 - 8pq - 10p + 16q \)
Answer:
\( 5p^2 - 8pq - 10p + 16q = 5p^2 - 10p - 8pq + 16q = 5p(p - 2) - 8q(p - 2) = (5p - 8q)(p - 2) \)

Therefore, \( 5p^2 - 8pq - 10p + 16q = (5p - 8q)(p - 2) \)
In simple words: Rearrange so that related terms sit together. Group them, factor each group, and pull out the common bracket.

Exam Tip: Check that both groups yield the same bracketed term when factored - if they do not, rearrange and try again.

Question 3(i). Factorise the following:
\( a^2b - ab^2 + 3a - 3b \)
Answer:
\( a^2b - ab^2 + 3a - 3b = ab(a - b) + 3(a - b) = (ab + 3)(a - b) \)

Therefore, \( a^2b - ab^2 + 3a - 3b = (ab + 3)(a - b) \)
In simple words: The first two terms share \( ab \), and the last two share 3. Both pairs leave \( (a - b) \) inside their brackets. Factor this out.

Exam Tip: When grouping, ensure the bracketed expressions are identical - this signals you have correctly identified the common factor.

Question 3(ii). Factorise the following:
\( x^3 - 3x^2 + x - 3 \)
Answer:
\( x^3 - 3x^2 + x - 3 = x^2(x - 3) + 1(x - 3) = (x^2 + 1)(x - 3) \)

Therefore, \( x^3 - 3x^2 + x - 3 = (x^2 + 1)(x - 3) \)
In simple words: The first two terms have \( x^2 \) in common, and the last two terms have 1 in common. Both leave \( (x - 3) \) in the brackets.

Exam Tip: Remember that 1 is a valid common factor - do not skip extracting it from a pair like \( x - 3 \).

Question 4(i). Factorise the following:
\( 6xy^2 - 3xy - 10y + 5 \)
Answer:
\( 6xy^2 - 3xy - 10y + 5 = 3xy(2y - 1) - 5(2y - 1) = (2y - 1)(3xy - 5) \)

Therefore, \( 6xy^2 - 3xy - 10y + 5 = (2y - 1)(3xy - 5) \)
In simple words: Factor \( 3xy \) from the first two terms and 5 from the last two. Both groups have \( (2y - 1) \) left in their brackets.

Exam Tip: If the first grouping does not produce identical brackets, try pairing the terms differently before concluding the expression cannot be factored this way.

Question 4(ii). Factorise the following:
\( 3ax - 6ay - 8by + 4bx \)
Answer: Rearranging the terms:

\( 3ax - 6ay - 8by + 4bx = 3ax - 6ay + 4bx - 8by = 3a(x - 2y) + 4b(x - 2y) = (x - 2y)(3a + 4b) \)

Therefore, \( 3ax - 6ay - 8by + 4bx = (x - 2y)(3a + 4b) \)
In simple words: Rearrange so terms with \( a \) and \( 3 \) come together, and terms with \( b \) and \( 4 \) come together. Both groups will have \( (x - 2y) \) left inside the brackets.

Exam Tip: When the original order does not group nicely, always consider rearranging - the expression will often factor once terms are in a better sequence.

Question 5(i). Factorise the following:
\( 5px - 8qy + 4qx - 10py \)
Answer:
\( 5px - 8qy + 4qx - 10py = 5px + 4qx - 8qy - 10py = x(5p + 4q) - 2y(4q + 5p) = (5p + 4q)(x - 2y) \)

Therefore, \( 5px - 8qy + 4qx - 10py = (5p + 4q)(x - 2y) \)
In simple words: Rearrange and group terms with \( x \) together and terms with \( y \) together. Both groups will share the factor \( (5p + 4q) \).

Exam Tip: Notice that \( 4q + 5p \) is the same as \( 5p + 4q \) - recognizing this allows the final factorization to emerge cleanly.

Question 5(ii). Factorise the following:
\( 9a^2y - 9ay + 6a - 6 \)
Answer:
\( 9a^2y - 9ay + 6a - 6 = 9ay(a - 1) + 6(a - 1) = (a - 1)(9ay + 6) = (a - 1) \cdot 3 \cdot (3ay + 2) = 3(a - 1)(3ay + 2) \)

Therefore, \( 9a^2y - 9ay + 6a - 6 = 3(a - 1)(3ay + 2) \)
In simple words: Group the first two terms to pull out \( 9ay \), and the last two to pull out 6. Both leave \( (a - 1) \). Then take out the 3 that divides 9 and 6.

Exam Tip: After grouping, check if the resulting brackets can be factored further - often an additional common factor remains that must be extracted.

Question 6(i). Factorise the following:
\( 1 - a - b + ab \)
Answer:
\( 1 - a - b + ab = 1(1 - a) - b(1 - a) = (1 - a)(1 - b) \)

Therefore, \( 1 - a - b + ab = (1 - a)(1 - b) \)
In simple words: Group 1 and \( -a \) together (their common factor is 1), and \( -b \) and \( ab \) together. Both pairs leave \( (1 - a) \) inside.

Exam Tip: Be careful with negative signs during grouping - ensure you write \( -b(1 - a) \), not \( +b(1 - a) \), to maintain correctness.

Question 6(ii). Factorise the following:
\( a(a - 2b - c) + 2bc \)
Answer:
\( a(a - 2b - c) + 2bc = a^2 - 2ab - ac + 2bc = a^2 - ac - 2ab + 2bc = a(a - c) - 2b(a - c) = (a - 2b)(a - c) \)

Therefore, \( a(a - 2b - c) + 2bc = (a - 2b)(a - c) \)
In simple words: Expand the bracket first. Then rearrange and group so that \( (a - c) \) appears in both pairs. Factor this out along with what remains.

Exam Tip: When given a product plus extra terms, expand everything first - this often reveals the correct grouping order for factoring.

Question 7(i). Factorise the following:
\( x^2 + xy(1 + y) + y^3 \)
Answer:
\( x^2 + xy(1 + y) + y^3 = x^2 + xy + xy^2 + y^3 = x(x + y) + y^2(x + y) = (x + y)(x + y^2) \)

Therefore, \( x^2 + xy(1 + y) + y^3 = (x + y)(x + y^2) \)
In simple words: Expand the bracket in the middle term. Then group the first two terms and the last two. Both groups will share \( (x + y) \).

Exam Tip: When a term is a product involving brackets, always expand it as the first step before attempting to group and factor.

Question 7(ii). Factorise the following:
\( y^2 - xy(1 - x) - x^3 \)
Answer:
\( y^2 - xy(1 - x) - x^3 = y^2 - xy + x^2y - x^3 = y(y - x) + x^2(y - x) = (y - x)(y + x^2) \)

Therefore, \( y^2 - xy(1 - x) - x^3 = (y - x)(y + x^2) \)
In simple words: Expand the middle term. Rearrange if needed. Group the terms, and pull out the common bracket that both groups share.

Exam Tip: Pay close attention to signs when expanding - a negative sign in front of a bracketed term changes the sign of each term inside.

Question 8(i). Factorise the following:
\( ab^2 + (a - 1)b - 1 \)
Answer:
\( ab^2 + (a - 1)b - 1 = ab^2 + ab - b - 1 = ab(b + 1) - 1(b + 1) = (b + 1)(ab - 1) \)

Therefore, \( ab^2 + (a - 1)b - 1 = (b + 1)(ab - 1) \)
In simple words: Expand the \( (a - 1)b \) term. Group the first two terms and the last two. Both will have \( (b + 1) \) left in the brackets.

Exam Tip: After expanding any bracketed term, check whether grouping becomes simpler - usually the factors will emerge once the expression is fully written out.

Question 8(ii). Factorise the following:
\( 2a - 4b - xa + 2bx \)
Answer: Rearranging the terms:

\( 2a - 4b - xa + 2bx = 2a - xa - 4b + 2bx = a(2 - x) - 2b(2 - x) = (2 - x)(a - 2b) \)

Therefore, \( 2a - 4b - xa + 2bx = (2 - x)(a - 2b) \)
In simple words: Rearrange so that terms with \( a \) come first, then terms with \( b \). Factor each pair. Both will have \( (2 - x) \) inside the brackets.

Exam Tip: When rearranging, place terms strategically so that after factoring each group, an identical bracket emerges - this is the sign you have grouped correctly.

Question 8(ii). Factorise the following: 4a² - b² + 2a + b
Answer: Start by recognizing that \( 4a^2 - b^2 = (2a)^2 - b^2 \). Apply the difference of squares identity \( a^2 - b^2 = (a - b)(a + b) \) to get \( (2a - b)(2a + b) \). The expression becomes \( (2a - b)(2a + b) + 2a + b \). Factor out the common term \( (2a + b) \) from the last two terms: \( (2a - b)(2a + b) + (2a + b) = (2a + b)[(2a - b) + 1] = (2a + b)(2a - b + 1) \). Hence, \( 4a^2 - b^2 + 2a + b = (2a + b)(2a - b + 1) \).
In simple words: Break the expression into a difference of squares plus extra terms. Factor and find a common binomial to pull out.

Exam Tip: Look for the difference of squares first, then check if remaining terms share a factor with one of the resulting binomials - this often unlocks a clean final factorization.

Question 9(i). Factorise the following: a(a - 2) - b(b - 2)
Answer: Expand and rearrange the given expression as \( a^2 - 2a - b^2 + 2b \). Rewrite this by grouping the difference of squares and the remaining terms: \( a^2 - b^2 - 2a + 2b \). Using the identity \( a^2 - b^2 = (a - b)(a + b) \), factor the first part, then extract the common factor - 2 from the remaining terms: \( (a - b)(a + b) - 2(a - b) \). Finally, pull out the common factor (a - b) from both parts to arrive at the result.
In simple words: Group the terms into pairs, apply the difference of squares to one pair, and then find the common factor across the whole expression.

Exam Tip: Always look for ways to group terms that produce a difference of squares or a perfect square, then extract common factors systematically.

Question 9(ii). Factorise the following: a(a - 1) - b(b - 1)
Answer: Expand the expression to get \( a^2 - a - b^2 + b \). Reorganize by collecting the perfect square difference and the linear terms: \( a^2 - b^2 - a + b \). Apply the identity \( a^2 - b^2 = (a - b)(a + b) \) to the first portion. Rewrite - a + b as - (a - b) to identify the common factor. Extract (a - b) from the entire expression to get the final factored form.
In simple words: First expand, then rearrange to create a difference of squares plus a common term that you can pull out from the whole expression.

Exam Tip: When expanding products, immediately look to reorganize terms into groups where you can spot familiar algebraic identities.

Question 10(i). Factorise the following: 9 - x² + 2xy - y²
Answer: Break down the expression by splitting 2xy into two separate xy terms. Add and subtract 3x and 3y strategically to enable regrouping. Collect the terms into meaningful clusters: 9 - 3x + 3y combined with x and y cross-terms. Factor out 3 from the first group, x from the second group, and - y from the third group. Each group will contain the factor (3 - x + y), which can then be extracted. This yields the product of two binomials with one containing the sum and the other the difference.
In simple words: Rearrange terms by adding and subtracting helper values, then group them to create repeated factors you can pull out.

Exam Tip: When factoring complex expressions with multiple variables, look for opportunities to create identical factors by clever regrouping and helper terms.

Question 10(ii). Factorise the following: 9x⁴ - (x² + 2x + 1)
Answer: Recognize that the subtracted expression in parentheses equals \( (x + 1)^2 \). Rewrite the entire expression as \( (3x^2)^2 - (x + 1)^2 \). Apply the difference of squares identity \( a^2 - b^2 = (a - b)(a + b) \) with \( a = 3x^2 \) and \( b = x + 1 \). This produces the product of \( (3x^2 - x - 1) \) and \( (3x^2 + x + 1) \).
In simple words: First rewrite the binomial inside the parentheses as a perfect square, then use the difference of squares formula.

Exam Tip: Always check whether expressions inside parentheses or brackets match perfect square patterns - recognizing these immediately simplifies the problem.

Question 11(i). Factorise the following: 9x⁴ - x² - 12x - 36
Answer: Regroup the expression as \( 9x^4 - (x^2 + 12x + 36) \). Notice that \( x^2 + 12x + 36 \) matches the perfect square pattern \( (x + 6)^2 \) where the middle coefficient is \( 2 \times 6 \times x \). This allows you to rewrite the whole expression as \( (3x^2)^2 - (x + 6)^2 \). Using the difference of squares identity, factor this into two binomials: one combining both terms with a positive sign between them, and the other with a negative sign.
In simple words: Group the last three terms as a perfect square, then apply the difference of squares to factor the whole expression.

Exam Tip: When subtracting a polynomial expression, immediately check if the subtracted part is a perfect trinomial - this transforms the problem into a simple difference of squares.

Question 11(ii). Factorise the following: x³ - 5x² - x + 5
Answer: Factor by grouping. Extract \( x^2 \) from the first two terms and - 1 from the last two terms, producing \( x^2(x - 5) - 1(x - 5) \). Identify the common factor \( (x - 5) \) and extract it to get \( (x^2 - 1)(x - 5) \). Next, apply the difference of squares identity to \( x^2 - 1 \), which becomes \( (x - 1)(x + 1) \). The complete factorization is the product of three linear factors.
In simple words: Group pairs of terms, extract common factors from each pair, then factor the resulting expression further using the difference of squares.

Exam Tip: For cubic and higher-degree polynomials, always try grouping pairs of terms first - this often reveals hidden common factors and simpler binomials to work with.

Question 12(i). Factorise the following: a⁴ - b⁴ + 2b² - 1
Answer: Rearrange as \( a^4 - (b^4 - 2b^2 + 1) \). Observe that the expression in parentheses matches the perfect square pattern \( (b^2 - 1)^2 \). This rewrites the entire expression as \( (a^2)^2 - (b^2 - 1)^2 \). Apply the difference of squares identity to get \( (a^2 + b^2 - 1)(a^2 - b^2 + 1) \).
In simple words: Regroup terms so a perfect square emerges in the parentheses, then use the difference of squares on the overall expression.

Exam Tip: When rearranging terms by grouping, look for perfect squares inside parentheses or brackets - these allow you to apply the difference of squares identity to the whole expression.

Question 12(ii). Factorise the following: x³ - 25x
Answer: Extract the common factor x from all terms to obtain \( x(x^2 - 25) \). Recognize that \( x^2 - 25 \) is a difference of perfect squares: \( x^2 - 5^2 \). Apply the identity \( a^2 - b^2 = (a - b)(a + b) \) to factor this binomial into \( (x - 5)(x + 5) \). The complete factorization is the product of three factors.
In simple words: First pull out any common factors, then apply the difference of squares to the remaining binomial expression.

Exam Tip: Always extract the greatest common factor first - this simplifies the remaining expression and often reveals standard factoring patterns.

Question 13(i). Factorise the following: 2x⁴ - 32
Answer: Begin by factoring out the common factor of 2 to get \( 2(x^4 - 16) \). Recognize that \( x^4 - 16 \) can be written as \( (x^2)^2 - 4^2 \), a difference of perfect squares. Apply the identity to obtain \( 2(x^2 - 4)(x^2 + 4) \). The binomial \( x^2 - 4 \) is itself a difference of squares, which factors further to \( (x - 2)(x + 2) \). The expression \( x^2 + 4 \) cannot be factored over the reals. The complete factorization involves the constant 2, two linear factors from the second difference of squares, and the irreducible quadratic.
In simple words: Extract the common factor first, then apply the difference of squares to the fourth-degree term, and factor the resulting quadratic if possible.

Exam Tip: When you have a fourth-degree binomial, it often factors as a difference of two squares twice - first when \( a^4 - b^4 \) factors to \( (a^2 - b^2)(a^2 + b^2) \), and the first part factors again.

Question 13(ii). Factorise the following: a²(b + c) - (b + c)³
Answer: Extract the common factor \( (b + c) \) from both terms to get \( (b + c)[a^2 - (b + c)^2] \). The expression inside the brackets is a difference of perfect squares. Apply the identity \( a^2 - b^2 = (a - b)(a + b) \) with \( a = a \) and \( b = (b + c) \). This yields \( (b + c)(a - (b + c))(a + (b + c)) \). Simplify the second and third factors to \( (a - b - c) \) and \( (a + b + c) \) respectively.
In simple words: Pull out the repeated factor (b + c) first, then use the difference of squares formula on what remains inside the brackets.

Exam Tip: Whenever the same multi-term expression appears in different parts of a problem, immediately extract it as a common factor.

Question 14(i). Factorise the following: (a + b)³ - a - b
Answer: Rewrite the second part as \( (a + b) \) to get \( (a + b)^3 - (a + b) \). Extract the common factor \( (a + b) \) to obtain \( (a + b)[(a + b)^2 - 1] \). Apply the difference of squares identity to \( (a + b)^2 - 1 \), factoring it as \( ((a + b) - 1)((a + b) + 1) \) or \( (a + b - 1)(a + b + 1) \). The complete factorization involves the original binomial and two linear factors.
In simple words: Treat (a + b) as a single unit, factor out that unit, then apply the difference of squares formula.

Exam Tip: Always look for opportunities to treat multi-term expressions as single units - this simplifies extraction of common factors significantly.

Question 14(ii). Factorise the following: x² - 2xy + y² - a² - 2ab - b²
Answer: Recognize that the first three terms form the perfect square \( (x - y)^2 \) and the last three terms form \( (a + b)^2 \). Rewrite the expression as \( (x - y)^2 - (a + b)^2 \). Apply the difference of squares identity with \( a = (x - y) \) and \( b = (a + b) \). This produces \( ((x - y) - (a + b))((x - y) + (a + b)) \). Simplify the factors by removing parentheses and combining like terms to get \( (x - y - a - b)(x - y + a + b) \).
In simple words: Recognize the perfect square trinomials in the expression, then treat their difference as a difference of perfect squares.

Exam Tip: When an expression has more than four terms, look for groups that form perfect squares - combining these groups often leads to a simple difference of squares.

Question 15(i). Factorise the following: (a² - b²)(c² - d²) - 4abcd
Answer: Expand the first part by distributing to obtain \( a^2c^2 - a^2d^2 - b^2c^2 + b^2d^2 - 4abcd \). Rearrange by grouping into two perfect square trinomials. The first trinomial \( a^2c^2 + b^2d^2 - 2abcd \) is \( (ac - bd)^2 \), and the second trinomial \( a^2d^2 + b^2c^2 + 2abcd \) is \( (ad + bc)^2 \). Rewrite as the difference \( (ac - bd)^2 - (ad + bc)^2 \). Apply the difference of squares identity to obtain two factors, each obtained by taking the sum and difference of the two trinomial square roots.
In simple words: Expand and rearrange the terms to form two perfect square expressions, then use the difference of squares formula.

Exam Tip: For products of differences of squares with subtracted cross-terms, expand fully first, then look for hidden perfect square trinomials in the rearranged form.

Question 15(ii). Factorise the following: 4x² - y² - 3xy + 2x - 2y
Answer: Rewrite 4x² as x² + 3x² to help with regrouping. Organize the first part as \( (x^2 - y^2) + (3x^2 - 3xy) + (2x - 2y) \). Apply the difference of squares to the first group: \( (x - y)(x + y) \). Factor out 3x from the second group: \( 3x(x - y) \). Factor out 2 from the third group: \( 2(x - y) \). Now extract the common factor \( (x - y) \) from all three terms, leaving \( (x + y) + 3x + 2 \) which simplifies to \( (4x + y + 2) \).
In simple words: Add and subtract terms to enable clever regrouping, then factor out common elements from each group.

Exam Tip: When an expression mixes quadratic and linear terms, try regrouping the quadratic and linear parts separately to find hidden common factors.

Question 16(i). Factorise the following: x² + \(\frac{1}{x^2}\) - 11
Answer: Rearrange as \( x^2 + \frac{1}{x^2} - 2 - 9 \). Recognize that \( x^2 + \frac{1}{x^2} - 2 \) equals \( (x - \frac{1}{x})^2 \) by the perfect square formula. This transforms the expression to \( (x - \frac{1}{x})^2 - 3^2 \). Apply the difference of squares identity with \( a = (x - \frac{1}{x}) \) and \( b = 3 \). The result is \( ((x - \frac{1}{x}) - 3)((x - \frac{1}{x}) + 3) \), which simplifies to \( (x - \frac{1}{x} - 3)(x - \frac{1}{x} + 3) \).
In simple words: Rewrite the expression using a perfect square trinomial, then apply the difference of squares to the result.

Exam Tip: For expressions involving \(x + \frac{1}{x}\) or \(x - \frac{1}{x}\), always check whether part of the expression matches the perfect square formula for these terms.

Question 16(ii). Factorise the following: x⁴ + 5x² + 9
Answer: Rewrite the expression as \( x^4 + 6x^2 + 9 - x^2 \) by adding and subtracting x². Recognize that \( x^4 + 6x^2 + 9 \) is the perfect square \( (x^2 + 3)^2 \) because it matches the pattern \( (a + b)^2 = a^2 + 2ab + b^2 \) with \( a = x^2 \) and \( b = 3 \). The entire expression becomes \( (x^2 + 3)^2 - x^2 \). Apply the difference of squares identity to obtain \( (x^2 + 3 - x)(x^2 + 3 + x) \), which rearranges to \( (x^2 - x + 3)(x^2 + x + 3) \).
In simple words: Add and subtract x² to create a perfect square, then use the difference of squares formula.

Exam Tip: For expressions like \(x^4 + px^2 + q\), try writing it as \(x^4 + rx^2 + q \pm (r-p)x^2\) to form a perfect square trinomial.

Question 17. Factorise the following:
(i) x² + \(\frac{4}{x^2}\) - 5
(ii) x⁴ + 12x² + 11
Answer:
(i) Rearrange as \( x^2 + \frac{4}{x^2} - 4 - 1 \). The first three terms fit the perfect square pattern: \( x^2 + \frac{4}{x^2} - 2 \times x \times \frac{2}{x} = (x - \frac{2}{x})^2 \). Rewrite the entire expression as \( (x - \frac{2}{x})^2 - 1^2 \). Apply the difference of squares to get \( ((x - \frac{2}{x}) - 1)((x - \frac{2}{x}) + 1) \), which simplifies to \( (x - \frac{2}{x} - 1)(x - \frac{2}{x} + 1) \).

(ii) Rewrite as \( x^4 + 11x^2 + x^2 + 11 \). Group into two pairs: \( (x^4 + 11x^2) + (x^2 + 11) \). Factor out \( x^2 \) from the first pair and 1 from the second to get \( x^2(x^2 + 11) + 1(x^2 + 11) \). Extract the common factor \( (x^2 + 11) \) to obtain \( (x^2 + 11)(x^2 + 1) \).
In simple words: (i) Use the perfect square pattern for expressions with reciprocals, then apply difference of squares. (ii) Group terms carefully to find a repeated factor that can be extracted.

Exam Tip: For quartic expressions, try regrouping as two pairs of terms - often one pair factors as a product, and a common factor emerges across both groups.

Question 18(i). Factorise the following: a⁴ + b⁴ - 7a²b²
Answer: Rewrite as \( a^4 + b^4 + 2a^2b^2 - 9a^2b^2 \). The first three terms match the perfect square pattern \( (a^2)^2 + (b^2)^2 + 2(a^2)(b^2) = (a^2 + b^2)^2 \). The entire expression becomes \( (a^2 + b^2)^2 - (3ab)^2 \). Apply the difference of squares identity to obtain \( (a^2 + b^2 + 3ab)(a^2 + b^2 - 3ab) \).
In simple words: Add and subtract 2a²b² to form a perfect square trinomial, then use the difference of squares formula.

Exam Tip: For fourth-degree expressions in two variables, look for ways to rearrange into a perfect square trinomial minus another perfect square, which factors by the difference of squares identity.

Question 18(ii). Factorise the following: \( x^4 - 14x^2 + 1 \)
Answer: Start by rewriting the expression. Add and subtract \( 2x^2 \) to get \( x^4 + 2x^2 - 16x^2 + 1 \). Group the first three terms: \( x^4 + 2x^2 + 1 - 16x^2 = (x^2 + 1)^2 - (4x)^2 \). Using the difference of squares formula \( a^2 - b^2 = (a - b)(a + b) \), we get \( (x^2 + 1 - 4x)(x^2 + 1 + 4x) \). Rearranging gives us \( (x^2 - 4x + 1)(x^2 + 4x + 1) \).
In simple words: Break up the middle term cleverly to form a perfect square minus another square, then apply the difference of squares rule.

Exam Tip: Look for ways to create perfect square trinomials within the expression - this often unlocks the factorisation using difference of squares.

Question 19(i). Express each of the following as the difference of two squares: \( (x^2 - 5x + 7)(x^2 + 5x + 7) \)
Answer: Rearrange the terms by grouping: \( [(x^2 + 7) - 5x][(x^2 + 7) + 5x] \). This fits the form \( (a - b)(a + b) \) where \( a = x^2 + 7 \) and \( b = 5x \). Applying the difference of squares identity gives \( (x^2 + 7)^2 - (5x)^2 \).
In simple words: Reorder the parts so you can spot the pattern \( (a - b)(a + b) \), which always equals \( a^2 - b^2 \).

Exam Tip: Always look for a common grouping in both brackets - if one bracket has \( +5x \) and the other has \( -5x \), they can be rearranged as \( (a+b)(a-b) \).

Question 19(ii). Express each of the following as the difference of two squares: \( (x^2 - 5x + 7)(x^2 - 5x - 7) \)
Answer: Group the common part: \( [(x^2 - 5x) + 7][(x^2 - 5x) - 7] \). Set \( a = x^2 - 5x \) and \( b = 7 \). Using \( (a + b)(a - b) = a^2 - b^2 \), we get \( (x^2 - 5x)^2 - 7^2 \).
In simple words: When both brackets share the same terms with opposite signs attached, treat the shared part as your main piece \( a \).

Exam Tip: Identify the identical part that appears in both brackets first, then apply the formula - this makes the problem straightforward.

Question 19(iii). Express each of the following as the difference of two squares: \( (x^2 + 5x - 7)(x^2 - 5x + 7) \)
Answer: Rewrite as \( [x^2 + (5x - 7)][x^2 - (5x - 7)] \). This gives the form \( (a + b)(a - b) \) where \( a = x^2 \) and \( b = 5x - 7 \). Therefore, \( (x^2)^2 - (5x - 7)^2 \).
In simple words: Group the variable terms separately from the number terms, then match the pattern where one bracket adds and the other subtracts.

Exam Tip: When brackets look different at first glance, try bracketing portions together to reveal the hidden \( (a+b)(a-b) \) structure.

Question 20(i). Evaluate the following by using factors: \( (979)^2 - (21)^2 \)
Answer: Apply the difference of squares formula: \( (979)^2 - (21)^2 = (979 - 21)(979 + 21) = 958 \times 1000 = 958000 \).
In simple words: Use the shortcut formula instead of squaring the big numbers - just subtract and add them, then multiply.

Exam Tip: Always check if a numerical problem fits the difference of squares pattern - it saves calculation time and reduces errors.

Question 20(ii). Evaluate the following by using factors: \( (99.9)^2 - (0.1)^2 \)
Answer: Using \( a^2 - b^2 = (a - b)(a + b) \): \( (99.9)^2 - (0.1)^2 = (99.9 - 0.1)(99.9 + 0.1) = 99.8 \times 100 = 9980 \).
In simple words: Decimal numbers are just as easy - subtract to get one number, add to get another, then multiply them together.

Exam Tip: The difference of squares formula works perfectly for decimal and fractional values too - no special tricks needed.

Exercise 4.4

Question 1(i). Factorise the following: \( x^2 + 5x + 6 \)
Answer: Look for two numbers that add to 5 and multiply to 6. These are 2 and 3 since \( 2 + 3 = 5 \) and \( 2 \times 3 = 6 \). Split the middle term: \( x^2 + 2x + 3x + 6 = x(x + 2) + 3(x + 2) = (x + 3)(x + 2) \).
In simple words: Find two numbers that fit both conditions, split the middle term using them, then group to pull out common brackets.

Exam Tip: Always check your factorisation by expanding - if you get the original expression back, you're correct.

Question 1(ii). Factorise the following: \( x^2 - 8x + 7 \)
Answer: Find two numbers that sum to -8 and have a product of 7. These are -7 and -1 since \( -7 + (-1) = -8 \) and \( -7 \times (-1) = 7 \). Rewrite: \( x^2 - 7x - x + 7 = x(x - 7) - 1(x - 7) = (x - 1)(x - 7) \).
In simple words: When both the sum and product are negative, look for two negative numbers that work.

Exam Tip: Pay careful attention to signs - if the constant is positive but the middle term is negative, both numbers must be negative.

Question 2(i). Factorise the following: \( x^2 + 6x - 7 \)
Answer: Search for two numbers that add to 6 and multiply to -7. These are 7 and -1 since \( 7 + (-1) = 6 \) and \( 7 \times (-1) = -7 \). Split the expression: \( x^2 + 7x - x - 7 = x(x + 7) - 1(x + 7) = (x - 1)(x + 7) \).
In simple words: When the product is negative, one number must be positive and one negative - pick the pair that gives the right sum.

Exam Tip: The larger number in your pair should be positive if the middle coefficient is positive.

Question 2(ii). Factorise the following: \( y^2 + 7y - 18 \)
Answer: Find two numbers summing to 7 with a product of -18. These are 9 and -2 since \( 9 + (-2) = 7 \) and \( 9 \times (-2) = -18 \). Rewrite and factor: \( y^2 + 9y - 2y - 18 = y(y + 9) - 2(y + 9) = (y - 2)(y + 9) \).
In simple words: Use the same two-number search method - once you find them, split the middle term and group in pairs.

Exam Tip: Double-check by multiplying out: \( (y - 2)(y + 9) = y^2 + 9y - 2y - 18 = y^2 + 7y - 18 \). ✓

Question 3(i). Factorise the following: \( y^2 - 7y - 18 \)
Answer: Look for numbers that sum to -7 and give a product of -18. These are -9 and 2 since \( -9 + 2 = -7 \) and \( -9 \times 2 = -18 \). Break it down: \( y^2 - 9y + 2y - 18 = y(y - 9) + 2(y - 9) = (y - 9)(y + 2) \).
In simple words: With a negative product, expect one positive and one negative number in your pair.

Exam Tip: When the middle term is negative and the constant is negative, the larger (in absolute value) of your two numbers should be negative.

Question 3(ii). Factorise the following: \( a^2 - 3a - 54 \)
Answer: Find numbers adding to -3 that multiply to -54. These are -9 and 6 since \( -9 + 6 = -3 \) and \( -9 \times 6 = -54 \). Split and group: \( a^2 - 9a + 6a - 54 = a(a - 9) + 6(a - 9) = (a - 9)(a + 6) \).
In simple words: Always check: does the sum match the middle term and does the product match the last term?

Exam Tip: Verify: \( (a - 9)(a + 6) = a^2 + 6a - 9a - 54 = a^2 - 3a - 54 \). ✓

Question 4(i). Factorise the following: \( 2x^2 - 7x + 6 \)
Answer: For quadratics where the leading coefficient is not 1, find two numbers that sum to -7 and have a product of \( 2 \times 6 = 12 \). These are -4 and -3 since \( -4 + (-3) = -7 \) and \( -4 \times (-3) = 12 \). Rewrite: \( 2x^2 - 4x - 3x + 6 = 2x(x - 2) - 3(x - 2) = (x - 2)(2x - 3) \).
In simple words: Multiply the first and last coefficients to find the product you need - the process then stays the same.

Exam Tip: For \( ax^2 + bx + c \), use product = \( a \times c \) - this is the key difference from simpler quadratics.

Question 4(ii). Factorise the following: \( 6x^2 + 13x - 5 \)
Answer: Find numbers summing to 13 with a product of \( 6 \times (-5) = -30 \). These are 15 and -2 since \( 15 + (-2) = 13 \) and \( 15 \times (-2) = -30 \). Rewrite and factor: \( 6x^2 + 15x - 2x - 5 = 3x(2x + 5) - 1(2x + 5) = (2x + 5)(3x - 1) \).
In simple words: The process is identical - find your pair, split the middle, group by common factors, and simplify.

Exam Tip: Expand to verify: \( (2x + 5)(3x - 1) = 6x^2 - 2x + 15x - 5 = 6x^2 + 13x - 5 \). ✓

Question 5(i). Factorise the following: \( 6x^2 + 11x - 10 \)
Answer: The numbers that sum to 11 and multiply to \( 6 \times (-10) = -60 \) are 15 and -4 since \( 15 + (-4) = 11 \) and \( 15 \times (-4) = -60 \). Break it into parts: \( 6x^2 + 15x - 4x - 10 = 3x(2x + 5) - 2(2x + 5) = (2x + 5)(3x - 2) \).
In simple words: This method works for any quadratic - whether the leading coefficient is 1 or much larger.

Exam Tip: If factorisation seems tricky, remember: the middle term always splits into two parts using your discovered pair of numbers.

Question 5(ii). Factorise the following: \( 6x^2 - 7x - 3 \)
Answer: Look for numbers that sum to -7 and multiply to \( 6 \times (-3) = -18 \). These are -9 and 2 since \( -9 + 2 = -7 \) and \( -9 \times 2 = -18 \). Separate and group: \( 6x^2 - 9x + 2x - 3 = 3x(2x - 3) + 1(2x - 3) = (2x - 3)(3x + 1) \).
In simple words: Even when numbers are larger, the same steps apply - stay consistent with your method.

Exam Tip: Always extract the greatest common factor from each pair before finalising your brackets.

Question 6(i). Factorise the following: \( 2x^2 - x - 6 \)
Answer: Find numbers that sum to -1 and multiply to \( 2 \times (-6) = -12 \). These are -4 and 3 since \( -4 + 3 = -1 \) and \( -4 \times 3 = -12 \). Rewrite: \( 2x^2 - 4x + 3x - 6 = 2x(x - 2) + 3(x - 2) = (x - 2)(2x + 3) \).
In simple words: A sum of -1 means the numbers are close in value but opposite in sign - one is just one more than the other.

Exam Tip: Check by multiplying: \( (x - 2)(2x + 3) = 2x^2 + 3x - 4x - 6 = 2x^2 - x - 6 \). ✓

Question 6(ii). Factorise the following: \( 1 - 18y - 63y^2 \)
Answer: Rearrange in standard order if helpful. Find numbers summing to -18 with a product of \( 1 \times (-63) = -63 \). These are -21 and 3 since \( -21 + 3 = -18 \) and \( -21 \times 3 = -63 \). Break it down: \( 1 - 21y + 3y - 63y^2 = 1(1 - 21y) + 3y(1 - 21y) = (1 - 21y)(1 + 3y) \).
In simple words: Even when the constant term is first, the method stays the same - look at leading and constant coefficients to find your product.

Exam Tip: Verify the factorisation by expanding in any order - the result should match the original expression.

Question 7(i). Factorise the following: \( 2y^2 + y - 45 \)
Answer: Find two numbers that sum to 1 and give a product of \( 2 \times (-45) = -90 \). These are 10 and -9 since \( 10 + (-9) = 1 \) and \( 10 \times (-9) = -90 \). Split the middle: \( 2y^2 + 10y - 9y - 45 = 2y(y + 5) - 9(y + 5) = (y + 5)(2y - 9) \).
In simple words: A middle term of +1 (or just one of the variable) still needs the two-number approach - don't treat it as a special case.

Exam Tip: The pair must differ by 1 when the middle coefficient is exactly 1 - this is a quick check while searching.

Question 7(ii). Factorise the following: \( 5 - 4x - 12x^2 \)
Answer: Look for numbers summing to -4 that multiply to \( 5 \times (-12) = -60 \). These are -10 and 6 since \( -10 + 6 = -4 \) and \( -10 \times 6 = -60 \). Rewrite: \( 5 - 10x + 6x - 12x^2 = 5(1 - 2x) + 6x(1 - 2x) = (1 - 2x)(5 + 6x) \).
In simple words: When terms aren't in the usual order, rearrange mentally or on paper to help identify the pattern.

Exam Tip: Grouping is easier when you keep like terms together - write out the split middle term clearly.

Question 8(i). Factorise the following: \( x(12x + 7) - 10 \)
Answer: Expand first: \( x(12x + 7) - 10 = 12x^2 + 7x - 10 \). Now find numbers that sum to 7 with a product of \( 12 \times (-10) = -120 \). These are 15 and -8 since \( 15 + (-8) = 7 \) and \( 15 \times (-8) = -120 \). Rewrite: \( 12x^2 + 15x - 8x - 10 = 3x(4x + 5) - 2(4x + 5) = (4x + 5)(3x - 2) \).
In simple words: Always expand brackets first, then apply your usual factorisation steps to the resulting quadratic.

Exam Tip: After expanding, double-check your quadratic before searching for the number pair.

Question 8(ii). Factorise the following: \( (4 - x)^2 - 2x \)
Answer: Expand the square using \( (a - b)^2 = a^2 - 2ab + b^2 \): \( (4 - x)^2 - 2x = 16 - 8x + x^2 - 2x = x^2 - 10x + 16 \). Find numbers summing to -10 with a product of 16. These are -8 and -2 since \( -8 + (-2) = -10 \) and \( -8 \times (-2) = 16 \). Break it down: \( x^2 - 8x - 2x + 16 = x(x - 8) - 2(x - 8) = (x - 8)(x - 2) \).
In simple words: Expand everything first to get a clean quadratic, then use your standard factorisation method.

Exam Tip: Remember the perfect square formula when you see brackets squared - expand carefully to avoid sign errors.

Question 9(i). Factorise the following: \( 60x^2 - 70x - 30 \)
Answer: First, pull out the greatest common factor. All terms share a factor of 10: \( 60x^2 - 70x - 30 = 10(6x^2 - 7x - 3) \). Now factorise the quadratic inside. Find numbers summing to -7 with a product of \( 6 \times (-3) = -18 \). These are -9 and 2 since \( -9 + 2 = -7 \) and \( -9 \times 2 = -18 \). Rewrite: \( 6x^2 - 9x + 2x - 3 = 3x(2x - 3) + 1(2x - 3) = (2x - 3)(3x + 1) \). Therefore, \( 60x^2 - 70x - 30 = 10(2x - 3)(3x + 1) \).
In simple words: Always check if all terms share a common number or variable first - factor that out before tackling the quadratic.

Exam Tip: Factoring out a common term reduces the size of the numbers you're working with, making the rest easier and less error-prone.

Question 9(i). Factorise the following: 60x² - 70x - 30
Answer: Start by extracting the common factor: 60x² - 70x - 30 = 10(6x² - 7x - 3). To factor 6x² - 7x - 3, find two numbers that add to -7 and multiply to 6 × (-3) = -18. Testing values, (-9) and 2 work: (-9) + 2 = -7 and (-9) × 2 = -18.
∴ 6x² - 7x - 3 = 6x² - 9x + 2x - 3 = 3x(2x - 3) + 1(2x - 3) = (2x - 3)(3x + 1).
∴ 10(6x² - 7x - 3) = 10(2x - 3)(3x + 1).
Hence, 60x² - 70x - 30 = 10(2x - 3)(3x + 1).
In simple words: Pull out the common factor 10 first. Then break down what remains by finding two numbers that combine correctly.

Exam Tip: Always extract the GCF before applying the AC method - this simplifies the factoring process and reduces arithmetic errors.

Question 9(ii). Factorise the following: x² - 6xy - 7y²
Answer: To factor x² - 6xy - 7y², locate two numbers whose sum is -6 and product is -7. By inspection, (-7) and 1 work: (-7) + 1 = -6 and (-7) × 1 = -7.
∴ x² - 6xy - 7y² = x² - 7xy + xy - 7y² = x(x - 7y) + y(x - 7y) = (x - 7y)(x + y).
Hence, x² - 6xy - 7y² = (x - 7y)(x + y).
In simple words: Find two numbers that add to the middle term and multiply to give the first times the last term. Split the middle and group.

Exam Tip: When working with two variables, treat them together when calculating the product (first × last) - this keeps the process consistent.

Question 10(i). Factorise the following: 2x² + 13xy - 24y²
Answer: To factor 2x² + 13xy - 24y², find two numbers whose sum is 13 and product is 2 × (-24) = -48. Testing: 16 and (-3) work since 16 + (-3) = 13 and 16 × (-3) = -48.
∴ 2x² + 13xy - 24y² = 2x² + 16xy - 3xy - 24y² = 2x(x + 8y) - 3y(x + 8y) = (x + 8y)(2x - 3y).
Hence, 2x² + 13xy - 24y² = (x + 8y)(2x - 3y).
In simple words: Identify two numbers matching the sum and product requirements. Rewrite the middle term and factor by grouping.

Exam Tip: Show the split of the middle term explicitly - examiners look for this step as proof of your method.

Question 10(ii). Factorise the following: 6x² - 5xy - 6y²
Answer: To factor 6x² - 5xy - 6y², determine two numbers that sum to -5 and have a product of 6 × (-6) = -36. By trial, (-9) and 4 satisfy these conditions: (-9) + 4 = -5 and (-9) × 4 = -36.
∴ 6x² - 5xy - 6y² = 6x² - 9xy + 4xy - 6y² = 3x(2x - 3y) + 2y(2x - 3y) = (2x - 3y)(3x + 2y).
Hence, 6x² - 5xy - 6y² = (2x - 3y)(3x + 2y).
In simple words: Find the correct pair of numbers, split the middle term, and extract common factors from each group.

Exam Tip: Double-check your factoring by multiplying the two brackets back together - this catches errors immediately.

Question 11(i). Factorise the following: 5x² + 17xy - 12y²
Answer: To factor 5x² + 17xy - 12y², locate two numbers that add to 17 and multiply to 5 × (-12) = -60. Testing reveals 20 and (-3) work: 20 + (-3) = 17 and 20 × (-3) = -60.
∴ 5x² + 17xy - 12y² = 5x² + 20xy - 3xy - 12y² = 5x(x + 4y) - 3y(x + 4y) = (x + 4y)(5x - 3y).
Hence, 5x² + 17xy - 12y² = (x + 4y)(5x - 3y).
In simple words: Identify the pair of numbers matching both conditions. Break the middle term and factor out common parts from each pair.

Exam Tip: Verify by expanding - if your factors are correct, multiplying them yields the original expression.

Question 11(ii). Factorise the following: x²y² - 8xy - 48
Answer: To factor x²y² - 8xy - 48, locate two numbers that sum to -8 and multiply to -48. By inspection, (-12) and 4 satisfy: (-12) + 4 = -8 and (-12) × 4 = -48.
∴ x²y² - 8xy - 48 = x²y² - 12xy + 4xy - 48 = xy(xy - 12) + 4(xy - 12) = (xy - 12)(xy + 4).
Hence, x²y² - 8xy - 48 = (xy - 12)(xy + 4).
In simple words: Treat xy as a single unit. Find the numbers and rewrite the middle term, then factor out the common expression.

Exam Tip: Recognize when a product of variables can be treated as a single entity - this simplifies the entire process.

Question 12(i). Factorise the following: 2a²b² - 7ab - 30
Answer: To factor 2a²b² - 7ab - 30, identify two numbers that sum to -7 and multiply to 2 × (-30) = -60. Testing: (-12) and 5 work since (-12) + 5 = -7 and (-12) × 5 = -60.
∴ 2a²b² - 7ab - 30 = 2a²b² - 12ab + 5ab - 30 = 2ab(ab - 6) + 5(ab - 6) = (ab - 6)(2ab + 5).
Hence, 2a²b² - 7ab - 30 = (ab - 6)(2ab + 5).
In simple words: Treat ab as a single variable. Find your number pair and split the middle term for factoring.

Exam Tip: This pattern uses a composite variable (ab) - recognizing such substitutions saves time and reduces errors.

Question 12(ii). Factorise the following: a(2a - b) - b²
Answer: Start by expanding: a(2a - b) - b² = 2a² - ab - b². To factor 2a² - ab - b², find two numbers that sum to -1 and multiply to 2 × (-1) = -2. By trial, (-2) and 1 work: (-2) + 1 = -1 and (-2) × 1 = -2.
∴ 2a² - ab - b² = 2a² - 2ab + ab - b² = 2a(a - b) + b(a - b) = (a - b)(2a + b).
Hence, a(2a - b) - b² = (a - b)(2a + b).
In simple words: Expand first, then use the standard AC method. Find the pair of numbers and proceed as usual.

Exam Tip: Always expand bracket expressions before applying factoring techniques - this avoids confusion and keeps the method consistent.

Question 13(i). Factorise the following: (x - y)² - 6(x - y) + 5
Answer: Let (x - y) = a. Then the expression becomes a² - 6a + 5. To factor this, locate two numbers that sum to -6 and multiply to 5. Testing: (-5) and (-1) work since (-5) + (-1) = -6 and (-5) × (-1) = 5.
∴ a² - 6a + 5 = a² - 5a - a + 5 = a(a - 5) - 1(a - 5) = (a - 5)(a - 1).
Substituting back: (a - 5)(a - 1) = (x - y - 5)(x - y - 1).
Hence, (x - y)² - 6(x - y) + 5 = (x - y - 5)(x - y - 1).
In simple words: Replace the complicated expression with a single letter. Factor normally, then substitute back to get your final answer.

Exam Tip: Substitution simplifies expressions with repeated brackets - always reverse the substitution in your final answer.

Question 13(ii). Factorise the following: (2x - y)² - 11(2x - y) + 28
Answer: Let (2x - y) = a. The expression becomes a² - 11a + 28. Find two numbers that sum to -11 and multiply to 28. By inspection, (-7) and (-4) work: (-7) + (-4) = -11 and (-7) × (-4) = 28.
∴ a² - 11a + 28 = a² - 7a - 4a + 28 = a(a - 7) - 4(a - 7) = (a - 7)(a - 4).
Substituting back: (a - 7)(a - 4) = (2x - y - 7)(2x - y - 4).
Hence, (2x - y)² - 11(2x - y) + 28 = (2x - y - 7)(2x - y - 4).
In simple words: Use substitution to turn a complex bracket expression into a simpler quadratic. Factor and then replace the letter with the original expression.

Exam Tip: Always verify by checking one term after substituting back - this confirms your work is correct.

Question 14(i). Factorise the following: 4(a - 1)² - 4(a - 1) - 3
Answer: Let (a - 1) = t. The expression becomes 4t² - 4t - 3. To factor this, determine two numbers that sum to -4 and multiply to 4 × (-3) = -12. Testing: (-6) and 2 work since (-6) + 2 = -4 and (-6) × 2 = -12.
∴ 4t² - 4t - 3 = 4t² - 6t + 2t - 3 = 2t(2t - 3) + 1(2t - 3) = (2t - 3)(2t + 1).
Substituting back: [2(a - 1) - 3][2(a - 1) + 1] = (2a - 5)(2a - 1).
Hence, 4(a - 1)² - 4(a - 1) - 3 = (2a - 5)(2a - 1).
In simple words: Replace the bracket with a letter to make the quadratic simpler. After factoring, substitute the original expression back in.

Exam Tip: When substituting back, expand the brackets carefully to avoid sign errors.

Question 14(ii). Factorise the following: 1 - 2a - 2b - 3(a + b)²
Answer: Rewrite as 1 - 2(a + b) - 3(a + b)². Let (a + b) = t. The expression becomes 1 - 2t - 3t². To factor, locate two numbers that sum to -2 and multiply to 1 × (-3) = -3. By trial, 1 and (-3) work: 1 + (-3) = -2 and 1 × (-3) = -3.
∴ 1 - 2t - 3t² = 1 - 3t + t - 3t² = 1(1 - 3t) + t(1 - 3t) = (1 + t)(1 - 3t).
Substituting back: (1 + a + b)(1 - 3a - 3b).
Hence, 1 - 2a - 2b - 3(a + b)² = (1 + a + b)(1 - 3a - 3b).
In simple words: Group terms with the same bracket, replace the bracket with a letter, factor the result, and substitute back.

Exam Tip: Factor out the GCF from related terms before substituting - this makes the algebra cleaner.

Question 15(i). Factorise the following: 3 - 5a - 5b - 12(a + b)²
Answer: Rewrite as 3 - 5(a + b) - 12(a + b)². Let (a + b) = t. The expression becomes 3 - 5t - 12t². To factor, identify two numbers that sum to -5 and multiply to 3 × (-12) = -36. Testing: (-9) and 4 work since (-9) + 4 = -5 and (-9) × 4 = -36.
∴ 3 - 5t - 12t² = 3 - 9t + 4t - 12t² = 3(1 - 3t) + 4t(1 - 3t) = (1 - 3t)(3 + 4t).
Substituting back: [1 - 3(a + b)][3 + 4(a + b)] = (1 - 3a - 3b)(3 + 4a + 4b).
Hence, 3 - 5a - 5b - 12(a + b)² = (1 - 3a - 3b)(3 + 4a + 4b).
In simple words: Collect the same bracket expression, substitute with a letter, apply the AC method, and reverse the substitution.

Exam Tip: Be careful with sign changes when expanding brackets after substitution - a common source of mistakes.

Question 15(ii). Factorise the following: a⁴ - 11a² + 10
Answer: To factor a⁴ - 11a² + 10, identify two numbers that sum to -11 and multiply to 10. By trial, (-10) and (-1) work: (-10) + (-1) = -11 and (-10) × (-1) = 10.
∴ a⁴ - 11a² + 10 = a⁴ - 10a² - a² + 10 = a²(a² - 10) - 1(a² - 10) = (a² - 10)(a² - 1).
The term (a² - 1) is a difference of squares: a² - 1 = (a - 1)(a + 1).
Hence, a⁴ - 11a² + 10 = (a² - 10)(a - 1)(a + 1).
In simple words: Treat a² as a single unit and apply factoring. Then recognize that one of your factors is itself a difference of squares.

Exam Tip: Always check if your resulting factors can be factored further - difference of squares and other patterns may apply.

Question 16(i). Factorise the following: (x + 4)² - 5xy - 20y - 6y²
Answer: Rewrite as (x + 4)² - 5y(x + 4) - 6y². Let (x + 4) = t. The expression becomes t² - 5yt - 6y². To factor, find two numbers that sum to -5 and multiply to -6. By inspection, (-6) and 1 work: (-6) + 1 = -5 and (-6) × 1 = -6.
∴ t² - 5yt - 6y² = t² - 6yt + yt - 6y² = t(t - 6y) + y(t - 6y) = (t - 6y)(t + y).
Substituting back: (x + 4 - 6y)(x + 4 + y) = (x - 6y + 4)(x + y + 4).
Hence, (x + 4)² - 5xy - 20y - 6y² = (x + y + 4)(x - 6y + 4).
In simple words: Group terms that contain the same bracket. Substitute the bracket with a letter and factor as a quadratic in two variables.

Exam Tip: When reordering factors, ensure each contains the bracket expression - this is crucial for correctness.

Question 16(ii). Factorise the following: (x² - 2x)² - 23(x² - 2x) + 120
Answer: Let (x² - 2x) = t. The expression becomes t² - 23t + 120. To factor, locate two numbers that sum to -23 and multiply to 120. Testing: (-15) and (-8) work since (-15) + (-8) = -23 and (-15) × (-8) = 120.
∴ t² - 23t + 120 = t² - 15t - 8t + 120 = t(t - 15) - 8(t - 15) = (t - 15)(t - 8).
Substituting back: (x² - 2x - 15)(x² - 2x - 8).
For x² - 2x - 15: Find two numbers summing to -2 and multiplying to -15. These are (-5) and 3.
∴ x² - 2x - 15 = (x - 5)(x + 3).
For x² - 2x - 8: Find two numbers summing to -2 and multiplying to -8. These are (-4) and 2.
∴ x² - 2x - 8 = (x - 4)(x + 2).
Hence, (x² - 2x)² - 23(x² - 2x) + 120 = (x - 5)(x + 3)(x - 4)(x + 2).
In simple words: Use substitution for the complex part, factor the resulting quadratic, then factor each resulting quadratic separately.

Exam Tip: After substitution and initial factoring, check if each remaining factor can be broken down further - often your final answer has four linear factors.

Question 17. Factorise the following: 4(2a - 3)² - 3(2a - 3)(a - 1) - 7(a - 1)²
Answer: Let (2a - 3) = x and (a - 1) = y. The expression becomes 4x² - 3xy - 7y². To factor, identify two numbers that sum to -3 and multiply to 4 × (-7) = -28. Testing: (-7) and 4 work since (-7) + 4 = -3 and (-7) × 4 = -28.
∴ 4x² - 3xy - 7y² = 4x² - 7xy + 4xy - 7y² = x(4x - 7y) + y(4x - 7y) = (x + y)(4x - 7y).
Substituting back: [(2a - 3) + (a - 1)][4(2a - 3) - 7(a - 1)] = (3a - 4)(8a - 12 - 7a + 7) = (3a - 4)(a - 5).
Hence, 4(2a - 3)² - 3(2a - 3)(a - 1) - 7(a - 1)² = (3a - 4)(a - 5).
In simple words: Replace both bracketed parts with different letters. Factor the resulting two-variable quadratic, then substitute both expressions back.

Exam Tip: Use different substitution letters for different bracketed expressions - this keeps the algebra organized and reduces mistakes.

Question 18. Factorise the following: (2x² + 5x)(2x² + 5x - 19) + 84
Answer: Let 2x² + 5x = y. The expression becomes y(y - 19) + 84 = y² - 19y + 84. To factor, locate two numbers that sum to -19 and multiply to 84. Testing: (-12) and (-7) work since (-12) + (-7) = -19 and (-12) × (-7) = 84.
∴ y² - 19y + 84 = y² - 12y - 7y + 84 = y(y - 12) - 7(y - 12) = (y - 12)(y - 7).
Substituting back: (2x² + 5x - 12)(2x² + 5x - 7).
For 2x² + 5x - 12: Two numbers summing to 5 and multiplying to 2 × (-12) = -24 are 8 and (-3).
∴ 2x² + 5x - 12 = 2x² + 8x - 3x - 12 = 2x(x + 4) - 3(x + 4) = (2x - 3)(x + 4).
For 2x² + 5x - 7: Two numbers summing to 5 and multiplying to 2 × (-7) = -14 are 7 and (-2).
∴ 2x² + 5x - 7 = 2x² + 7x - 2x - 7 = x(2x + 7) - 1(2x + 7) = (2x + 7)(x - 1).
Hence, (2x² + 5x)(2x² + 5x - 19) + 84 = (2x - 3)(x + 4)(2x + 7)(x - 1).
In simple words: Notice the repeated expression, substitute it with a letter, expand and factor the result, then factor each remaining quadratic into linear factors.

Exam Tip: This technique of substituting repeated expressions dramatically simplifies seemingly complex products - always look for patterns.

Exercise 4.5

Question 1(i). Factorise the following: 8x³ + y³
Answer: Rewrite as (2x)³ + (y)³. Recognize this as a sum of cubes. Apply the formula a³ + b³ = (a + b)(a² - ab + b²).
∴ (2x)³ + (y)³ = (2x + y)[(2x)² - 2xy + (y)²] = (2x + y)(4x² - 2xy + y²).
Hence, 8x³ + y³ = (2x + y)(4x² - 2xy + y²).
In simple words: Express each term as a perfect cube. Use the sum of cubes formula to break it into a linear and a quadratic factor.

Exam Tip: Always check if each term is a perfect cube first - this tells you which formula (sum or difference) to apply.

Question 1(ii). Factorise the following: 64x³ - 125y³
Answer: Rewrite as (4x)³ - (5y)³. This matches the difference of cubes pattern. Apply the formula a³ - b³ = (a - b)(a² + ab + b²).
∴ (4x)³ - (5y)³ = (4x - 5y)[(4x)² + 4x(5y) + (5y)²] = (4x - 5y)(16x² + 20xy + 25y²).
Hence, 64x³ - 125y³ = (4x - 5y)(16x² + 20xy + 25y²).
In simple words: Identify perfect cubes in each term. Use the difference of cubes formula to create a linear factor and a quadratic factor.

Exam Tip: In the difference of cubes formula, the middle term of the quadratic is always positive, even though the original was a difference - don't make a sign error here.

Question 1. Factorise 64x³ + 1
Answer: We express 64x³ + 1 as (4x)³ + (1)³. Using the sum of cubes formula \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \), we get: \( (4x)^3 + (1)^3 = (4x + 1)[(4x)^2 - 4x \cdot 1 + 1^2] = (4x + 1)(16x^2 - 4x + 1) \). Therefore, 64x³ + 1 = (4x + 1)(16x² - 4x + 1).
In simple words: Write 64x³ + 1 as two cubes added together, then use the sum of cubes rule to break it into two factors.

Exam Tip: Always recognize perfect cube coefficients (64 = 4³, 8 = 2³, 27 = 3³) and rewrite the expression before applying the sum or difference formula.

Question 2(ii). Factorise 7a³ + 56b³
Answer: First, factor out the common term: 7a³ + 56b³ = 7(a³ + 8b³) = 7[(a)³ + (2b)³]. Apply the sum of cubes formula: \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \). Thus: \( 7[(a)^3 + (2b)^3] = 7(a + 2b)[a^2 - a \cdot 2b + (2b)^2] = 7(a + 2b)(a^2 - 2ab + 4b^2) \). Therefore, 7a³ + 56b³ = 7(a + 2b)(a² - 2ab + 4b²).
In simple words: Pull out the common factor 7 first. Then treat what remains as two cubes added together and use the sum of cubes formula.

Exam Tip: Always look for a common factor before applying factorization formulas - it often simplifies the algebra significantly.

Question 3(i). Factorise \( \frac{x^6}{343} + \frac{343}{x^6} \)
Answer: Rewrite the expression as \( \frac{x^6}{343} + \frac{343}{x^6} = \left(\frac{x^2}{7}\right)^3 + \left(\frac{7}{x^2}\right)^3 \). Using the sum of cubes formula \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \): \( \left(\frac{x^2}{7}\right)^3 + \left(\frac{7}{x^2}\right)^3 = \left(\frac{x^2}{7} + \frac{7}{x^2}\right)\left[\left(\frac{x^2}{7}\right)^2 - \frac{x^2}{7} \cdot \frac{7}{x^2} + \left(\frac{7}{x^2}\right)^2\right] = \left(\frac{x^2}{7} + \frac{7}{x^2}\right)\left(\frac{x^4}{49} - 1 + \frac{49}{x^4}\right) \). Therefore, \( \frac{x^6}{343} + \frac{343}{x^6} = \left(\frac{x^2}{7} + \frac{7}{x^2}\right)\left(\frac{x^4}{49} - 1 + \frac{49}{x^4}\right) \).
In simple words: Recognize that 343 = 7³. Rewrite both fractions as perfect cubes, then use the sum of cubes rule.

Exam Tip: When fractions appear in cube form, express them using the numerator and denominator as separate cubes - this reveals the factorization pattern.

Question 3(ii). Factorise \( 8x^3 - \frac{1}{27y^3} \)
Answer: Express as cubes: \( 8x^3 - \frac{1}{27y^3} = (2x)^3 - \left(\frac{1}{3y}\right)^3 \). Use the difference of cubes formula \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \): \( (2x)^3 - \left(\frac{1}{3y}\right)^3 = \left(2x - \frac{1}{3y}\right)\left[(2x)^2 + 2x \cdot \frac{1}{3y} + \left(\frac{1}{3y}\right)^2\right] = \left(2x - \frac{1}{3y}\right)\left(4x^2 + \frac{2x}{3y} + \frac{1}{9y^2}\right) \). Therefore, \( 8x^3 - \frac{1}{27y^3} = \left(2x - \frac{1}{3y}\right)\left(4x^2 + \frac{2x}{3y} + \frac{1}{9y^2}\right) \).
In simple words: Write 8x³ as (2x)³ and the fraction as a perfect cube. Then apply the difference of cubes formula.

Exam Tip: For difference of cubes, remember the middle term in the second factor is positive - this is the key difference from sum of cubes.

Question 4(i). Factorise x² + x⁵
Answer: Factor out x²: x² + x⁵ = x²(1 + x³) = x²[(1)³ + (x)³]. Use the sum of cubes formula: \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \). Therefore: \( x^2[(1)^3 + (x)^3] = x^2(1 + x)(1^2 - 1 \cdot x + x^2) = x^2(1 + x)(1 - x + x^2) \). Thus, x² + x⁵ = x²(1 + x)(1 - x + x²).
In simple words: Pull out x² as a common factor. The remaining part is 1 + x³, which factors using the sum of cubes rule.

Exam Tip: Always extract the highest common factor first before identifying cube patterns - it makes the remaining algebra cleaner.

Question 4(ii). Factorise 32x⁴ - 500x
Answer: Extract the common factor: 32x⁴ - 500x = 4x(8x³ - 125) = 4x[(2x)³ - 5³]. Apply the difference of cubes formula: \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \). Thus: \( 4x[(2x)^3 - 5^3] = 4x(2x - 5)[(2x)^2 + 2x \cdot 5 + 5^2] = 4x(2x - 5)(4x^2 + 10x + 25) \). Therefore, 32x⁴ - 500x = 4x(2x - 5)(4x² + 10x + 25).
In simple words: Factor out 4x first. Then recognize 8x³ = (2x)³ and 125 = 5³, and use the difference of cubes formula.

Exam Tip: For large coefficients, always check for common factors - they often reveal cube or power patterns in the remaining terms.

Question 5(i). Factorise 27x³y³ - 8
Answer: Recognize the cubes: 27x³y³ - 8 = (3xy)³ - (2)³. Apply the difference of cubes formula \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \): \( (3xy)^3 - (2)^3 = (3xy - 2)[(3xy)^2 + 3xy \cdot 2 + 2^2] = (3xy - 2)(9x^2y^2 + 6xy + 4) \). Therefore, 27x³y³ - 8 = (3xy - 2)(9x²y² + 6xy + 4).
In simple words: Notice that 27x³y³ is the cube of 3xy and 8 is the cube of 2. Use the difference of cubes to factor.

Exam Tip: When a product of variables is raised to the third power (like x³y³), recognize it as (xy)³ and apply the cube formulas.

Question 5(ii). Factorise 27(x + y)³ + 8(2x - y)³
Answer: Rewrite as cubes: 27(x + y)³ + 8(2x - y)³ = [3(x + y)]³ + [2(2x - y)]³. Using the sum of cubes formula: \( [3(x + y)]^3 + [2(2x - y)]^3 = [3(x + y) + 2(2x - y)]\{[3(x + y)]^2 - 3(x + y) \times 2(2x - y) + [2(2x - y)]^2\} \). Simplify the first factor: 3x + 3y + 4x - 2y = 7x + y. Expand the second factor: 9(x² + y² + 2xy) - 6(x + y)(2x - y) + 4(4x² + y² - 4xy). After careful algebraic simplification, this becomes: (7x + y)(13x² + 19y² - 4xy). Therefore, 27(x + y)³ + 8(2x - y)³ = (7x + y)(13x² + 19y² - 4xy).
In simple words: Recognize 27 = 3³ and 8 = 2³. Rewrite the expression with these coefficients as cubes of the binomial factors, then apply the sum of cubes formula.

Exam Tip: When coefficients are in front of binomial cubes, extract them as cube roots first - this reveals the true cube form needed for the formula.

Question 6(i). Factorise a³ + b³ + a + b
Answer: Use the sum of cubes formula: a³ + b³ = (a + b)(a² - ab + b²). Rewrite the expression: a³ + b³ + a + b = (a + b)(a² - ab + b²) + (a + b). Factor out (a + b): (a + b)(a² - ab + b² + 1). Therefore, a³ + b³ + a + b = (a + b)(a² - ab + b² + 1).
In simple words: Factor a³ + b³ using the sum of cubes rule. Then observe that (a + b) appears as a common factor in both terms of the result.

Exam Tip: Look for binomial factors that appear in multiple terms - these can be extracted as common factors after applying cube formulas.

Question 6(ii). Factorise a³ - b³ - a + b
Answer: Use the difference of cubes formula: a³ - b³ = (a - b)(a² + ab + b²). Rewrite: a³ - b³ - a + b = (a - b)(a² + ab + b²) - 1(a - b). Factor out (a - b): (a - b)(a² + ab + b² - 1). Therefore, a³ - b³ - a + b = (a - b)(a² + ab + b² - 1).
In simple words: Apply the difference of cubes formula to a³ - b³. Then extract (a - b) as a common factor from the entire expression.

Exam Tip: When terms follow a pattern like "cube term" - "linear term", always apply the cube formula first, then look for common binomial factors.

Question 7(i). Factorise x³ + x + 2
Answer: Rewrite by adding and subtracting 1: x³ + x + 2 = x³ + x + 1 + 1 = x³ + 1 + x + 1 = x³ + 1³ + x + 1. Apply the sum of cubes formula to x³ + 1³: \( x^3 + 1^3 = (x + 1)(x^2 - x + 1) \). Then: \( (x + 1)(x^2 - x + 1) + (x + 1) = (x + 1)(x^2 - x + 1 + 1) = (x + 1)(x^2 - x + 2) \). Therefore, x³ + x + 2 = (x + 1)(x² - x + 2).
In simple words: Split the constant 2 into 1 + 1. This lets you use the sum of cubes on x³ + 1, which creates a common factor with the remaining terms.

Exam Tip: When a cubic has no obvious cube pattern, try decomposing constants strategically to create one - this often reveals a hidden factorization.

Question 7(ii). Factorise a³ - a - 120
Answer: Rewrite by breaking the constant: a³ - a - 120 = a³ - a - 125 + 5 = a³ - 125 - a + 5 = (a³ - 5³) - (a - 5). Apply the difference of cubes: \( a^3 - 5^3 = (a - 5)(a^2 + 5a + 25) \). Thus: \( (a - 5)(a^2 + 5a + 25) - (a - 5) = (a - 5)(a^2 + 5a + 25 - 1) = (a - 5)(a^2 + 5a + 24) \). Therefore, a³ - a - 120 = (a - 5)(a² + 5a + 24).
In simple words: Rearrange the terms to form a³ - 5³, which creates a cube difference. This reveals (a - 5) as a factor that appears in both parts.

Exam Tip: For cubics minus a linear term, try finding which cube number is close to the constant - the difference often factors neatly.

Question 7(iii). Factorise a³ - 2a - 115
Answer: Rewrite the constant: a³ - 2a - 115 = a³ - 2a - 125 + 10 = a³ - 125 - 2a + 10 = (a³ - 5³) - 2(a - 5). Apply the difference of cubes: \( a^3 - 5^3 = (a - 5)(a^2 + 5a + 25) \). Thus: \( (a - 5)(a^2 + 5a + 25) - 2(a - 5) = (a - 5)[(a^2 + 5a + 25) - 2] = (a - 5)(a^2 + 5a + 23) \). Therefore, a³ - 2a - 115 = (a - 5)(a² + 5a + 23).
In simple words: Break 115 into 125 - 10. This allows you to create a difference of cubes and factor out (a - 5) from both remaining terms.

Exam Tip: The key is choosing the right constant decomposition - look for values whose cube appears nearby in the original expression.

Question 8(i). Factorise x³ + 6x² + 12x + 16
Answer: Rewrite: x³ + 6x² + 12x + 16 = x³ + 6x² + 12x + 8 + 8 = [x³ + 3(2)(x²) + 3(2²)(x) + 2³] + 8. Recognize this as the binomial cube expansion \( (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \) with a = x and b = 2: [(x + 2)³] + 8 = (x + 2)³ + 2³. Apply the sum of cubes formula: \( (x + 2)^3 + 2^3 = [(x + 2) + 2]\{[(x + 2)^2 - (x + 2)(2) + 2^2]\} = (x + 4)[x^2 + 4x + 4 - 2x - 4 + 4] = (x + 4)(x^2 + 2x + 4) \). Therefore, x³ + 6x² + 12x + 16 = (x + 4)(x² + 2x + 4).
In simple words: Recognize that the first four terms match the expansion of (x + 2)³. The 8 left over is 2³, so you have two cubes to add together.

Exam Tip: Learn the binomial cube expansion pattern - spotting it in a quartic helps you reduce the problem to a sum or difference of cubes.

Question 8(ii). Factorise a³ - 3a²b + 3ab² - 2b³
Answer: Rewrite by separating the last term: a³ - 3a²b + 3ab² - 2b³ = a³ - 3a²b + 3ab² - b³ - b³ = (a - b)³ - b³. Recognize the binomial cube \( (a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 \). Now apply the difference of cubes: \( (a - b)^3 - b^3 = [(a - b) - b]\{[(a - b)^2 + (a - b)(b) + b^2]\} = (a - 2b)[a^2 - 2ab + b^2 + ab - b^2 + b^2] = (a - 2b)(a^2 + b^2 - ab) \). Therefore, a³ - 3a²b + 3ab² - 2b³ = (a - 2b)(a² + b² - ab).
In simple words: The first four terms form (a - b)³. Separate 2b³ into b³ - b³ to leave a difference of two cubes, which then factors.

Exam Tip: Recognize when terms follow the binomial cube pattern - this transforms a seemingly complex expression into a difference of cubes.

Question 9(i). Factorise 2a³ + 16b³ - 5a - 10b
Answer: Group and factor: 2a³ + 16b³ - 5a - 10b = 2(a³ + 8b³) - 5(a + 2b) = 2[a³ + (2b)³] - 5(a + 2b). Apply the sum of cubes: \( 2[a^3 + (2b)^3] = 2(a + 2b)(a^2 - 2ab + 4b^2) \). Thus: \( 2(a + 2b)(a^2 - 2ab + 4b^2) - 5(a + 2b) = (a + 2b)[2(a^2 - 2ab + 4b^2) - 5] = (a + 2b)(2a^2 - 4ab + 8b^2 - 5) \). Therefore, 2a³ + 16b³ - 5a - 10b = (a + 2b)(2a² - 4ab + 8b² - 5).
In simple words: Separate the cubic terms from the linear terms. Factor each group, then extract the common binomial factor (a + 2b).

Exam Tip: For mixed expressions with cubic and linear terms, always try grouping by degree first - this often reveals a common binomial factor.

Question 9(ii). Factorise \( a^3 - \frac{1}{a^3} - 2a + \frac{2}{a} \)
Answer: Group the terms: \( a^3 - \frac{1}{a^3} - 2\left(a - \frac{1}{a}\right) \). Apply the difference of cubes to the first part: \( a^3 - \frac{1}{a^3} = \left(a - \frac{1}{a}\right)\left(a^2 + a \cdot \frac{1}{a} + \frac{1}{a^2}\right) = \left(a - \frac{1}{a}\right)\left(a^2 + 1 + \frac{1}{a^2}\right) \). Thus: \( \left(a - \frac{1}{a}\right)\left(a^2 + 1 + \frac{1}{a^2}\right) - 2\left(a - \frac{1}{a}\right) = \left(a - \frac{1}{a}\right)\left[a^2 + 1 + \frac{1}{a^2} - 2\right] = \left(a - \frac{1}{a}\right)\left(a^2 + \frac{1}{a^2} - 1\right) \). Therefore, \( a^3 - \frac{1}{a^3} - 2a + \frac{2}{a} = \left(a - \frac{1}{a}\right)\left(a^2 + \frac{1}{a^2} - 1\right) \).
In simple words: Recognize that \( a^3 - \frac{1}{a^3} \) is a difference of cubes. Factor it to reveal \( \left(a - \frac{1}{a}\right) \) as a common factor with the remaining terms.

Exam Tip: When reciprocal fractions appear, apply cube formulas carefully - the middle term in the factorization is always 1 when the second term is \( \frac{1}{a} \).

Question 10(i). Factorise a⁶ - b⁶
Answer: Express as a difference of squares and cubes: a⁶ - b⁶ = (a²)³ - (b²)³. Apply the difference of cubes: \( (a^2)^3 - (b^2)^3 = (a^2 - b^2)[(a^2)^2 + a^2b^2 + (b^2)^2] = (a^2 - b^2)(a^4 + a^2b^2 + b^4) \). Factor the first part as a difference of squares: a² - b² = (a - b)(a + b). For the second part, rewrite: a⁴ + a²b² + b⁴ = (a²)² + 2a²b² + (b²)² - a²b² = (a² + b²)² - (ab)². Apply difference of squares: \( (a^2 + b^2)^2 - (ab)^2 = (a^2 + b^2 + ab)(a^2 + b^2 - ab) \). Therefore, a⁶ - b⁶ = (a - b)(a + b)(a² + b² + ab)(a² + b² - ab).
In simple words: First treat it as a difference of cubes with a² and b² as the bases. Then factor the resulting quadratic in terms of a² and b² by completing the square pattern.

Exam Tip: Sixth-degree differences can be approached as both (difference of cubes) and (difference of squares) - choosing the right approach reveals all four linear factors.

Question 10(ii). Factorise x⁶ - 1
Answer: Express as a difference of squares: x⁶ - 1 = (x³)² - (1)² = (x³ - 1)(x³ + 1). Apply the difference and sum of cubes formulas: \( x^3 - 1 = (x - 1)(x^2 + x + 1) \) and \( x^3 + 1 = (x + 1)(x^2 - x + 1) \). Therefore: \( (x^3 - 1)(x^3 + 1) = (x - 1)(x^2 + x + 1)(x + 1)(x^2 - x + 1) = (x - 1)(x + 1)(x^2 + x + 1)(x^2 - x + 1) \). Thus, x⁶ - 1 = (x - 1)(x + 1)(x² + x + 1)(x² - x + 1).
In simple words: Write x⁶ - 1 as (x³)² - 1², a difference of squares. Then factor each cubic separately using sum and difference of cube rules.

Exam Tip: The expression x⁶ - 1 factors into four distinct factors - two linear and two irreducible quadratics. This is a standard sixth-degree factorization to memorize.

Question 11(i). Factorise 64x⁶ - 729y⁶
Answer: Express as perfect cubes: 64x⁶ - 729y⁶ = [(2x)³]² - [(3y)³]² = (8x³)² - (27y³)². Apply the difference of squares formula: \( (8x^3)^2 - (27y^3)^2 = (8x^3 - 27y^3)(8x^3 + 27y^3) \). Apply the difference of cubes to the first factor: \( 8x^3 - 27y^3 = (2x)^3 - (3y)^3 = (2x - 3y)[(2x)^2 + 2x(3y) + (3y)^2] = (2x - 3y)(4x^2 + 6xy + 9y^2) \). Apply the sum of cubes to the second factor: \( 8x^3 + 27y^3 = (2x)^3 + (3y)^3 = (2x + 3y)[(2x)^2 - 2x(3y) + (3y)^2] = (2x + 3y)(4x^2 - 6xy + 9y^2) \). Therefore, 64x⁶ - 729y⁶ = (2x - 3y)(4x² + 6xy + 9y²)(2x + 3y)(4x² - 6xy + 9y²).
In simple words: Recognize 64 = 4³ = 2⁶ and 729 = 9³ = 3⁶. Rewrite as a difference of squares using cubes, then apply difference and sum of cube formulas to each square factor.

Exam Tip: For sixth-degree binomials with large coefficients, always check if they are perfect sixth powers - this approach factorizes them completely into four factors.

Question 11(ii). Factorise the following:
\( x^2 - \frac{8}{x} \)
Answer: We can rewrite the expression as \( x^2 - \frac{8}{x} = \frac{1}{x}(x^3 - 8) = \frac{1}{x}(x^3 - 2^3) \). Using the difference of cubes formula \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \), we get \( \frac{1}{x}(x - 2)(x^2 + 2x + 4) \). Therefore, \( x^2 - \frac{8}{x} = \frac{1}{x}(x - 2)(x^2 + 2x + 4) \).
In simple words: First, take out \( \frac{1}{x} \) as a common factor. Then use the difference of cubes rule to break down \( x^3 - 8 \) into three factors.

Exam Tip: Always check if you can factor out a common term first. Recognizing difference of cubes is key - remember the formula \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \).

Question 12(i). Factorise the following:
\( 250(a - b)^3 + 2 \)
Answer: We can write this as \( 2[125(a - b)^3 + 1] = 2[{5(a - b)}^3 + 1^3] \). Using the sum of cubes formula \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \), we get \( 2[5(a - b) + 1][{5(a - b)}^2 - 5(a - b) + 1] = 2(5a - 5b + 1)[(5a - 5b)^2 - (5a - 5b) + 1] = 2(5a - 5b + 1)(25a^2 + 25b^2 - 50ab - 5a + 5b + 1) \). Therefore, \( 250(a - b)^3 + 2 = 2(5a - 5b + 1)(25a^2 + 25b^2 - 50ab - 5a + 5b + 1) \).
In simple words: Pull out the common factor 2, then recognize the sum of two perfect cubes and apply the sum of cubes formula.

Exam Tip: Factor out common factors first, then look for perfect cubes. The sum of cubes formula is essential here - write it out clearly to avoid errors.

Question 12(ii). Factorise the following:
\( 32a^2x^3 - 8b^2x^3 - 4a^2y^3 + b^2y^3 \)
Answer: We can group the terms as \( 8x^3(4a^2 - b^2) - y^3(4a^2 - b^2) = (4a^2 - b^2)(8x^3 - y^3) \). Now \( 4a^2 - b^2 = (2a)^2 - b^2 = (2a - b)(2a + b) \) and \( 8x^3 - y^3 = (2x)^3 - y^3 = (2x - y)[(2x)^2 + 2xy + y^2] = (2x - y)(4x^2 + 2xy + y^2) \). Therefore, \( 32a^2x^3 - 8b^2x^3 - 4a^2y^3 + b^2y^3 = (2a - b)(2a + b)(2x - y)(4x^2 + 2xy + y^2) \).
In simple words: Group terms to find a common factor. Then apply the difference of squares rule to one part and the difference of cubes rule to the other.

Exam Tip: Look for common factors by grouping strategically. Show all four factors in the final answer - this is what marks are awarded for.

Question 13(i). Factorise the following:
\( x^9 + y^9 \)
Answer: We rewrite this as \( (x^3)^3 + (y^3)^3 \). Using the sum of cubes formula \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \), we get \( (x^3 + y^3)[(x^3)^2 - x^3y^3 + (y^3)^2] = (x^3 + y^3)(x^6 - x^3y^3 + y^6) \). We can factor \( x^3 + y^3 \) further using the sum of cubes: \( x^3 + y^3 = (x + y)(x^2 - xy + y^2) \). Therefore, \( x^9 + y^9 = (x + y)(x^2 - xy + y^2)(x^6 - x^3y^3 + y^6) \).
In simple words: Rewrite \( x^9 \) and \( y^9 \) as cubes. Apply the sum of cubes formula twice - once to the whole expression and once to the first factor you obtain.

Exam Tip: When you see high powers, look for ways to express them as perfect cubes or squares. Apply formulas step by step and simplify at each stage.

Question 13(ii). Factorise the following:
\( x^6 - 7x^3 - 8 \)
Answer: We split the middle term to get \( x^6 - 8x^3 + x^3 - 8 = x^3(x^3 - 8) + 1(x^3 - 8) = (x^3 - 8)(x^3 + 1) \). Now using the factorization formulas: \( x^3 - 8 = (x)^3 - (2)^3 = (x - 2)(x^2 + 2x + 4) \) and \( x^3 + 1 = (x)^3 + (1)^3 = (x + 1)(x^2 - x + 1) \). Therefore, \( x^6 - 7x^3 - 8 = (x - 2)(x^2 + 2x + 4)(x + 1)(x^2 - x + 1) \).
In simple words: Treat \( x^3 \) as a single variable and split the middle term like a normal quadratic. Then factor the two cubic expressions that result.

Exam Tip: Recognize patterns - here the expression is quadratic in \( x^3 \). Factor it as a quadratic first, then apply cube formulas to each resulting factor.

Multiple Choice Questions

Question 1. Factorisation of \( 12a^2b + 15ab^2 \) is
(a) \( 3a(4ab + 5b^2) \)
(b) \( 3b(4a^2 + 5ab) \)
(c) \( 3ab(4a + 5b) \)
(d) None of the options
Answer: (c) \( 3ab(4a + 5b) \)
In simple words: Find the greatest common factor of both terms - it is \( 3ab \). Divide each term by this factor and write the result in parentheses.

Exam Tip: Always start by finding the H.C.F. of all terms. This method works for any expression with multiple terms.

Question 2. Factorisation of \( 6xy - 4y + 6 - 9x \) is
(a) \( (3y - 2)(2x - 3) \)
(b) \( (3x - 2)(2y - 3) \)
(c) \( (2y - 3)(2 - 3x) \)
(d) None of the options
Answer: (b) \( (3x - 2)(2y - 3) \)
In simple words: Rearrange the terms strategically so you can group pairs. Factor each pair, then pull out the common binomial.

Exam Tip: Grouping works best when you rearrange terms to make common factors visible. Write out the rearrangement step clearly.

Question 3. Factorisation of \( 49p^3q - 36pq \) is
(a) \( p(7p + 6q)(7p - 6q) \)
(b) \( q(7p - 6)(7p + 6) \)
(c) \( pq(7p + 6)(7p - 6) \)
(d) None of the options
Answer: (c) \( pq(7p + 6)(7p - 6) \)
In simple words: First extract the common factor \( pq \). What remains is a difference of squares that can be split using the formula \( a^2 - b^2 = (a + b)(a - b) \).

Exam Tip: Always check for a common factor before looking for special patterns like difference of squares.

Question 4. Factorisation of \( y(y - z) + 9(z - y) \) is
(a) \( (y - z)(y + 9) \)
(b) \( (y - z)(y - 9) \)
(c) \( (z - y)(y + 9) \)
(d) None of the options
Answer: (b) \( (y - z)(y - 9) \)
In simple words: Notice that \( z - y = -(y - z) \). Rewrite the second term to show this. Then both groups share the common factor \( (y - z) \).

Exam Tip: Look for signs that are opposite - you can often rewrite one term to create a common factor.

Question 5. Factorisation of \( (lm + l) + m + 1 \) is
(a) \( (lm + 1)(m + l) \)
(b) \( (lm + m)(l + 1) \)
(c) \( l(m + 1) \)
(d) \( (l + 1)(m + 1) \)
Answer: (d) \( (l + 1)(m + 1) \)
In simple words: Rearrange as \( lm + m + l + 1 \). Group the first two and last two terms, then factor each group to reveal a common binomial.

Exam Tip: Grouping in pairs is powerful - rearrange terms to make the grouping obvious and the common factor clear.

Question 6. Factorisation of \( 63x^2 - 112y^2 \) is
(a) \( 63(x - 2y)(x + 2y) \)
(b) \( 7(3x + 2y)(3x - 2y) \)
(c) \( 7(3x + 4y)(3x - 4y) \)
(d) None of the options
Answer: (c) \( 7(3x + 4y)(3x - 4y) \)
In simple words: Take out 7 as a common factor. What remains is a difference of squares - \( 9x^2 - 16y^2 = (3x)^2 - (4y)^2 \) - which splits into two binomials.

Exam Tip: Be careful with the perfect square parts - identify them correctly, then apply the difference of squares rule accurately.

Question 7. Factorisation of \( p^4 - 81 \) is
(a) \( (p^2 - 9)(p^2 + 9) \)
(b) \( (p - 3)(p + 3)(p^2 + 9) \)
(c) \( (p - 3)^2(p + 3)^2 \)
(d) None of the options
Answer: (b) \( (p - 3)(p + 3)(p^2 + 9) \)
In simple words: First write as \( (p^2)^2 - 9^2 \) and apply the difference of squares formula. Then apply it again to \( p^2 - 9 \).

Exam Tip: When you get a difference of squares factor, check if it can be factored further - always fully factor.

Question 8. One of the factors of \( (25x^2 - 1) + (1 + 5x)^2 \) is
(a) \( 5 + x \)
(b) \( 5 - x \)
(c) \( 5x - 1 \)
(d) \( 10x \)
Answer: (d) \( 10x \)
In simple words: Expand the squared term and combine like terms. You will get \( 50x^2 + 10x \), which factors as \( 10x(5x + 1) \).

Exam Tip: Always expand and simplify completely before attempting to factor. Watch for common factors in the simplified form.

Question 9. Factorisation of \( x^2 - 4x - 12 \) is
(a) \( (x + 6)(x - 2) \)
(b) \( (x - 6)(x + 2) \)
(c) \( (x - 6)(x - 2) \)
(d) \( (x + 6)(x + 2) \)
Answer: (b) \( (x - 6)(x + 2) \)
In simple words: Split the middle term into two parts whose coefficients multiply to give \( -12 \) and add to give \( -4 \). Group and factor.

Exam Tip: For quadratics, split the middle term strategically - the two numbers must multiply to give the constant and add to give the middle coefficient.

Question 10. Factorisation of \( 3x^2 + 7x - 6 \) is
(a) \( (3x - 2)(x + 3) \)
(b) \( (3x + 2)(x - 3) \)
(c) \( (3x - 2)(x - 3) \)
(d) \( (3x + 2)(x + 3) \)
Answer: (a) \( (3x - 2)(x + 3) \)
In simple words: Split the middle term \( 7x \) into two terms whose coefficients multiply to \( 3 \times (-6) = -18 \) and add to 7. Then group and factor.

Exam Tip: When the leading coefficient is not 1, multiply it by the constant term to find the numbers for splitting the middle term.

Question 11. Factorisation of \( 4x^2 + 8x + 3 \) is
(a) \( (x + 1)(x + 3) \)
(b) \( (2x + 1)(2x + 3) \)
(c) \( (2x + 2)(2x + 5) \)
(d) \( (2x - 1)(2x - 3) \)
Answer: (b) \( (2x + 1)(2x + 3) \)
In simple words: Split \( 8x \) into \( 6x + 2x \) since these multiply to give \( 4 \times 3 = 12 \) and add to 8. Group and extract common factors.

Exam Tip: Check your answer by expanding - it should give you back the original quadratic.

Question 12. Factorisation of \( 16x^2 + 40x + 25 \) is
(a) \( (4x + 5)(4x + 5) \)
(b) \( (4x + 5)(4x - 5) \)
(c) \( (4x - 5)(4x - 5) \)
(d) \( (4x + 5)(4x + 7) \)
Answer: (a) \( (4x + 5)(4x + 5) \)
In simple words: This is a perfect square trinomial. The first term is \( (4x)^2 \), the last is \( 5^2 \), and the middle term is \( 2 \times 4x \times 5 \).

Exam Tip: Recognize perfect square trinomials - they factor as \( (a + b)^2 \) or \( (a - b)^2 \) and save time in factorization.

Question 13. Factorisation of \( x^2 - 4xy + 4y^2 \) is
(a) \( (x + 2y)(x - 2y) \)
(b) \( (x + 2y)(x + 2y) \)
(c) \( (x - 2y)(x - 2y) \)
(d) \( (2x - y)(2x + y) \)
Answer: (c) \( (x - 2y)(x - 2y) \)
In simple words: This is a perfect square trinomial of the form \( a^2 - 2ab + b^2 = (a - b)^2 \) where \( a = x \) and \( b = 2y \).

Exam Tip: Learn to spot perfect square trinomials instantly - check that the first and last terms are perfect squares and the middle term is twice their product.

Question 14. Which of the following is a factor of \( (x + y)^3 - (x^3 + y^3) \)?
(a) \( x^2 + xy + 2xy \)
(b) \( x^2 + y^2 - xy \)
(c) \( xy^2 \)
(d) \( 3xy \)
Answer: (d) \( 3xy \)
In simple words: Expand \( (x + y)^3 = x^3 + y^3 + 3xy(x + y) \). When you subtract \( x^3 + y^3 \), you are left with \( 3xy(x + y) \), which has \( 3xy \) as a factor.

Exam Tip: Remember the expansion of \( (a + b)^3 = a^3 + b^3 + 3ab(a + b) \) - it is useful for many factorization problems.

Question 15. If \( \frac{x}{y} + \frac{y}{x} = -1 \) (where \( x \neq 0, y \neq 0 \)), then the value of \( x^3 - y^3 \) is
(a) \( 1 \)
(b) \( -1 \)
(c) \( 0 \)
(d) \( \frac{1}{2} \)
Answer: (c) \( 0 \)
In simple words: From the given condition, combine the fractions to get \( \frac{x^2 + y^2}{xy} = -1 \), which gives \( x^2 + y^2 = -xy \). Substitute this into the formula \( x^3 - y^3 = (x - y)(x^2 + xy + y^2) \) to find that \( x^2 + xy + y^2 = 0 \).

Exam Tip: When given a condition involving fractions, clear denominators and simplify to find relationships. Use these relationships in the target expression.

Question 16. If a + b + c = 0, then the value of a³ + b³ + c³ is
(a) 0
(b) abc
(c) 2abc
(d) 3abc
Answer: (d) 3abc
In simple words: When three numbers add up to zero, the sum of their cubes always equals three times the product of all three numbers.

Exam Tip: Remember the identity a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca). When (a + b + c) = 0, the entire right side becomes zero, leaving a³ + b³ + c³ = 3abc.

Question 17. If \( x^{\frac{1}{3}} + y^{\frac{1}{3}} + z^{\frac{1}{3}} = 0 \), then
(a) x³ + y³ + z³ = 0
(b) x³ + y³ + z³ = 27xyz
(c) (x + y + z)³ = 27xyz
(d) x + y + z = 3xyz
Answer: (c) (x + y + z)³ = 27xyz
In simple words: If the cube roots of three numbers sum to zero, then when you cube the sum of the original three numbers, you get 27 times their product.

Exam Tip: Recognize that \( x^{\frac{1}{3}} + y^{\frac{1}{3}} + z^{\frac{1}{3}} = 0 \) allows you to use the sum-of-cubes identity. Cube both sides strategically to isolate (x + y + z)³.

Question 18. Consider the following two statements. Statement 1: The factorisation of x² + 2x + 1 is (x - 1)². Statement 2: (a - b)² = a² + 2ab + b². Which of the following is valid?
(a) Both the statements are true.
(b) Both the statements are false.
(c) Statement 1 is true, and Statement 2 is false.
(d) Statement 1 is false, and Statement 2 is true.
Answer: (b) Both the statements are false
In simple words: Statement 1 factors to (x + 1)², not (x - 1)². Statement 2 should say a² - 2ab + b², not a² + 2ab + b². Both statements contain errors.

Exam Tip: Always expand your factorisation to check it. For (a - b)², multiply it out carefully: (a - b)(a - b) = a² - 2ab + b². The middle term is negative, not positive.

Assertion Reason Type Questions

Question. Assertion (A): For the trinomial 2x² + 8x - 9 cannot be factorised. Reason (R): For trinomial ax² + bx + c to be factorised, b² - 4ac must be a perfect square.
(a) Assertion (A) is true, Reason (R) is false.
(b) Assertion (A) is false, Reason (R) is true.
(c) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
(d) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
Answer: (c) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A)
To check Reason (R): A trinomial can break down into linear factors with rational coefficients only if its roots are rational. This occurs when the discriminant b² - 4ac is a perfect square.
Reason (R) is true.
For the trinomial 2x² + 8x - 9, we have a = 2, b = 8, c = -9.

D = b² - 4ac
\( \implies \) D = 8² - 4 × 2 × (-9)
\( \implies \) D = 64 + 72
\( \implies \) D = 136

Since 136 is not a perfect square, the trinomial 2x² + 8x - 9 cannot break down into linear factors.
Assertion (A) is true.
In simple words: The discriminant tells us if a trinomial can factor. When the discriminant is not a perfect square, factorisation is impossible.

Exam Tip: Always calculate the discriminant first when testing factorisability. If it is a perfect square (1, 4, 9, 16, 25...), factorisation is possible; otherwise, it is not.

Question. Assertion (A): Factorisation of 4x² + 9y² is (2x - 3y)(2x + 3y). Reason (R): a² - b² = (a - b)(a + b)
(a) Assertion (A) is true, Reason (R) is false.
(b) Assertion (A) is false, Reason (R) is true.
(c) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
(d) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
Answer: (b) Assertion (A) is false, Reason (R) is true
According to Reason (R): a² - b² = (a - b)(a + b)

Expanding: (a - b)(a + b)
\( \implies \) a(a + b) - b(a + b)
\( \implies \) a² + ab - ab - b²
\( \implies \) a² - b²

Thus, (a - b)(a + b) = a² - b²
Reason (R) is true.

For the expression: (2x - 3y)(2x + 3y)
\( \implies \) 2x(2x + 3y) - 3y(2x + 3y)
\( \implies \) 4x² + 6xy - 6xy - 9y²
\( \implies \) 4x² - 9y²

The expression 4x² + 9y² contains addition (sum of squares), not subtraction. A sum of squares cannot be factored using the difference-of-squares formula.
Assertion (A) is false.
In simple words: The rule a² - b² = (a - b)(a + b) works only for a difference (subtraction). A sum like 4x² + 9y² does not factor this way.

Exam Tip: Watch the sign carefully. The difference-of-squares rule applies only to minus, never to plus. A sum of two perfect squares is generally not factorable over the real numbers.

Question. Assertion (A): Factorisation of x³ - 27 is (x - 3)(x² + 3x + 9). Reason (R): a³ - b³ = (a - b)(a² + ab + b²).
(a) Assertion (A) is true, Reason (R) is false.
(b) Assertion (A) is false, Reason (R) is true.
(c) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
(d) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
Answer: (c) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A)
We know that a³ - b³ = (a - b)(a² + ab + b²)
Reason (R) is true.

For x³ - 27:
\( \implies \) x³ - 3³
\( \implies \) (x - 3)(x² + 3x + 3²)
\( \implies \) (x - 3)(x² + 3x + 9)
Assertion (A) is true.
In simple words: The difference-of-cubes formula breaks a cubic term minus another into a simple linear factor and a quadratic term.

Exam Tip: Recognise cubes in the form a³ and b³. Apply the difference-of-cubes identity directly. Remember: the middle term of the quadratic factor is ab (positive), not minus.

Chapter Test

Question 1(i). Factorise the following: 15(2x - 3)³ - 10(2x - 3)
Answer: The highest common factor of 15(2x - 3)³ and 10(2x - 3) is 5(2x - 3).

\( \implies \) 15(2x - 3)³ - 10(2x - 3) = 5(2x - 3)[3(2x - 3)² - 2]
In simple words: Take out the common factor 5(2x - 3) from both terms, then simplify what remains.

Exam Tip: Always find the highest common factor first. Look for repeated terms (like (2x - 3)) and numbers that divide all coefficients.

Question 1(ii). Factorise the following: a(b - c)(b + c) - d(c - b)
Answer: a(b - c)(b + c) - d(c - b) = a(b - c)(b + c) - d(-1)(b - c)
\( \implies \) a(b - c)(b + c) + d(b - c)
\( \implies \) (b - c)[a(b + c) + d]

Hence, a(b - c)(b + c) - d(c - b) = (b - c)[a(b + c) + d]
In simple words: Notice that (c - b) = (-1)(b - c). Rewrite the expression to reveal the common factor (b - c), then take it out.

Exam Tip: When you see opposite terms like (c - b) and (b - c), rewrite one as the negative of the other to expose the common factor.

Question 2(i). Factorise the following: 2a²x - bx + 2a² - b
Answer: Rearranging the terms, we get:

2a²x + 2a² - bx - b
\( \implies \) 2a²(x + 1) - b(x + 1)
\( \implies \) (x + 1)(2a² - b)

Hence, 2a²x - bx + 2a² - b = (x + 1)(2a² - b)
In simple words: Group the first and third terms together, then the second and fourth terms. Each group shares a common factor, which you can then extract.

Exam Tip: When there is no obvious common factor across all terms, rearrange and group. Look for pairs that share a factor.

Question 2(ii). Factorise the following: p² - (a + 2b)p + 2ab
Answer: p² - (a + 2b)p + 2ab = p² - ap - 2bp + 2ab
\( \implies \) p² - 2bp - ap + 2ab
\( \implies \) p(p - 2b) - a(p - 2b)
\( \implies \) (p - 2b)(p - a)

Hence, p² - (a + 2b)p + 2ab = (p - 2b)(p - a)
In simple words: Expand the middle term and regroup. Two of the resulting terms will have a common factor (p - 2b), which can be taken out.

Exam Tip: For a quadratic in one variable, split the middle term based on the factors of the constant term, then group and factorise.

Question 3(i). Factorise the following: (x² - y²)z + (y² - z²)x
Answer: (x² - y²)z + (y² - z²)x = x²z - y²z + y²x - z²x
\( \implies \) x²z - z²x - y²z + y²x
\( \implies \) xz(x - z) + y²(x - z)
\( \implies \) (x - z)(xz + y²)

Hence, (x² - y²)z + (y² - z²)x = (x - z)(xz + y²)
In simple words: Expand the expression, then group the terms in a way that reveals a common factor across pairs.

Exam Tip: When expanding brackets, look for ways to reorder the resulting terms so that pairs share a visible common factor.

Question 3(ii). Factorise the following: 5a⁴ - 5a³ + 30a² - 30a
Answer: 5a⁴ - 5a³ + 30a² - 30a = 5a(a³ - a² + 6a - 6)
\( \implies \) 5a[a²(a - 1) + 6(a - 1)]
\( \implies \) 5a(a - 1)(a² + 6)

Hence, 5a⁴ - 5a³ + 30a² - 30a = 5a(a - 1)(a² + 6)
In simple words: First, take out the common number and variable factor (5a). Then group the remaining terms in pairs that share a factor.

Exam Tip: Always extract the greatest common factor from all terms before attempting to group. This simplifies the remaining expression significantly.

Question 4(i). Factorise the following: b(c - d)² + a(d - c) + 3c - 3d
Answer: b(c - d)² + a(d - c) + 3c - 3d = b(c - d)² + a(-1)(c - d) + 3(c - d)
\( \implies \) (c - d)[b(c - d) - a + 3]
\( \implies \) (c - d)(bc - bd - a + 3)

Hence, b(c - d)² + a(d - c) + 3c - 3d = (c - d)(bc - bd - a + 3)
In simple words: Rewrite (d - c) and (3c - 3d) as negatives and multiples of (c - d), then extract this common factor.

Exam Tip: Notice patterns like (d - c) = -(c - d) and 3c - 3d = 3(c - d). Use these to expose the common binomial factor.

Question 4(ii). Factorise the following: x³ - x² - xy + x + y - 1
Answer: Rearranging, we get:

x³ - x² - xy + y + x - 1
\( \implies \) x²(x - 1) - y(x - 1) + 1(x - 1)
\( \implies \) (x - 1)(x² - y + 1)

Hence, x³ - x² - xy + x + y - 1 = (x - 1)(x² - y + 1)
In simple words: Group pairs of terms that share a factor. Once you identify (x - 1) as a common factor, take it out and simplify.

Exam Tip: For longer expressions, reorder terms strategically before grouping. The goal is to find at least two pairs with an obvious common factor.

Question 5(i). Factorise the following: x(x + z) - y(y + z)
Answer: x(x + z) - y(y + z) = x² + xz - y² - yz

Rearranging:
x² - y² + xz - yz

We know that a² - b² = (a + b)(a - b).
\( \implies \) (x + y)(x - y) + z(x - y)
\( \implies \) (x - y)(x + y + z)

Hence, x(x + z) - y(y + z) = (x - y)(x + y + z)
In simple words: Expand, regroup to get a difference of squares plus an extra term, then factorise using the difference-of-squares formula followed by grouping.

Exam Tip: When you see products with sums (like x(x + z)), always expand first, then look for recognisable patterns like differences of squares.

Question 5(ii). Factorise the following: a¹²x⁴ - a⁴x¹²
Answer: a¹²x⁴ - a⁴x¹² = a⁴x⁴(a⁸ - x⁸)
\( \implies \) a⁴x⁴[(a⁴)² - (x⁴)²]

We know that a² - b² = (a + b)(a - b).
\( \implies \) a⁴x⁴(a⁴ + x⁴)(a⁴ - x⁴)
\( \implies \) a⁴x⁴(a⁴ + x⁴)[(a²)² - (x²)²]
\( \implies \) a⁴x⁴(a⁴ + x⁴)(a² + x²)(a² - x²)
\( \implies \) a⁴x⁴(a⁴ + x⁴)(a² + x²)(a + x)(a - x)

Hence, a¹²x⁴ - a⁴x¹² = a⁴x⁴(a⁴ + x⁴)(a² + x²)(a + x)(a - x)
In simple words: Take out the common factor first. Then repeatedly apply the difference-of-squares rule as the exponents keep halving.

Exam Tip: For very large exponents, extract the greatest common monomial factor first. Then recognise nested differences of squares—each factorisation step reduces the exponent by half.

Question 6(i). Factorise the following: 9x² + 12x + 4 - 16y²
Answer: 9x² + 12x + 4 - 16y² = (3x)² + (2 × 3x × 2) + 2² - (4y)²

We know that (a + b)² = a² + b² + 2ab.
\( \implies \) (3x + 2)² - (4y)²

We know that a² - b² = (a + b)(a - b).
\( \implies \) (3x + 2 + 4y)(3x + 2 - 4y)

Hence, 9x² + 12x + 4 - 16y² = (3x + 2 + 4y)(3x + 2 - 4y)
In simple words: Recognise that the first three terms form a perfect square trinomial. Then apply difference-of-squares to the result.

Exam Tip: Watch for perfect square trinomials hidden in longer expressions. Once you identify them, you can apply further factorisation rules to complete the problem.

Question 6(ii). Factorise the following: x⁴ + 3x² + 4
Answer: To factorise x⁴ + 3x² + 4, we use a technique called "completing the square" in the context of the variable x².

Let u = x². Then the expression becomes u² + 3u + 4.

We want to write this in the form (u + a)² - b² (difference of squares).

(u + a)² = u² + 2au + a². For this to match u² + 3u + ..., we need 2a = 3, so a = 3/2.

Now, (u + 3/2)² = u² + 3u + 9/4.

Thus, u² + 3u + 4 = (u + 3/2)² + 4 - 9/4 = (u + 3/2)² + 7/4

\( \implies \) (x² + 3/2)² + 7/4
\( \implies \) (x² + 3/2)² - (-7/4)

Since -7/4 is negative, we write this as: (x² + 3/2)² - (i√(7/4))², but this leads to complex factors.

Alternatively, the expression x⁴ + 3x² + 4 does not factorise over the reals into linear or simple quadratic factors with rational coefficients. However, it can be written as a product of two irreducible quadratics with irrational coefficients.

The real factorisation is: (x² + 2x + 2)(x² - 2x + 2)
In simple words: This quartic does not break down neatly using simple integer methods. It factors into two quadratic expressions, each with no real roots.

Exam Tip: Not all polynomials factorise over the integers. If standard methods fail, check whether the expression can be written as a product of two quadratics by expanding (x² + ax + b)(x² + cx + d) and comparing coefficients.

Question 7(i). Factorise the following: \( 21x^2 - 59xy + 40y^2 \)
Answer: Start by breaking the middle term. Split \( -59xy \) into two parts such that their product matches the outer coefficients. We get \( 21x^2 - 35xy - 24xy + 40y^2 \). Factor by grouping: \( 7x(3x - 5y) - 8y(3x - 5y) \). Pull out the common binomial to get \( (3x - 5y)(7x - 8y) \).
In simple words: Break the middle term smartly, group the terms in pairs, then pull out what both pairs share.

Exam Tip: Look for two numbers that multiply to give the product of the first and last coefficients and add up to the middle coefficient—this is the key to splitting the middle term correctly.

Question 7(ii). Factorise the following: \( 4x^3y - 44x^2y + 112xy \)
Answer: First, pull out the common factor \( 4xy \), leaving \( x^2 - 11x + 28 \) inside the brackets. Now factor the quadratic by splitting \( -11x \) into \( -7x - 4x \). This gives \( 4xy[x(x - 7) - 4(x - 7)] \). Extract the common binomial \( (x - 7) \) to get the final result \( 4xy(x - 7)(x - 4) \).
In simple words: Always take out any common factors first. Then factor what remains using the same splitting method.

Exam Tip: Never forget to check for a greatest common factor before attempting any other factoring method—it simplifies the problem right away.

Question 8(i). Factorise the following: \( x^2y^2 - xy - 72 \)
Answer: Treat \( xy \) as a single unit. Split the middle term \( -xy \) into \( -9xy + 8xy \). Rewrite as \( x^2y^2 - 9xy + 8xy - 72 \). Group: \( xy(xy - 9) + 8(xy - 9) \). Factor out the common expression to get \( (xy - 9)(xy + 8) \).
In simple words: When you see products like \( xy \), think of them as one block, then split and group just like a regular quadratic.

Exam Tip: Recognizing composite terms like \( xy \) or \( (x + 1) \) as single units is crucial for quick factoring of these expressions.

Question 8(ii). Factorise the following: \( 9x^3y + 41x^2y^2 + 20xy^3 \)
Answer: Pull out the common factor \( xy \) to get \( xy(9x^2 + 41xy + 20y^2) \). Now factor the quadratic inside by splitting \( 41xy \) into \( 36xy + 5xy \). This gives \( xy[9x(x + 4y) + 5y(x + 4y)] \). Extract \( (x + 4y) \) to reach \( xy(x + 4y)(9x + 5y) \).
In simple words: Take out common factors, then use splitting to break down what's left inside the brackets.

Exam Tip: After extracting a common factor, treat the remaining quadratic as a standard problem and apply the splitting method carefully.

Question 9(i). Factorise the following: \( (3a - 2b)^2 + 3(3a - 2b) - 10 \)
Answer: Let \( t = (3a - 2b) \). The expression becomes \( t^2 + 3t - 10 \). Split \( 3t \) into \( 5t - 2t \): this gives \( t^2 + 5t - 2t - 10 \). Group: \( t(t + 5) - 2(t + 5) = (t + 5)(t - 2) \). Substitute back \( t = (3a - 2b) \) to get \( (3a - 2b + 5)(3a - 2b - 2) \).
In simple words: When you see the same expression repeated, call it \( t \) to make it simpler. Factor the resulting quadratic, then swap \( t \) back.

Exam Tip: Substitution (using a variable like \( t \)) is a powerful tool for expressions that repeat—it turns complex-looking problems into basic ones.

Question 9(ii). Factorise the following: \( (x^2 - 3x)(x^2 - 3x + 7) + 10 \)
Answer: Let \( t = x^2 - 3x \). Then the expression is \( t(t + 7) + 10 = t^2 + 7t + 10 \). Split \( 7t \) into \( 5t + 2t \): giving \( t^2 + 5t + 2t + 10 \). Group: \( t(t + 5) + 2(t + 5) = (t + 5)(t + 2) \). Replace \( t \) to get \( (x^2 - 3x + 5)(x^2 - 3x + 2) \). Now factor \( x^2 - 3x + 2 \) further by splitting to get \( (x - 1)(x - 2) \). Final answer: \( (x^2 - 3x + 5)(x - 1)(x - 2) \).
In simple words: Spot the repeating part, substitute it as \( t \), factor the result, put \( t \) back, and check if what you get can be factored even more.

Exam Tip: After substitution and factoring, always check whether each factor can be broken down further—do not stop early.

Question 10(i). Factorise the following: \( (x^2 - x)(4x^2 - 4x - 5) - 6 \)
Answer: Let \( p = x^2 - x \). Then \( 4x^2 - 4x = 4(x^2 - x) = 4p \), so the expression becomes \( p(4p - 5) - 6 = 4p^2 - 5p - 6 \). Split \( -5p \) into \( -8p + 3p \): giving \( 4p^2 - 8p + 3p - 6 \). Group: \( 4p(p - 2) + 3(p - 2) = (p - 2)(4p + 3) \). Substitute back: \( (x^2 - x - 2)[4(x^2 - x) + 3] = (x^2 - x - 2)(4x^2 - 4x + 3) \). Factor \( x^2 - x - 2 \) by splitting to get \( (x - 2)(x + 1) \). Final answer: \( (x - 2)(x + 1)(4x^2 - 4x + 3) \).
In simple words: Find what repeats, substitute it, factor, then replace it back and factor further if possible.

Exam Tip: After back-substitution, carefully check each factor to see whether it breaks down into linear factors—this determines your final form.

Question 10(ii). Factorise the following: \( x^4 + 9x^2y^2 + 81y^4 \)
Answer: Rewrite by adding and subtracting \( 9x^2y^2 \): \( x^4 + 18x^2y^2 + 81y^4 - 9x^2y^2 \). The first three terms form a perfect square: \( (x^2 + 9y^2)^2 - 9x^2y^2 = (x^2 + 9y^2)^2 - (3xy)^2 \). Apply the difference of squares formula \( a^2 - b^2 = (a + b)(a - b) \): \( (x^2 + 9y^2 + 3xy)(x^2 + 9y^2 - 3xy) \).
In simple words: Add and subtract a term to create a perfect square trinomial. Then use the difference of squares on what remains.

Exam Tip: Recognize patterns like perfect squares and differences of squares—they are hidden in many expressions and reveal the factoring path.

Question 11(i). Factorise the following: \( \frac{8}{27}x^3 - \frac{1}{8}y^3 \)
Answer: Recognize this as a difference of cubes: \( \left(\frac{2}{3}x\right)^3 - \left(\frac{1}{2}y\right)^3 \). Apply the formula \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \) with \( a = \frac{2}{3}x \) and \( b = \frac{1}{2}y \). First factor: \( \frac{2}{3}x - \frac{1}{2}y \). Second factor: \( \left(\frac{2}{3}x\right)^2 + \left(\frac{2}{3}x\right)\left(\frac{1}{2}y\right) + \left(\frac{1}{2}y\right)^2 = \frac{4}{9}x^2 + \frac{1}{3}xy + \frac{1}{4}y^2 \). Final answer: \( \left(\frac{2}{3}x - \frac{1}{2}y\right)\left(\frac{4}{9}x^2 + \frac{1}{3}xy + \frac{1}{4}y^2\right) \).
In simple words: Write each fraction as a perfect cube, then apply the difference of cubes rule.

Exam Tip: With fractions, always express them as clean cube roots first—this makes the formula application much clearer and mistakes less likely.

Question 11(ii). Factorise the following: \( x^6 + 63x^3 - 64 \)
Answer: Rewrite by splitting the middle term: \( x^6 + 64x^3 - x^3 - 64 \). Group: \( x^3(x^3 + 64) - 1(x^3 + 64) = (x^3 + 64)(x^3 - 1) \). Recognize \( 64 = 4^3 \) and \( 1 = 1^3 \), so factor each using the sum and difference of cubes formulas. For \( x^3 + 4^3 \): \( (x + 4)(x^2 - 4x + 16) \). For \( x^3 - 1^3 \): \( (x - 1)(x^2 + x + 1) \). Final answer: \( (x + 4)(x^2 - 4x + 16)(x - 1)(x^2 + x + 1) \).
In simple words: Split the middle term cleverly so grouping gives you recognizable cube expressions, then apply cube formulas.

Exam Tip: Always check if a number is a perfect cube—many problems hide cube expressions that become obvious once you spot them.

Question 12(i). Factorise the following: \( x^3 + x^2 - \frac{1}{x^2} + \frac{1}{x^3} \)
Answer: Rearrange by grouping related terms: \( x^3 + \frac{1}{x^3} + x^2 - \frac{1}{x^2} \). For the first pair, note that \( x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)^3 - 3\left(x + \frac{1}{x}\right) \). However, a simpler approach is to factor by recognizing the pattern directly. Apply the formula for sum/difference: \( x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)\left(x^2 - 1 + \frac{1}{x^2}\right) \) and \( x^2 - \frac{1}{x^2} = \left(x + \frac{1}{x}\right)\left(x - \frac{1}{x}\right) \). Extracting the common factor \( \left(x + \frac{1}{x}\right) \): \( \left(x + \frac{1}{x}\right)\left(x^2 - 1 + \frac{1}{x^2} + x - \frac{1}{x}\right) \).
In simple words: Group terms that fit sum/difference formulas, apply those formulas, and pull out what appears in both groups.

Exam Tip: When fractions with variables appear, look for patterns that let you factor out common expressions—the algebra often simplifies dramatically.

Question 12(ii). Factorise the following: \( (x + 1)^6 - (x - 1)^6 \)
Answer: Write as \( [(x + 1)^3]^2 - [(x - 1)^3]^2 \), a difference of squares. Apply \( a^2 - b^2 = (a + b)(a - b) \): \( [(x + 1)^3 + (x - 1)^3][(x + 1)^3 - (x - 1)^3] \). For each bracket, apply sum and difference of cubes formulas. \( (x + 1)^3 + (x - 1)^3 = [(x + 1) + (x - 1)][(x + 1)^2 - (x + 1)(x - 1) + (x - 1)^2] = 2x[x^2 + 2x + 1 - (x^2 - 1) + x^2 - 2x + 1] = 2x(x^2 + 3) \). \( (x + 1)^3 - (x - 1)^3 = [(x + 1) - (x - 1)][(x + 1)^2 + (x + 1)(x - 1) + (x - 1)^2] = 2[(x^2 + 2x + 1) + (x^2 - 1) + (x^2 - 2x + 1)] = 2(3x^2 + 1) \). Final answer: \( 4x(x^2 + 3)(3x^2 + 1) \).
In simple words: Recognize the sixth powers as squares of cubes. Use difference of squares, then apply cube formulas to simplify each part.

Exam Tip: Break high powers into manageable pieces—write \( a^6 = (a^3)^2 = (a^2)^3 \) depending on which factoring formula fits best.

Question 13. Factorise \( (x + 1)(x - 3) + (x + 1)(x + 4) \)
Answer: Both terms share the common factor \( (x + 1) \). Extract it: \( (x + 1)[(x - 3) + (x + 4)] = (x + 1)(x - 3 + x + 4) = (x + 1)(2x + 1) \).
In simple words: When you see the same bracket in two terms, pull it out. Then combine what's left inside.

Exam Tip: Always scan for common binomial factors first—they are quick wins and often the intended factoring route.

Question 14. Show that \( 97^3 + 14^3 \) is divisible by 111.
Answer: Apply the sum of cubes formula \( a^3 + b^3 = (a + b)(a^2 + b^2 - ab) \) with \( a = 97 \) and \( b = 14 \). This gives \( 97^3 + 14^3 = (97 + 14)(97^2 + 14^2 - 97 \times 14) = 111(9409 + 196 - 1358) = 111 \times 8247 \). Since \( 97^3 + 14^3 \) equals \( 111 \times 8247 \), it is clearly divisible by 111.
In simple words: Factor using the cube formula. The sum of the two numbers is 111, which is our divisor.

Exam Tip: The sum (or difference) of the cube roots often reveals a factor—this is a neat trick for divisibility proofs.

Question 15. If \( a + b = 8 \) and \( ab = 15 \), find the value of \( a^4 + a^2b^2 + b^4 \)
Answer: From the condition \( a + b = 8 \), square both sides: \( a^2 + b^2 + 2ab = 64 \). Substitute \( ab = 15 \): \( a^2 + b^2 = 64 - 30 = 34 \). Now, rewrite the target expression: \( a^4 + a^2b^2 + b^4 = a^4 + 2a^2b^2 - a^2b^2 + b^4 = (a^2 + b^2)^2 - a^2b^2 \). Substitute: \( 34^2 - 15^2 = 1156 - 225 = 931 \). Alternatively, use the factorization \( (a^2 + b^2)^2 - a^2b^2 = (a^2 + b^2 + ab)(a^2 + b^2 - ab) = (34 + 15)(34 - 15) = 49 \times 19 = 931 \).
In simple words: Use the given conditions to find \( a^2 + b^2 \) first. Then reshape the target expression to use this value.

Exam Tip: Given conditions often force you to compute intermediate values like \( a^2 + b^2 \)—do not skip these steps, as they unlock the final answer.