Selina Concise Solutions for ICSE Class 9 Mathematics Chapter 5 Factorisation

Exercise 5(A)

Question 1. Factorise: \( 3a^2 - 9ab \)
Answer:
To factorise the given expression, we find the greatest common factor of both terms, which is \( 3a \):
\( 3a^2 - 9ab = 3a(a) - 3a(3b) \)

\( \implies 3a(a - 3b) \)
In simple words: Find what both parts of the expression have in common and pull it out. Here, both terms can be divided by \( 3a \).

Exam Tip: Always look for the greatest common monomial factor first before trying other factorisation techniques.

Question 2. Factorise: \( 2(x + y)^3 - 6(x + y) \)
Answer:
We begin by extracting the common factor \( 2(x + y) \) from both terms:
\( = 2(x + y)[(x + y)^2 - 3] \)
Expanding the squared term inside the brackets:
\( = 2(x + y)(x^2 + y^2 + 2xy - 3) \)
In simple words: First, pull out the common factor \( 2(x + y) \). Then, expand the rest of the expression inside the brackets.

Exam Tip: Do not forget to expand \( (x+y)^2 \) to \( x^2 + 2xy + y^2 \) inside the square brackets for a fully simplified answer.

Question 3. Factorise: \( x^3(2x - 3y) - x^2(2x - 3y)^2 \)
Answer:
Let us factor out the term \( x^2(2x - 3y) \) which is common to both parts:
\( = x^2(2x - 3y)[x - (2x - 3y)] \)
Now, simplify the terms within the square brackets:
\( = x^2(2x - 3y)[x - 2x + 3y] \)
\( = x^2(2x - 3y)[-x + 3y] \)
Rearranging the terms in the second bracket:
\( = x^2(2x - 3y)(3y - x) \)
In simple words: Pull out \( x^2(2x - 3y) \) from both parts. Then, combine the leftover terms inside the bracket and simplify them.

Exam Tip: Be careful with the negative sign when expanding \( -(2x - 3y) \) to avoid simple calculation errors.

Question 4. Factorise: \( 2(3x + 4y)(2x - 5y) - 6(x - y)(2x - 5y) \)
Answer:
We can factor out the common term \( 2(2x - 5y) \) from both groups:
\( = 2(2x - 5y)[(3x + 4y) - 3(x - y)] \)
Simplify the terms inside the square brackets:
\( = 2(2x - 5y)(3x + 4y - 3x + 3y) \)
\( = 2(2x - 5y)(7y) \)
\( = 14y(2x - 5y) \)
In simple words: Take out the common parts, which are \( 2 \) and \( (2x - 5y) \). Then simplify what is left inside the bracket.

Exam Tip: Grouping common terms early saves you from doing long and complex polynomial multiplications.

Question 5. Factorise: \( a^3 + a - 3a^2 - 3 \)
Answer:
Let us group the terms to find common binomial factors:
\( a^3 + a - 3a^2 - 3 = a(a^2 + 1) - 3(a^2 + 1) \)
Now, take \( (a^2 + 1) \) as the common factor:
\( = (a^2 + 1)(a - 3) \)
In simple words: Group the terms in pairs and find what is common in each pair. Then, factor out the common bracketed term.

Exam Tip: Rearrange the terms if grouping them in their original order does not immediately show a common factor.

Question 6. Factorise: \( 16(a + b)^2 - 4a - 4b \)
Answer:
First, group the last two terms together by factorising out \( -4 \):
\( 16(a+b)^2 - 4(a+b) \)
Now, pull out the common factor \( 4(a + b) \) from both parts:
\( = 4(a + b)[4(a + b) - 1] \)
\( = 4(a + b)(4a + 4b - 1) \)
In simple words: Turn the last part into \( -4(a+b) \) first. After that, pull out \( 4(a+b) \) because it is common to both halves.

Exam Tip: Always look for hidden groups by factoring out negative signs or constants from a subset of terms.

Question 7. Factorise: \( a^4 - 2a^3 - 4a + 8 \)
Answer:
Group the terms into two pairs to find common factors:
\( a^4 - 2a^3 - 4a + 8 = a^3(a - 2) - 4(a - 2) \)
Now, extract the common factor \( (a - 2) \):
\( = (a^3 - 4)(a - 2) \)
In simple words: Group the first two terms and the last two terms. Pull out the common parts from each group, then combine them.

Exam Tip: Make sure the sign of the constant term inside both brackets is identical before taking the binomial common.

Question 8. Factorise: \( ab - 2b + a^2 - 2a \)
Answer:
Group the terms to extract common factors:
\( ab - 2b + a^2 - 2a = b(a - 2) + a(a - 2) \)
Now, factor out the common binomial \( (a - 2) \):
\( = (a + b)(a - 2) \)
In simple words: Group the first two terms together and the last two terms together, find their common parts, and then factor them out.

Exam Tip: Grouping can be done in different ways - for instance, grouping \( ab + a^2 \) and \( -2b - 2a \) also yields the same result.

Question 9. Factorise: \( ab(x^2 + 1) + x(a^2 + b^2) \)
Answer:
First, expand the given expression entirely:
\( abx^2 + ab + a^2x + b^2x \)
Now, rearrange the terms to facilitate grouping:
\( = abx^2 + a^2x + b^2x + ab \)
Group the terms and find the common factors:
\( = ax(bx + a) + b(bx + a) \)
Factor out the common term \( (bx + a) \):
\( = (ax + b)(bx + a) \)
In simple words: Multiply out the brackets first, mix the terms around, and then group them in pairs to find the factors.

Exam Tip: If direct grouping is not obvious, expand the terms first and then try rearranging them.

Question 10. Factorise: \( a^2 + b - ab - a \)
Answer:
Rearrange the terms to make grouping easier:
\( a^2 - a + b - ab \)
Factor out common terms from each pair:
\( = a(a - 1) + b(1 - a) \)
Change the sign of the second term to make the brackets match:
\( = a(a - 1) - b(a - 1) \)
Now, factor out the common binomial \( (a - 1) \):
\( = (a - 1)(a - b) \)
In simple words: Rearrange the terms, group them, and pull out the common parts. Turn \( (1-a) \) into \( -(a-1) \) so both brackets are the same.

Exam Tip: Remember that \( (y - x) = -(x - y) \). This identity is very useful for matching binomial terms.

Question 11. Factorise: \( (ax + by)^2 + (bx - ay)^2 \)
Answer:
Expand both squared terms using algebraic identities:
\( = a^2x^2 + b^2y^2 + 2axby + b^2x^2 + a^2y^2 - 2bxay \)
The middle terms cancel each other out:
\( = a^2x^2 + b^2y^2 + b^2x^2 + a^2y^2 \)
Rearrange and group the terms:
\( = x^2(a^2 + b^2) + y^2(a^2 + b^2) \)
Factor out the common bracket \( (a^2 + b^2) \):
\( = (x^2 + y^2)(a^2 + b^2) \)
In simple words: Expand both parts completely. The middle parts will cancel out, leaving four terms that can be grouped and factorised.

Exam Tip: Cross out terms of equal value but opposite signs to keep your work tidy and avoid counting them twice.

Question 12. Factorise: \( a^2x^2 + (ax^2 + 1)x + a \)
Answer:
Rearrange the terms to establish a pattern:
\( = a^2x^2 + a + (ax^2 + 1)x \)
Factorise the first two terms by extracting \( a \):
\( = a(ax^2 + 1) + x(ax^2 + 1) \)
Now, take out the common term \( (ax^2 + 1) \):
\( = (a + x)(ax^2 + 1) \)
In simple words: Swap the order of terms so you can factor \( a \) out of the first two. Then, pull out the common binomial bracket.

Exam Tip: Recognising that \( ax^2+1 \) is already part of the middle term is key to choosing the correct rearrangement.

Question 13. Factorise: \( (2a - b)^2 - 10a + 5b \)
Answer:
Group the last two terms by factoring out \( -5 \):
\( = (2a - b)^2 - 5(2a - b) \)
Now, extract the common binomial factor \( (2a - b) \):
\( = (2a - b)(2a - b - 5) \)
In simple words: Factor \( -5 \) out of the last two terms. This gives you a common binomial which you can then pull out.

Exam Tip: Be mindful of signs when factoring a negative number out of multiple terms.

Question 14. Factorise: \( a(a - 4) - a + 4 \)
Answer:
Group the last two terms together by taking out a factor of \( -1 \):
\( = a(a - 4) - 1(a - 4) \)
Extract the common binomial factor \( (a - 4) \):
\( = (a - 4)(a - 1) \)
In simple words: Rewrite the second part as \( -1(a-4) \). Now you have \( (a-4) \) in both parts, so factor it out.

Exam Tip: Treating a standalone negative binomial group as \( -1(bracket) \) is a simple trick to avoid grouping mistakes.

Question 15. Factorise: \( y^2 - (a + b)y + ab \)
Answer:
Expand the term with the bracket:
\( = y^2 - ay - by + ab \)
Group the terms and factor them in pairs:
\( = y(y - a) - b(y - a) \)
Take out the common factor \( (y - a) \):
\( = (y - a)(y - b) \)
In simple words: Multiply the \( y \) through the middle bracket. Then group the four terms in pairs and factorise.

Exam Tip: Expanding the brackets completely is the best approach when direct factoring isn't clear.

Question 16. Factorise: \( a^2 + \frac{1}{a^2} - 2 - 3a + \frac{3}{a} \)
Answer:
Group the terms into two separate sections:
\( = \left(a^2 + \frac{1}{a^2} - 2\right) - 3\left(a - \frac{1}{a}\right) \)
Recognise that the first group is a perfect square trinomial:
\( = \left(a - \frac{1}{a}\right)^2 - 3\left(a - \frac{1}{a}\right) \)
Now, take out the common factor \( \left(a - \frac{1}{a}\right) \):
\( = \left(a - \frac{1}{a}\right)\left[\left(a - \frac{1}{a}\right) - 3\right] \)
\( = \left(a - \frac{1}{a}\right)\left(a - \frac{1}{a} - 3\right) \)
In simple words: Turn the first three terms into a square of \( \left(a - \frac{1}{a}\right) \). After that, pull out \( \left(a - \frac{1}{a}\right) \) as a common factor.

Exam Tip: Learn to identify algebraic identities such as \( (x - y)^2 = x^2 + y^2 - 2xy \) in fraction forms like \( a^2 + \frac{1}{a^2} - 2 \).

Question 17. Factorise: \( x^2 + y^2 + 2xy + x + y \)
Answer:
Group the first three terms together, as they form a perfect square trinomial:
\( = (x^2 + y^2 + 2xy) + (x + y) \)
Write the first group as a square:
\( = (x + y)^2 + (x + y) \)
Now, pull out the common binomial factor \( (x + y) \):
\( = (x + y)(x + y + 1) \)
In simple words: Notice that the first three terms are just \( (x+y)^2 \). Then, factor out \( (x+y) \) from the whole expression.

Exam Tip: Spotting perfect square trinomials like \( x^2+2xy+y^2 \) early makes complex factoring much simpler.

Question 18. Factorise: \( a^2 + 4b^2 - 4ab - 3a + 6b \)
Answer:
Rearrange and group the terms into a perfect square and a simple linear group:
\( = [a^2 + (2b)^2 - 2 \times a \times 2b] - 3(a - 2b) \)
Convert the first part into a squared binomial:
\( = (a - 2b)^2 - 3(a - 2b) \)
Factor out the common term \( (a - 2b) \):
\( = (a - 2b)(a - 2b - 3) \)
In simple words: Turn the first three terms into a squared group: \( (a - 2b)^2 \). Then factor out \( (a - 2b) \) from the rest.

Exam Tip: Write squared terms as \( (something)^2 \) to clearly see the matching values when grouping.

Question 19. Factorise: \( m(x - 3y)^2 - n(x - 3y) + 5(x - 3y) \)
Answer:
Since \( (x - 3y) \) is common to all terms, let's extract it:
\( = (x - 3y)[m(x - 3y) - n + 5] \)
Expand and simplify inside the square brackets:
\( = (x - 3y)(mx - 3my - n + 5) \)
In simple words: Pull out the common factor \( (x - 3y) \) from all three parts, and expand what is left inside.

Exam Tip: Do not expand the entire polynomial first if a common binomial factor is already present across all terms.

Question 20. Factorise: \( x(6x - 5y) - 4(6x - 5y)^2 \)
Answer:
Extract the common binomial factor \( (6x - 5y) \):
\( = (6x - 5y)[x - 4(6x - 5y)] \)
Simplify the terms within the square brackets:
\( = (6x - 5y)(x - 24x + 20y) \)
\( = (6x - 5y)(-23x + 20y) \)
Rearranging the terms inside the second bracket:
\( = (6x - 5y)(20y - 23x) \)
In simple words: Pull out \( (6x - 5y) \). Then multiply the remaining terms inside the bracket and simplify them.

Exam Tip: Keeping binomial factors intact while factoring is much faster than expanding everything from the beginning.

Exercise 5(B)

Question 1. Factorise: \( a^2 + 10a + 24 \)
Answer:
Let us split the middle term \( 10a \) into two terms whose sum is \( 10a \) and product is \( 24a^2 \):
\( = a^2 + 6a + 4a + 24 \)
Group the terms and factorise them:
\( = a(a + 6) + 4(a + 6) \)
\( = (a + 6)(a + 4) \)
In simple words: Find two numbers that add up to \( 10 \) and multiply to \( 24 \). Those numbers are \( 6 \) and \( 4 \). Use them to split the middle term and factor.

Exam Tip: For trinomials of the form \( x^2 + bx + c \), identify factors of \( c \) that sum to \( b \).

Question 2. Factorise: \( a^2 - 3a - 40 \)
Answer:
We split the middle term \( -3a \) into two parts with a sum of \( -3a \) and a product of \( -40a^2 \):
\( = a^2 - 8a + 5a - 40 \)
Group the terms and extract common factors:
\( = a(a - 8) + 5(a - 8) \)
\( = (a - 8)(a + 5) \)
In simple words: Find two numbers that multiply to \( -40 \) and add up to \( -3 \). These are \( -8 \) and \( 5 \). Use them to group and factor.

Exam Tip: When the constant term is negative, the two split factors must have opposite signs.

Question 3. Factorise: \( 1 - 2a - 3a^2 \)
Answer:
Split the middle term \( -2a \) into terms whose product is \( -3a^2 \):
\( = 1 - 3a + a - 3a^2 \)
Group and factor the expression:
\( = 1(1 - 3a) + a(1 - 3a) \)
\( = (1 + a)(1 - 3a) \)
In simple words: Split the middle term \( -2a \) into \( -3a + a \). Then group and pull out common terms.

Exam Tip: You can factor quadratic expressions even when the terms are written in ascending power order.

Question 4. Factorise: \( x^2 - 3ax - 88a^2 \)
Answer:
Find two terms that sum to \( -3ax \) and multiply to \( -88a^2x^2 \):
\( = x^2 - 11ax + 8ax - 88a^2 \)
Group the terms and pull out common factors:
\( = x(x - 11a) + 8a(x - 11a) \)
\( = (x + 8a)(x - 11a) \)
In simple words: Split the middle term into \( -11ax \) and \( 8ax \), then group and factor out the common brackets.

Exam Tip: Treat the variable \( a \) as a constant factor while splitting the middle term.

Question 5. Factorise: \( 6a^2 - a - 15 \)
Answer:
Find two factors of \( 6 \times (-15) = -90 \) that sum to \( -1 \). These are \( -10 \) and \( 9 \):
\( = 6a^2 - 10a + 9a - 15 \)
Group the terms and factor them:
\( = 2a(3a - 5) + 3(3a - 5) \)
\( = (2a + 3)(3a - 5) \)
In simple words: Split \( -a \) into \( -10a + 9a \). Then find what is common in each pair and factor.

Exam Tip: When the coefficient of \( a^2 \) is not \( 1 \), multiply it by the constant term before finding the factor pair.

Question 6. Factorise: \( 24a^3 + 37a^2 - 5a \)
Answer:
First, pull out the common factor \( a \) from all terms:
\( = a(24a^2 + 37a - 5) \)
Now, split the middle term inside the brackets. We need factors of \( 24 \times (-5) = -120 \) that add up to \( 37 \), which are \( 40 \) and \( -3 \):
\( = a[24a^2 + 40a - 3a - 5] \)
Factor by grouping inside the square brackets:
\( = a[8a(3a + 5) - 1(3a + 5)] \)
\( = a(8a - 1)(3a + 5) \)
In simple words: Take out \( a \) first. Then, split \( 37a \) inside the bracket into \( 40a - 3a \) and factorise the rest.

Exam Tip: Never leave out the common variable factor \( a \) extracted in the first step from your final answer.

Question 7. Factorise: \( a(3a - 2) - 1 \)
Answer:
Expand the bracket first to form a quadratic expression:
\( = 3a^2 - 2a - 1 \)
Split the middle term \( -2a \) into \( -3a + a \):
\( = 3a^2 - 3a + a - 1 \)
Group and factor:
\( = 3a(a - 1) + 1(a - 1) \)
\( = (3a + 1)(a - 1) \)
In simple words: Multiply out the first part to get \( 3a^2 - 2a - 1 \). Then split the middle term and factor it.

Exam Tip: Always expand products of variables before attempting middle-term splitting.

Question 8. Factorise: \( a^2b^2 + 8ab - 9 \)
Answer:
Split the middle term \( 8ab \) into \( 9ab - ab \):
\( = a^2b^2 + 9ab - ab - 9 \)
Group and factor the pairs:
\( = ab(ab + 9) - 1(ab + 9) \)
\( = (ab + 9)(ab - 1) \)
In simple words: Treat \( ab \) like a single variable. Split \( 8ab \) into \( 9ab - ab \) and then factorise by grouping.

Exam Tip: Substituting a single variable like \( y = ab \) can help simplify the steps if dual-variable expressions are confusing.

Question 9. Factorise: \( 3 - a(4 + 7a) \)
Answer:
First, expand the expression:
\( = 3 - 4a - 7a^2 \)
Now, split the middle term \( -4a \) into \( -7a + 3a \):
\( = 3 - 7a + 3a - 7a^2 \)
Group and factor:
\( = 1(3 - 7a) + a(3 - 7a) \)
\( = (3 - 7a)(a + 1) \)
In simple words: Expand the bracket to get \( 3 - 4a - 7a^2 \). Split \( -4a \) into \( -7a + 3a \) to group and factor.

Exam Tip: Pay close attention to negative signs when expanding variables outside of a bracket.

Question 10. Factorise: \( (2a + b)^2 - 6a - 3b - 4 \)
Answer:
First, rewrite the expression by factoring out \( -3 \) from the middle terms:
\( = (2a + b)^2 - 3(2a + b) - 4 \)
Let us substitute \( x = 2a + b \). This gives:
\( = x^2 - 3x - 4 \)
Split the middle term of this quadratic expression:
\( = x^2 - 4x + x - 4 \)
\( = x(x - 4) + 1(x - 4) \)
\( = (x + 1)(x - 4) \)
Now, substitute \( x = 2a + b \) back into the expression:
\( = (2a + b + 1)(2a + b - 4) \)
In simple words: Replace the repeating group \( (2a + b) \) with \( x \). Solve the simple quadratic equation, and then put \( 2a + b \) back in place of \( x \).

Exam Tip: Substituting a complex term with a single variable makes quadratic factorisation much easier to manage.

Question 11. Factorise: \( 1 - 2(a + b) - 3(a + b)^2 \)
Answer:
Let us substitute \( x = a + b \) to simplify the expression:
\( = 1 - 2x - 3x^2 \)
Split the middle term \( -2x \) into \( -3x + x \):
\( = 1 - 3x + x - 3x^2 \)
Group the terms and factorise:
\( = 1(1 - 3x) + x(1 - 3x) \)
\( = (1 - 3x)(1 + x) \)
Now, substitute \( x = a + b \) back into the expression:
\( = [1 - 3(a + b)][1 + (a + b)] \)
\( = (1 - 3a - 3b)(1 + a + b) \)
In simple words: Substitute \( x \) for \( a+b \). Factorise the quadratic expression, and then change \( x \) back to \( a+b \) at the end.

Exam Tip: Remember to multiply the negative sign through when replacing \( x \) with \( (a+b) \) in terms like \( -3x \).

Question 12. Factorise: \( 3a^2 - 1 - 2a \)
Answer:
First, write the terms in standard descending order:
\( = 3a^2 - 2a - 1 \)
Split the middle term \( -2a \) into \( -3a + a \):
\( = 3a^2 - 3a + a - 1 \)
Factor by grouping:
\( = 3a(a - 1) + 1(a - 1) \)
\( = (3a + 1)(a - 1) \)
In simple words: Rearrange the terms into standard quadratic form. Split \( -2a \) into \( -3a + a \), then group and factor.

Exam Tip: Rewriting polynomials in standard form \( ax^2 + bx + c \) is a good habit to prevent errors during splitting.

Question 13. Factorise: \( x^2 + 3x + 2 + ax + 2a \)
Answer:
Factorise the quadratic portion \( x^2 + 3x + 2 \) by splitting the middle term:
\( = x^2 + 2x + x + 2 + ax + 2a \)
Now group the terms in pairs and pull out common factors:
\( = x(x + 2) + 1(x + 2) + a(x + 2) \)
Extract the common binomial factor \( (x + 2) \):
\( = (x + 2)(x + 1 + a) \)
In simple words: Factorise the first three terms as a simple quadratic, then group all terms together to pull out the common bracket \( (x + 2) \).

Exam Tip: Grouping terms with common factors like \( (x+2) \) simplifies expressions containing different variables.

Question 14. Factorise: \( (3x - 2y)^2 + 3(3x - 2y) - 10 \)
Answer:
Let us substitute \( a = 3x - 2y \) to simplify:
\( = a^2 + 3a - 10 \)
Split the middle term \( 3a \) into \( 5a - 2a \):
\( = a^2 + 5a - 2a - 10 \)
Group and factor:
\( = a(a + 5) - 2(a + 5) \)
\( = (a + 5)(a - 2) \)
Substitute \( a = 3x - 2y \) back into the result:
\( = (3x - 2y + 5)(3x - 2y - 2) \)
In simple words: Replace \( (3x - 2y) \) with a single letter \( a \) to factorise easily. Then put the original terms back in place of \( a \) at the end.

Exam Tip: Use brackets when substituting compound binomials back into factored forms to maintain proper sign order.

Question 15. Factorise: \( 5 - (3a^2 - 2a)(6 - 3a^2 + 2a) \)
Answer:
We can rewrite the second bracket to highlight a repeating pattern:
\( = 5 - (3a^2 - 2a)[6 - (3a^2 - 2a)] \)
Let us substitute \( x = 3a^2 - 2a \). This simplifies the expression to:
\( = 5 - x(6 - x) \)
\( = 5 - 6x + x^2 \)
\( = x^2 - 6x + 5 \)
Split the middle term \( -6x \) into \( -5x - x \):
\( = x^2 - 5x - x + 5 \)
\( = x(x - 5) - 1(x - 5) \)
\( = (x - 5)(x - 1) \)
Substitute \( x = 3a^2 - 2a \) back:
\( = (3a^2 - 2a - 5)(3a^2 - 2a - 1) \)
Now, factorise both of these quadratic terms individually:
First part:
\( 3a^2 - 2a - 5 = 3a^2 - 5a + 3a - 5 \)
\( = a(3a - 5) + 1(3a - 5) = (a + 1)(3a - 5) \)
Second part:
\( 3a^2 - 2a - 1 = 3a^2 - 3a + a - 1 \)
\( = 3a(a - 1) + 1(a - 1) = (3a + 1)(a - 1) \)
Combine both parts to get the final factorised form:
\( = (3a - 5)(a + 1)(3a + 1)(a - 1) \)
In simple words: Replace \( (3a^2 - 2a) \) with \( x \) first. Factorise that quadratic to get two brackets, substitute the \( a \) terms back in, and then factorise both of those brackets as well.

Exam Tip: Be prepared to perform multiple stages of factorisation if the substituted brackets yield factorisable quadratics.

Question 16. Determine if the following expressions are factorisable. If yes, factorise them:
(i) \( x^2 - 3x - 54 \)
(ii) \( 2x^2 - 7x - 15 \)
(iii) \( 2x^2 + 2x - 75 \)
(iv) \( 3x^2 + 4x - 10 \)
(v) \( x(2x - 1) - 1 \)
Answer:
To test if a quadratic expression \( ax^2 + bx + c \) is factorisable, we calculate the discriminant \( D = b^2 - 4ac \). If \( D \) is a perfect square, the expression is factorisable.

(i) For \( x^2 - 3x - 54 \):
Comparing with \( ax^2 + bx + c \), we have \( a = 1, b = -3, c = -54 \).
\( b^2 - 4ac = (-3)^2 - 4(1)(-54) = 9 + 216 = 225 \)
Since \( 225 = 15^2 \) is a perfect square, the expression is factorisable.
\( x^2 - 3x - 54 = x^2 - 9x + 6x - 54 \)
\( = x(x - 9) + 6(x - 9) = (x - 9)(x + 6) \)

(ii) For \( 2x^2 - 7x - 15 \):
Comparing with \( ax^2 + bx + c \), we have \( a = 2, b = -7, c = -15 \).
\( b^2 - 4ac = (-7)^2 - 4(2)(-15) = 49 + 120 = 169 \)
Since \( 169 = 13^2 \) is a perfect square, the expression is factorisable.
\( 2x^2 - 7x - 15 = 2x^2 - 10x + 3x - 15 \)
\( = 2x(x - 5) + 3(x - 5) = (2x + 3)(x - 5) \)

(iii) For \( 2x^2 + 2x - 75 \):
Comparing with \( ax^2 + bx + c \), we have \( a = 2, b = 2, c = -75 \).
\( b^2 - 4ac = (2)^2 - 4(2)(-75) = 4 + 600 = 604 \)
Since \( 604 \) is not a perfect square, the expression is not factorisable.

(iv) For \( 3x^2 + 4x - 10 \):
Comparing with \( ax^2 + bx + c \), we have \( a = 3, b = 4, c = -10 \).
\( b^2 - 4ac = (4)^2 - 4(3)(-10) = 16 + 120 = 136 \)
Since \( 136 \) is not a perfect square, the expression is not factorisable.

(v) For \( x(2x - 1) - 1 \):
Simplify to standard form: \( 2x^2 - x - 1 \).
Comparing with \( ax^2 + bx + c \), we have \( a = 2, b = -1, c = -1 \).
\( b^2 - 4ac = (-1)^2 - 4(2)(-1) = 1 + 8 = 9 \)
Since \( 9 = 3^2 \) is a perfect square, the expression is factorisable.
\( 2x^2 - x - 1 = 2x^2 - 2x + x - 1 \)
\( = 2x(x - 1) + 1(x - 1) = (2x + 1)(x - 1) \)
In simple words: Check if \( b^2 - 4ac \) is a perfect square. If it is, split the middle term and factorise. If not, it cannot be factorised.

Exam Tip: Calculating the discriminant \( b^2 - 4ac \) is a foolproof way to verify if an expression can be factored before doing any splitting work.

Exercise 5(C)

Question 1. Factorise: \( 25a^2 - 9b^2 \)
Answer:
We use the difference of squares identity \( x^2 - y^2 = (x - y)(x + y) \):
\( = (5a)^2 - (3b)^2 \)
\( = (5a - 3b)(5a + 3b) \)
In simple words: Write both terms as perfect squares and use the standard difference of squares formula.

Exam Tip: Always look to express both terms in the form \( (A)^2 - (B)^2 \) to apply this identity.

Question 2. Factorise: \( a^2 - (2a + 3b)^2 \)
Answer:
Apply the difference of squares identity \( A^2 - B^2 = (A - B)(A + B) \):
\( = [a - (2a + 3b)][a + (2a + 3b)] \)
\( = (a - 2a - 3b)(a + 2a + 3b) \)
\( = (-a - 3b)(3a + 3b) \)
Factor out \( -1 \) from the first bracket and \( 3 \) from the second bracket:
\( = -(a + 3b) \times 3(a + b) \)
\( = -3(a + 3b)(a + b) \)
In simple words: Treat \( a \) and \( (2a+3b) \) as our squared parts. Apply the formula, simplify the brackets, and pull out any common factors.

Exam Tip: Be sure to factor out any constants from the final brackets to completely simplify the expression.

Question 3. Factorise: \( a^2 - 81(b - c)^2 \)
Answer:
Express both terms as perfect squares:
\( = a^2 - [9(b - c)]^2 \)
Using the difference of squares identity:
\( = [a - 9(b - c)][a + 9(b - c)] \)
\( = (a - 9b + 9c)(a + 9b - 9c) \)
In simple words: Write \( 81(b-c)^2 \) as \( [9(b-c)]^2 \). Then apply the difference of squares formula and open the inner brackets.

Exam Tip: Distribute the constant factor (like \( 9 \)) through the binomial before completing the subtraction steps.

Question 4. Factorise: \( 25(2a - b)^2 - 81b^2 \)
Answer:
Write the terms as perfect squares:
\( = [5(2a - b)]^2 - (9b)^2 \)
Using the difference of squares identity:
\( = [5(2a - b) - 9b][5(2a - b) + 9b] \)
Expand inside the brackets:
\( = (10a - 5b - 9b)(10a - 5b + 9b) \)
\( = (10a - 14b)(10a + 4b) \)
Factor out common constants from both brackets:
\( = 2(5a - 7b) \times 2(5a + 2b) \)
\( = 4(5a - 7b)(5a + 2b) \)
In simple words: Convert both parts to squares, apply the formula, simplify the expressions inside, and then pull out the common numbers at the end.

Exam Tip: Always look for common numerical factors like \( 2 \) in the final linear factors to simplify the product.

Question 5. Factorise: \( 50a^3 - 2a \)
Answer:
First, extract the common factor \( 2a \):
\( = 2a(25a^2 - 1) \)
Apply the difference of squares identity to the term in the brackets:
\( = 2a[(5a)^2 - (1)^2] \)
\( = 2a(5a - 1)(5a + 1) \)
In simple words: Pull out \( 2a \) first. Then write the bracket as a difference of squares and factorise it.

Exam Tip: Factoring out the greatest common monomial first is often required to reveal the difference of squares form.

Question 6. Factorise: \( 4a^2b - 9b^3 \)
Answer:
Extract the common monomial factor \( b \):
\( = b(4a^2 - 9b^2) \)
Write the terms inside the bracket as perfect squares:
\( = b[(2a)^2 - (3b)^2] \)
Apply the difference of squares identity:
\( = b(2a - 3b)(2a + 3b) \)
In simple words: Take out \( b \) from both terms, then apply the difference of squares formula on the remaining part inside the bracket.

Exam Tip: Always ensure the variable left outside (like \( b \)) is included in the final factored expression.

Question 7. Factorise: \( 3a^5 - 108a^3 \)
Answer:
Factorise out the greatest common monomial \( 3a^3 \):
\( = 3a^3(a^2 - 36) \)
Recognise \( 36 \) as a perfect square \( 6^2 \):
\( = 3a^3(a^2 - 6^2) \)
Factorise using difference of squares:
\( = 3a^3(a - 6)(a + 6) \)
In simple words: Pull out \( 3a^3 \) first, leaving \( a^2 - 36 \). Then write \( 36 \) as \( 6^2 \) and apply the difference of squares.

Exam Tip: Watch for high-power variables (like \( a^5 \)) that can be reduced by factoring out common powers first.

Question 8. Factorise: \( 9(a - 2)^2 - 16(a + 2)^2 \)
Answer:
Express the terms as squared expressions:
\( = [3(a - 2)]^2 - [4(a + 2)]^2 \)
Apply the difference of squares identity:
\( = [3(a - 2) - 4(a + 2)][3(a - 2) + 4(a + 2)] \)
Expand inside the brackets:
\( = (3a - 6 - 4a - 8)(3a - 6 + 4a + 8) \)
\( = (-a - 14)(7a + 2) \)
Factor out the negative sign to write it in standard form:
\( = -(a + 14)(7a + 2) \)
In simple words: Convert the coefficients into squares inside the brackets. Then apply the difference of squares formula and simplify.

Exam Tip: Be careful to apply negative signs to both terms of the binomial when simplifying subtraction brackets.

Question 9. Factorise: \( a^4 - 1 \)
Answer:
Write \( a^4 \) as \( (a^2)^2 \):
\( = (a^2)^2 - (1)^2 \)
Apply difference of squares:
\( = (a^2 - 1)(a^2 + 1) \)
Further factorise the first term which is also a difference of squares:
\( = (a - 1)(a + 1)(a^2 + 1) \)
In simple words: Treat \( a^4 \) as \( (a^2)^2 \). Apply the difference of squares once, and then apply it again to the \( a^2 - 1 \) term.

Exam Tip: Always check if any of the remaining factors (like \( a^2 - 1 \)) can be factorised further.

Question 10. Factorise: \( a^3 + 2a^2 - a - 2 \)
Answer:
Group the terms into pairs:
\( = a^2(a + 2) - 1(a + 2) \)
Factor out the binomial \( (a + 2) \):
\( = (a^2 - 1)(a + 2) \)
Now, factorise the difference of squares term \( (a^2 - 1) \):
\( = (a - 1)(a + 1)(a + 2) \)
In simple words: Group the terms in pairs and factor out the common parts. Once you get \( (a^2 - 1) \), break it down using the difference of squares identity.

Exam Tip: Look for opportunities to apply the difference of squares identity even after performing grouping factorisation.

Question 11. Factorise: \( (a + b)^3 - a - b \)
Answer:
Group the last two terms by factoring out \( -1 \):
\( = (a + b)^3 - (a + b) \)
Extract the common term \( (a + b) \):
\( = (a + b)[(a + b)^2 - 1] \)
Factorise the term inside the bracket as a difference of squares:
\( = (a + b)(a + b - 1)(a + b + 1) \)
In simple words: Group the terms to find the common factor \( (a+b) \). Pull it out, and then apply the difference of squares formula to the remaining part.

Exam Tip: Treating the binomial \( (a+b) \) as a single entity helps clarify the difference of squares pattern.

Question 12. Factorise: \( a(a - 1) - b(b - 1) \)
Answer:
Expand the brackets first:
\( = a^2 - a - b^2 + b \)
Rearrange terms to group squares together:
\( = (a^2 - b^2) - (a - b) \)
Apply difference of squares to the first term and factor out \( -1 \) from the second:
\( = (a - b)(a + b) - 1(a - b) \)
Now, factorise out the common term \( (a - b) \):
\( = (a - b)(a + b - 1) \)
In simple words: Open the brackets and group the squares together. Factor the squares using the formula, then pull out the common bracket.

Exam Tip: Grouping squares together is a very effective technique when dealing with quadratic variables of different types.

Question 13. Factorise: \( 4a^2 - 4b^2 - 4bc - c^2 \)
Answer:
Rearrange the terms and group the last three terms inside a negative bracket:
\( = 4a^2 - (4b^2 + 4bc + c^2) \)
Recognise that the term in brackets is a perfect square trinomial:
\( = (2a)^2 - (2b + c)^2 \)
Apply the difference of squares identity:
\( = [2a - (2b + c)][2a + (2b + c)] \)
\( = (2a - 2b - c)(2a + 2b + c) \)
In simple words: Put the last three terms in a bracket with a minus sign in front, which makes it a perfect square \( (2b+c)^2 \). Then apply the difference of squares.

Exam Tip: Watch for a perfect square trinomial preceded by a minus sign, which allows you to create a difference of squares.

Question 14. Factorise: \( 4a^2 - 49b^2 + 2a - 7b \)
Answer:
Group the terms into two parts:
\( = (4a^2 - 49b^2) + (2a - 7b) \)
Factorise the first group as a difference of squares:
\( = (2a - 7b)(2a + 7b) + 1(2a - 7b) \)
Extract the common binomial factor \( (2a - 7b) \):
\( = (2a - 7b)(2a + 7b + 1) \)
In simple words: Split the expression into two pairs. Factorise the first pair using difference of squares, then pull out the common factor \( (2a-7b) \).

Exam Tip: Always search for common binomial factors between different groups of an expression.

Question 15. Factorise: \( 9a^2 + 3a - 8b - 64b^2 \)
Answer:
Rearrange the terms to group the squares together:
\( = (9a^2 - 64b^2) + (3a - 8b) \)
Factorise the first group as a difference of squares:
\( = (3a - 8b)(3a + 8b) + 1(3a - 8b) \)
Now, take out the common term \( (3a - 8b) \):
\( = (3a - 8b)(3a + 8b + 1) \)
In simple words: Swap terms around so the squares are together. Factorise the squares first, then pull out the common binomial.

Exam Tip: Rearranging terms based on their degrees (squares vs. linear terms) is a highly useful step in algebraic grouping.

Question 16. Factorise: \( 4a^2 - 12a + 9 - 49b^2 \)
Answer:
Group the first three terms, which form a perfect square trinomial:
\( = (4a^2 - 12a + 9) - 49b^2 \)
Write both parts as squares:
\( = (2a - 3)^2 - (7b)^2 \)
Apply the difference of squares identity:
\( = (2a - 3 - 7b)(2a - 3 + 7b) \)
In simple words: Group the first three terms into a squared bracket: \( (2a - 3)^2 \). Then apply the difference of squares formula with \( 7b \).

Exam Tip: Look out for trinomials that can be compressed into a perfect square before subtracting another squared term.

Question 17. Factorise: \( 4xy - x^2 - 4y^2 + z^2 \)
Answer:
Rearrange terms to place the positive square first:
\( = z^2 - (x^2 + 4y^2 - 4xy) \)
Identify the trinomial inside the brackets as a perfect square:
\( = z^2 - (x - 2y)^2 \)
Factorise using the difference of squares identity:
\( = [z - (x - 2y)][z + (x - 2y)] \)
\( = (z - x + 2y)(z + x - 2y) \)
In simple words: Rearrange the terms so you can put a minus sign in front of a perfect square group. Then apply the difference of squares.

Exam Tip: Always put terms inside parenthesis when subtracting a binomial to ensure correct sign application.

Question 18. Factorise: \( a^2 + b^2 - c^2 - d^2 + 2ab - 2cd \)
Answer:
Group terms to form two separate perfect square trinomials:
\( = (a^2 + b^2 + 2ab) - (c^2 + d^2 + 2cd) \)
Express both groups as perfect squares:
\( = (a + b)^2 - (c + d)^2 \)
Apply the difference of squares identity:
\( = [(a + b) - (c + d)][(a + b) + (c + d)] \)
\( = (a + b - c - d)(a + b + c + d) \)
In simple words: Group the terms into two perfect square brackets, one positive and one negative. Use the difference of squares formula to factorise.

Exam Tip: Recognise that factoring out a minus sign from a group changes all internal signs, which can form a perfect trinomial square like \( (c+d)^2 \).

Question 19. Factorise: \( 4x^2 - 12ax - y^2 - z^2 - 2yz + 9a^2 \)
Answer:
Rearrange terms to group the independent variables into two distinct perfect squares:
\( = (4x^2 - 12ax + 9a^2) - (y^2 + 2yz + z^2) \)
Express each trinomial as a perfect square:
\( = (2x - 3a)^2 - (y + z)^2 \)
Factorise using the difference of squares identity:
\( = [2x - 3a - (y + z)][2x - 3a + (y + z)] \)
\( = (2x - 3a - y - z)(2x - 3a + y + z) \)
In simple words: Sort the terms into two separate square groups: one for \( x \) and \( a \), and one for \( y \) and \( z \). Then apply the difference of squares.

Exam Tip: Group terms containing related variables together to search for perfect square components.

Question 20. Factorise: \( (a^2 - 1)(b^2 - 1) + 4ab \)
Answer:
Expand the multiplication first:
\( = a^2b^2 - a^2 - b^2 + 1 + 4ab \)
Split \( 4ab \) into \( 2ab + 2ab \) and rearrange terms to form two perfect square groups:
\( = (a^2b^2 + 2ab + 1) - (a^2 + b^2 - 2ab) \)
Write both groups as perfect squares:
\( = (ab + 1)^2 - (a - b)^2 \)
Factorise using the difference of squares identity:
\( = [ab + 1 - (a - b)][ab + 1 + (a - b)] \)
\( = (ab - a + b + 1)(ab + a - b + 1) \)
In simple words: Multiply out the brackets first. Split \( 4ab \) to make two perfect squares, then apply the difference of squares identity.

Exam Tip: Splitting terms (like \( 4ab \) into \( 2ab + 2ab \)) is a standard technique to solve trickier algebraic factorisation problems.

Question 21. Factorise: \( x^4 + x^2 + 1 \)
Answer:
Add and subtract \( x^2 \) to create a perfect square trinomial:
\( = x^4 + 2x^2 + 1 - x^2 \)
Write the first three terms as a perfect square:
\( = (x^2 + 1)^2 - (x)^2 \)
Factorise using the difference of squares identity:
\( = (x^2 + 1 - x)(x^2 + 1 + x) \)
Rearranging the terms in standard quadratic order:
\( = (x^2 - x + 1)(x^2 + x + 1) \)
In simple words: Add \( x^2 \) to make the first part a perfect square \( (x^2+1)^2 \), and subtract \( x^2 \) to keep the expression balanced. Then use the difference of squares.

Exam Tip: Adding and subtracting the same term (completing the square) is a classic algebraic method for factorising quartic expressions.

Question 22. Factorise:
\( (a^2 + b^2 - 4c^2)^2 - 4a^2b^2 \)
Answer:
We can rewrite the expression as:
\( = (a^2 + b^2 - 4c^2)^2 - (2ab)^2 \)
Applying the identity \( x^2 - y^2 = (x - y)(x + y) \):
\( = (a^2 + b^2 - 4c^2 - 2ab)(a^2 + b^2 - 4c^2 + 2ab) \)
Grouping the terms to form perfect squares:
\( = (a^2 + b^2 - 2ab - 4c^2)(a^2 + b^2 + 2ab - 4c^2) \)
\( = \left[(a - b)^2 - (2c)^2\right]\left[(a + b)^2 - (2c)^2\right] \)
Using the difference of squares formula once more on both parts:
\( = (a - b - 2c)(a - b + 2c)(a + b - 2c)(a + b + 2c) \)
In simple words: This expression is factored by using the difference of squares formula \( x^2 - y^2 = (x - y)(x + y) \) two times. Rearranging the terms helps reveal the hidden square patterns.

Exam Tip: When you see a fourth-degree expression or nested squares, always check if you can apply the difference of squares formula a second time to fully factorise it.

Question 23. Factorise:
\( (x^2 + 4y^2 - 9z^2)^2 - 16x^2y^2 \)
Answer:
Expressing the second term as a perfect square:
\( = (x^2 + 4y^2 - 9z^2)^2 - (4xy)^2 \)
Applying the difference of squares identity \( a^2 - b^2 = (a - b)(a + b) \):
\( = (x^2 + 4y^2 - 9z^2 - 4xy)(x^2 + 4y^2 - 9z^2 + 4xy) \)
Rearranging to highlight the perfect square trinomials:
\( = (x^2 - 4xy + 4y^2 - 9z^2)(x^2 + 4xy + 4y^2 - 9z^2) \)
\( = \left[(x - 2y)^2 - (3z)^2\right]\left[(x + 2y)^2 - (3z)^2\right] \)
Applying the difference of squares rule again on both parts:
\( = (x - 2y - 3z)(x - 2y + 3z)(x + 2y - 3z)(x + 2y + 3z) \)
In simple words: Express the terms as squares and apply the difference of squares formula. Grouping the trinomial parts reveals another set of perfect squares to factorise further.

Exam Tip: Keep your variables organized. Grouping \( x^2 \pm 4xy + 4y^2 \) as \( (x \pm 2y)^2 \) is key to simplifying the expression.

Question 24. Factorise:
\( (a + b)^2 - a^2 + b^2 \)
Answer:
Expanding the first term:
\( = a^2 + 2ab + b^2 - a^2 + b^2 \)
Combining like terms:
\( = 2ab + 2b^2 \)
Factoring out the common term \( 2b \):
\( = 2b(a + b) \)
In simple words: Expand the squared term first, then cancel out \( a^2 \) and add the \( b^2 \) terms. Lastly, take out the common factor \( 2b \).

Exam Tip: Be careful with signs when canceling terms. Notice that \( a^2 \) and \( -a^2 \) cancel out, but \( b^2 \) and \( b^2 \) add up to \( 2b^2 \).

Question 25. Factorise:
\( a^2 - b^2 - (a + b)^2 \)
Answer:
First, expand the term in parentheses:
\( = a^2 - b^2 - (a^2 + 2ab + b^2) \)
Distribute the negative sign:
\( = a^2 - b^2 - a^2 - 2ab - b^2 \)
Simplifying the expression:
\( = -2ab - 2b^2 \)
Taking out the common factor of \( -2b \):
\( = -2b(a + b) \)
In simple words: Expand the squared brackets and distribute the negative sign to all terms inside. Simplify by canceling \( a^2 \) and grouping the rest, then pull out the common factor \( -2b \).

Exam Tip: Don't forget to distribute the minus sign to every single term inside the expanded parenthesis, especially the middle term \( 2ab \).

Question 26. Factorise:
\( 9a^2 - (a^2 - 4)^2 \)
Answer:
Rewriting as a difference of two squares:
\( = (3a)^2 - (a^2 - 4)^2 \)
Applying the identity \( x^2 - y^2 = (x - y)(x + y) \):
\( = [3a - (a^2 - 4)][3a + (a^2 - 4)] \)
\( = [-a^2 + 3a + 4][a^2 + 3a - 4] \)
Splitting the middle term for both quadratic expressions:
\( = [-a^2 + 4a - a + 4][a^2 + 4a - a - 4] \)
\( = [a(-a + 4) + 1(-a + 4)][a(a + 4) - 1(a + 4)] \)
\( = [(a + 1)(4 - a)][(a - 1)(a + 4)] \)
\( = (a + 1)(4 - a)(a - 1)(a + 4) \)
In simple words: Treat this as a difference of squares first. Once you expand, you get two quadratic trinomials. Factor each one of them by splitting their middle terms.

Exam Tip: In problems like this, always check if the resulting factors can be factored further. Here, both trinomials can be split and factored completely.

Question 27. Factorise:
\( x^2 + \frac{1}{x^2} - 11 \)
Answer:
We can split the constant term \( -11 \) into \( -2 \) and \( -9 \):
\( = x^2 + \frac{1}{x^2} - 2 - 9 \)
We know that \( x^2 + \frac{1}{x^2} - 2 \times x \times \frac{1}{x} \) forms a perfect square:
\( = \left(x - \frac{1}{x}\right)^2 - (3)^2 \)
Using the difference of squares identity:
\( = \left(x - \frac{1}{x} + 3\right)\left(x - \frac{1}{x} - 3\right) \)
In simple words: Break \( -11 \) into \( -2 \) and \( -9 \). The first part becomes the perfect square \( \left(x - \frac{1}{x}\right)^2 \), and the remaining \( 9 \) is \( 3^2 \). Then apply the \( a^2 - b^2 \) formula.

Exam Tip: Splitting the constant term is a standard trick in algebra to create a perfect square trinomial and a remaining perfect square.

Question 28. Factorise:
\( 4x^2 + \frac{1}{4x^2} + 1 \)
Answer:
We can split the constant \( +1 \) into \( +2 - 1 \):
\( = 4x^2 + \frac{1}{4x^2} + 2 - 1 \)
Rewrite the first three terms as a perfect square:
\( = (2x)^2 + \frac{1}{(2x)^2} + 2 \times 2x \times \frac{1}{2x} - 1 \)
\( = \left(2x + \frac{1}{2x}\right)^2 - (1)^2 \)
Applying the difference of squares formula:
\( = \left(2x + \frac{1}{2x} + 1\right)\left(2x + \frac{1}{2x} - 1\right) \)
In simple words: Change \( +1 \) to \( +2 - 1 \) so that you can group the first part into a perfect square. Then factor using the difference of squares.

Exam Tip: Always look for terms of the form \( a^2 + b^2 \). Adding and subtracting the required constant can construct the necessary \( 2ab \) term.

Question 29. Factorise:
\( 4x^4 - x^2 - 12x - 36 \)
Answer:
Group the last three terms by factoring out a negative sign:
\( = 4x^4 - (x^2 + 12x + 36) \)
Expressing both parts as perfect squares:
\( = (2x^2)^2 - (x + 6)^2 \)
Applying the difference of squares formula:
\( = (2x^2 + x + 6)(2x^2 - (x + 6)) \)
\( = (2x^2 + x + 6)(2x^2 - x - 6) \)
Now, split the middle term of the second quadratic factor:
\( = (2x^2 + x + 6)(2x^2 - 4x + 3x - 6) \)
\( = (2x^2 + x + 6)[2x(x - 2) + 3(x - 2)] \)
\( = (2x^2 + x + 6)(x - 2)(2x + 3) \)
In simple words: Put a bracket around the last three terms by pulling out a minus sign. This gives a perfect square, allowing you to use the difference of squares method. Finally, factor the remaining quadratic part.

Exam Tip: Be careful with the signs when you factor out the negative sign from \( -x^2 - 12x - 36 \). All terms inside the parentheses become positive.

Question 30. Factorise:
\( a^2(b + c) - (b + c)^3 \)
Answer:
First, take out the common factor \( (b + c) \):
\( = (b + c)\left[a^2 - (b + c)^2\right] \)
Use the difference of squares formula inside the brackets:
\( = (b + c)[a - (b + c)][a + (b + c)] \)
Simplify the signs inside the brackets:
\( = (b + c)(a - b - c)(a + b + c) \)
In simple words: Pull out the common factor of \( (b + c) \) first. The leftover part in the bracket is a difference of two squares, which can be factored easily.

Exam Tip: Always look for common expressions to factor out first. It simplifies the remaining terms immensely.

Exercise 5(D)

Question 1. Factorise:
\( a^3 - 27 \)
Answer:
We can write this as a difference of two cubes:
\( = (a)^3 - (3)^3 \)
Applying the algebraic identity \( x^3 - y^3 = (x - y)(x^2 + xy + y^2) \):
\( = (a - 3)[(a)^2 + a \times 3 + (3)^2] \)
\( = (a - 3)(a^2 + 3a + 9) \)
In simple words: Write \( 27 \) as \( 3^3 \) and then use the difference of cubes formula to expand it.

Exam Tip: Memorize the sum and difference of cubes identities perfectly, as they are very frequently tested.

Question 2. Factorise:
\( 1 - 8a^3 \)
Answer:
Expressing as a difference of cubes:
\( = (1)^3 - (2a)^3 \)
Applying the formula \( x^3 - y^3 = (x - y)(x^2 + xy + y^2) \):
\( = (1 - 2a)[(1)^2 + 1 \times 2a + (2a)^2] \)
\( = (1 - 2a)(1 + 2a + 4a^2) \)
In simple words: Rewrite \( 1 \) as \( 1^3 \) and \( 8a^3 \) as \( (2a)^3 \). Then, use the standard formula for the difference of two cubes.

Exam Tip: When dealing with terms like \( 8a^3 \), make sure to cube the entire term \( (2a)^3 \), not just the variable.

Question 3. Factorise:
\( 64 - a^3b^3 \)
Answer:
Rewriting as a difference of cubes:
\( = (4)^3 - (ab)^3 \)
Using the identity \( x^3 - y^3 = (x - y)(x^2 + xy + y^2) \):
\( = (4 - ab)[(4)^2 + 4 \times ab + (ab)^2] \)
\( = (4 - ab)(16 + 4ab + a^2b^2) \)
In simple words: Write \( 64 \) as \( 4^3 \) and \( a^3b^3 \) as \( (ab)^3 \). Then plug these into the difference of cubes identity.

Exam Tip: Treat \( ab \) as a single entity when applying the identity to ensure that both \( a \) and \( b \) are squared in the final term.

Question 4. Factorise:
\( a^6 + 27b^3 \)
Answer:
Rewriting the expression as a sum of two cubes:
\( = (a^2)^3 + (3b)^3 \)
Using the identity \( x^3 + y^3 = (x + y)(x^2 - xy + y^2) \):
\( = (a^2 + 3b)[(a^2)^2 - a^2 \times 3b + (3b)^2] \)
\( = (a^2 + 3b)(a^4 - 3a^2b + 9b^2) \)
In simple words: Write \( a^6 \) as \( (a^2)^3 \) and \( 27b^3 \) as \( (3b)^3 \). Then use the sum of cubes formula to factor it.

Exam Tip: Be careful with the middle term sign in the sum of cubes formula; it is negative: \( -xy \).

Question 5. Factorise:
\( 3x^7y - 81x^4y^4 \)
Answer:
First, pull out the common factor \( 3xy \):
\( = 3xy(x^6 - 27x^3y^3) \)
Expressing the inner term as a difference of cubes:
\( = 3xy\left[(x^2)^3 - (3xy)^3\right] \)
Using the identity \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \):
\( = 3xy(x^2 - 3xy)\left[(x^2)^2 + x^2 \times 3xy + (3xy)^2\right] \)
\( = 3xy(x^2 - 3xy)(x^4 + 3x^3y + 9x^2y^2) \)
Factoring out \( x \) from the first bracket and \( x^2 \) from the second bracket:
\( = 3xy \cdot x(x - 3y) \cdot x^2(x^2 + 3xy + 9y^2) \)
Combining all the outer terms:
\( = 3x^4y(x - 3y)(x^2 + 3xy + 9y^2) \)
In simple words: Start by pulling out \( 3xy \). Use the difference of cubes on the rest. Finally, factor out \( x \) and \( x^2 \) from the resulting expressions and combine them with the front terms.

Exam Tip: Always look to factor out common terms at every stage of factorization to reach the simplest, fully-factored form.

Question 6. Factorise:
\( a^3 - \frac{27}{a^3} \)
Answer:
Rewriting as a difference of cubes:
\( = (a)^3 - \left(\frac{3}{a}\right)^3 \)
Using the difference of cubes identity \( x^3 - y^3 = (x - y)(x^2 + xy + y^2) \):
\( = \left(a - \frac{3}{a}\right)\left[(a)^2 + a \times \frac{3}{a} + \left(\frac{3}{a}\right)^2\right] \)
\( = \left(a - \frac{3}{a}\right)\left(a^2 + 3 + \frac{9}{a^2}\right) \)
In simple words: Convert the expression into the difference of cubes of \( a \) and \( \frac{3}{a} \). Then, apply the standard cubic formula and simplify the middle term.

Exam Tip: Notice that in the middle term, \( a \times \frac{3}{a} \) simplifies to just \( 3 \) because the \( a \)'s cancel out.

Question 7. Factorise:
\( a^3 + 0.064 \)
Answer:
Rewriting as a sum of cubes:
\( = (a)^3 + (0.4)^3 \)
Applying the identity \( x^3 + y^3 = (x + y)(x^2 - xy + y^2) \):
\( = (a + 0.4)\left[(a)^2 - a \times 0.4 + (0.4)^2\right] \)
\( = (a + 0.4)(a^2 - 0.4a + 0.16) \)
In simple words: Write \( 0.064 \) as \( 0.4^3 \). Use the sum of cubes formula and calculate the squares and products of decimal numbers carefully.

Exam Tip: Always double-check decimal cubes; for example, \( 0.4 \times 0.4 \times 0.4 = 0.064 \) is correct, whereas \( 0.04^3 \) would be \( 0.000064 \).

Question 8. Factorise:
\( a^4 - 343a \)
Answer:
First, take out the common factor \( a \):
\( = a(a^3 - 343) \)
Expressing \( 343 \) as \( 7^3 \):
\( = a(a^3 - 7^3) \)
Applying the difference of cubes identity:
\( = a(a - 7)[a^2 + a \times 7 + (7)^2] \)
\( = a(a - 7)(a^2 + 7a + 49) \)
In simple words: Factor out the common \( a \) first. Then, notice that \( 343 \) is \( 7 \) cubed and use the difference of cubes formula.

Exam Tip: Remembering the cubes of numbers from 1 to 10 is very useful. Recognizing \( 343 = 7^3 \) saves time in exams.

Question 9. Factorise:
\( (x - y)^3 - 8x^3 \)
Answer:
Expressing as a difference of cubes:
\( = (x - y)^3 - (2x)^3 \)
Applying the identity \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \):
\( = \left[(x - y) - 2x\right]\left[(x - y)^2 + (x - y)(2x) + (2x)^2\right] \)
Simplifying the first bracket and expanding the second bracket:
\( = (-x - y)\left[(x^2 - 2xy + y^2) + (2x^2 - 2xy) + 4x^2\right] \)
Combining like terms inside the second bracket:
\( = -(x + y)(7x^2 - 4xy + y^2) \)
In simple words: Treat \( (x - y) \) as the first term and \( 2x \) as the second. Apply the difference of cubes formula, then expand all bracket terms and simplify by grouping.

Exam Tip: Be methodical when expanding \( (x-y)^2 \) and multiplying \( (x-y)(2x) \). Group like terms carefully to avoid arithmetic errors.

Question 10. Factorise:
\( \frac{8a^3}{27} - \frac{b^3}{8} \)
Answer:
Writing both terms as cubes:
\( = \left(\frac{2a}{3}\right)^3 - \left(\frac{b}{2}\right)^3 \)
Using the identity for difference of cubes:
\( = \left(\frac{2a}{3} - \frac{b}{2}\right)\left[\left(\frac{2a}{3}\right)^2 + \frac{2a}{3} \times \frac{b}{2} + \left(\frac{b}{2}\right)^2\right] \)
Simplifying the terms:
\( = \left(\frac{2a}{3} - \frac{b}{2}\right)\left(\frac{4a^2}{9} + \frac{ab}{3} + \frac{b^2}{4}\right) \)
In simple words: Rewrite both fractions as cubed values. Then use the standard cubic formula, keeping track of the fractions.

Exam Tip: Make sure you simplify the product term \( \frac{2a}{3} \times \frac{b}{2} \). The \( 2 \) in the numerator and denominator cancel out, leaving \( \frac{ab}{3} \).

Question 11. Factorise:
\( a^6 - b^6 \)
Answer:
Rewrite as a difference of squares:
\( = (a^3)^2 - (b^3)^2 \)
Factoring as a difference of squares:
\( = (a^3 + b^3)(a^3 - b^3) \)
Now, apply both sum and difference of cubes identities:
We know that:
\( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \)
\( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \)
Substituting these into our expression:
\( = (a + b)(a^2 - ab + b^2)(a - b)(a^2 + ab + b^2) \)
Rearranging the factors:
\( = (a + b)(a - b)(a^2 - ab + b^2)(a^2 + ab + b^2) \)
In simple words: Use the difference of squares first to get two cubic parts. Then, expand both of these cubic parts using the sum and difference of cubes formulas.

Exam Tip: Starting with the difference of squares \( (a^3)^2 - (b^3)^2 \) is much easier than starting with the difference of cubes \( (a^2)^3 - (b^2)^3 \).

Question 12. Factorise:
\( a^6 - 7a^3 - 8 \)
Answer:
Splitting the middle term of the polynomial:
\( = a^6 - 8a^3 + a^3 - 8 \)
Factoring by grouping:
\( = a^3(a^3 - 8) + 1(a^3 - 8) \)
\( = (a^3 + 1)(a^3 - 8) \)
Expressing as sum and difference of cubes:
\( = (a^3 + 1^3)(a^3 - 2^3) \)
Applying the sum and difference of cubes formulas:
\( = (a + 1)(a^2 - a + 1)(a - 2)(a^2 + 2a + 4) \)
Rearranging the factors:
\( = (a + 1)(a - 2)(a^2 - a + 1)(a^2 + 2a + 4) \)
In simple words: Treat this as a quadratic equation in terms of \( a^3 \). Split the middle term to factor it into two cubic parts, then factor each cubic part completely.

Exam Tip: Recognizing \( a^6 - 7a^3 - 8 \) as a quadratic-style polynomial (substituting \( x = a^3 \)) makes it easy to factor by splitting the middle term.

Question 13. Factorise:
\( a^3 - 27b^3 + 2a^2b - 6ab^2 \)
Answer:
Group and rewrite the terms:
\( = (a)^3 - (3b)^3 + 2ab(a - 3b) \)
Applying the difference of cubes identity to the first part:
\( = (a - 3b)\left[a^2 + a \times 3b + (3b)^2\right] + 2ab(a - 3b) \)
\( = (a - 3b)(a^2 + 3ab + 9b^2) + 2ab(a - 3b) \)
Factoring out the common binomial \( (a - 3b) \):
\( = (a - 3b)\left[a^2 + 3ab + 9b^2 + 2ab\right] \)
Simplifying the expression inside the brackets:
\( = (a - 3b)(a^2 + 5ab + 9b^2) \)
In simple words: Group the first two terms as a difference of cubes, and factor the last two terms. You'll get a common bracket \( (a - 3b) \) that you can pull out to simplify the rest.

Exam Tip: Look for opportunities to create a common binomial factor by combining different factorization techniques (like grouping and cubing) in the same expression.

Question 14. Factorise:
\( 8a^3 - b^3 - 4ax + 2bx \)
Answer:
Grouping the terms into two sections:
\( = \left[(2a)^3 - (b)^3\right] - 2x(2a - b) \)
Factoring the difference of cubes in the first part:
\( = (2a - b)\left[(2a)^2 + 2a \times b + (b)^2\right] - 2x(2a - b) \)
\( = (2a - b)(4a^2 + 2ab + b^2) - 2x(2a - b) \)
Pulling out the common factor \( (2a - b) \):
\( = (2a - b)(4a^2 + 2ab + b^2 - 2x) \)
In simple words: Factor the first two terms as a difference of cubes, and factor out \( -2x \) from the last two. Then, pull out the common factor \( (2a - b) \).

Exam Tip: When factoring out \( -2x \) from \( -4ax + 2bx \), be careful with the signs. It becomes \( -2x(2a - b) \).

Question 15. Factorise:
\( a - b - a^3 + b^3 \)
Answer:
Grouping the terms:
\( = (a - b) - (a^3 - b^3) \)
Applying the difference of cubes identity to the second part:
\( = (a - b) - (a - b)(a^2 + ab + b^2) \)
Factoring out the common binomial \( (a - b) \):
\( = (a - b)\left[1 - (a^2 + ab + b^2)\right] \)
\( = (a - b)(1 - a^2 - ab - b^2) \)
In simple words: Group the terms to find a common bracket \( (a - b) \) and a difference of cubes. Pulling out \( (a-b) \) leaves a simple algebraic expression.

Exam Tip: Be very careful when subtracting the trinomial. Ensure you distribute the negative sign to all terms inside: \( -a^2 - ab - b^2 \).

Question 16. Factorise:
\( 2x^3 + 54y^3 - 2x - 6y \)
Answer:
Factor out common numerical factors from groups:
\( = 2(x^3 + 27y^3) - 2(x + 3y) \)
Rewriting as a sum of cubes in the first part:
\( = 2\left[(x)^3 + (3y)^3\right] - 2(x + 3y) \)
Applying the sum of cubes identity:
\( = 2\left[(x + 3y)(x^2 - 3xy + 9y^2)\right] - 2(x + 3y) \)
Taking out the common term \( 2(x + 3y) \):
\( = 2(x + 3y)(x^2 - 3xy + 9y^2 - 1) \)
In simple words: Factor out \( 2 \) to get a sum of cubes, and factor out \( -2 \) from the rest. Using the sum of cubes formula gives you a common binomial to factor out.

Exam Tip: Factoring out the overall common number \( 2 \) at the very beginning is highly recommended to make the algebraic terms smaller and easier to manage.

Question 17. Show that:
(i) \( 13^3 - 5^3 \) is divisible by 8
(ii) \( 35^3 + 27^3 \) is divisible by 62
Answer:
(i) Using the identity \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \):
\( 13^3 - 5^3 = (13 - 5)(13^2 + 13 \times 5 + 5^2) \)
\( = 8 \times (169 + 65 + 25) \)
Since \( 8 \) is a factor, the value is divisible by \( 8 \).

(ii) Using the identity \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \):
\( 35^3 + 27^3 = (35 + 27)(35^2 - 35 \times 27 + 27^2) \)
\( = 62 \times (35^2 - 35 \times 27 + 27^2) \)
Since \( 62 \) is a factor, the value is divisible by \( 62 \).
In simple words: Use the sum or difference of cubes formula. The first factor you get is either \( (13 - 5) = 8 \) or \( (35 + 27) = 62 \), which directly proves divisibility.

Exam Tip: You don't need to calculate the value of the large second bracket. Just showing that the required divisor is the first factor is sufficient for proof.

Exercise 5(E)

Question 1. Factorise:
\( x^2 + \frac{1}{4x^2} + 1 - 7x - \frac{7}{2x} \)
Answer:
We can rewrite the expression by grouping terms:
\( = \left[(x)^2 + \frac{1}{(2x)^2} + 2 \times x \times \frac{1}{2x}\right] - 7\left(x + \frac{1}{2x}\right) \)
Recognizing the first three terms as a perfect square:
\( = \left(x + \frac{1}{2x}\right)^2 - 7\left(x + \frac{1}{2x}\right) \)
Pulling out the common factor \( \left(x + \frac{1}{2x}\right) \):
\( = \left(x + \frac{1}{2x}\right)\left(x + \frac{1}{2x} - 7\right) \)
In simple words: Combine the first three terms to form a perfect square, and pull out a common factor of \( -7 \) from the last two. Then, pull out the common binomial.

Exam Tip: Notice that \( 2 \times x \times \frac{1}{2x} = 1 \), which matches the constant in the trinomial perfectly.

Question 2. Factorise:
\( 9a^2 + \frac{1}{9a^2} - 2 - 12a + \frac{4}{3a} \)
Answer:
Grouping terms to find perfect squares and common binomials:
\( = \left[(3a)^2 + \frac{1}{(3a)^2} - 2 \times 3a \times \frac{1}{3a}\right] - 4\left(3a - \frac{1}{3a}\right) \)
\( = \left(3a - \frac{1}{3a}\right)^2 - 4\left(3a - \frac{1}{3a}\right) \)
Factoring out the common binomial \( \left(3a - \frac{1}{3a}\right) \):
\( = \left(3a - \frac{1}{3a}\right)\left(3a - \frac{1}{3a} - 4\right) \)
In simple words: The first three terms form a perfect square with a negative sign. Pull out \( -4 \) from the last two terms to create a matching bracket, then factor it out.

Exam Tip: Ensure that when you factor out \( -4 \) from \( -12a + \frac{4}{3a} \), the second term's sign changes to negative inside the bracket.

Question 3. Factorise:
\( x^2 + \frac{a^2 + 1}{a}x + 1 \)
Answer:
Distribute \( x \) through the fractional coefficient:
\( = x^2 + ax + \frac{1}{a}x + 1 \)
Factor by grouping pairs:
\( = x(x + a) + \frac{1}{a}(x + a) \)
Taking out the common binomial factor:
\( = (x + a)\left(x + \frac{1}{a}\right) \)
In simple words: Expand the middle term by splitting it into two fractions. Then group and factor the first two terms and the last two terms separately.

Exam Tip: Remember that \( \frac{1}{a} \times a = 1 \). This helps you recognize that the last pair can be factored by pulling out \( \frac{1}{a} \).

Question 4. Factorise:
\( x^4 + y^4 - 27x^2y^2 \)
Answer:
Split the last term to create a perfect square trinomial:
\( = (x^2)^2 + (y^2)^2 - 2x^2y^2 - 25x^2y^2 \)
\( = (x^2 - y^2)^2 - (5xy)^2 \)
Using the difference of squares identity:
\( = (x^2 - y^2 + 5xy)(x^2 - y^2 - 5xy) \)
Rearranging terms for standard form:
\( = (x^2 + 5xy - y^2)(x^2 - 5xy - y^2) \)
In simple words: Break down \( -27x^2y^2 \) into \( -2x^2y^2 \) and \( -25x^2y^2 \). This lets you write the expression as a difference of two squares and factor it.

Exam Tip: If an expression does not seem factorable initially, try adding and subtracting a term to construct a perfect square trinomial.

Question 5. Factorise:
\( 4x^4 + 9y^4 + 11x^2y^2 \)
Answer:
To form a perfect square trinomial, rewrite \( 11x^2y^2 \) as \( 12x^2y^2 - x^2y^2 \):
\( = (2x^2)^2 + (3y^2)^2 + 12x^2y^2 - x^2y^2 \)
\( = (2x^2 + 3y^2)^2 - (xy)^2 \)
Applying the difference of squares formula:
\( = (2x^2 + 3y^2 - xy)(2x^2 + 3y^2 + xy) \)
In simple words: Add and subtract \( x^2y^2 \) to make the first part a perfect square trinomial, then use the difference of squares method to finish factoring.

Exam Tip: Choose your perfect square constant carefully. Adding \( 1 \) to \( 11 \) gives \( 12 \), which perfectly matches the required \( 2(2x^2)(3y^2) = 12x^2y^2 \) term.

Question 6. Factorise:
\( x^2 + \frac{1}{x^2} - 3 \)
Answer:
Splitting \( -3 \) into \( -2 - 1 \):
\( = x^2 + \frac{1}{x^2} - 2 \times x \times \frac{1}{x} - 1 \)
\( = \left(x - \frac{1}{x}\right)^2 - (1)^2 \)
Applying the difference of squares identity:
\( = \left(x - \frac{1}{x} - 1\right)\left(x - \frac{1}{x} + 1\right) \)
In simple words: Split \( -3 \) into \( -2 \) and \( -1 \). The first part is the perfect square of \( \left(x - \frac{1}{x}\right) \), which you can then factor against the \( 1^2 \) term.

Exam Tip: This is a very common algebraic pattern. Remember that splitting constants often reveals hidden perfect squares.

Question 7. Factorise:
\( a - b - 4a^2 + 4b^2 \)
Answer:
Group the terms and factor out the common coefficients:
\( = (a - b) - 4(a^2 - b^2) \)
Applying the difference of squares formula to the second term:
\( = (a - b) - 4(a - b)(a + b) \)
Factoring out the common binomial \( (a - b) \):
\( = (a - b)[1 - 4(a + b)] \)
Simplifying the terms inside the square brackets:
\( = (a - b)(1 - 4a - 4b) \)
In simple words: Group the last two terms by factoring out \( -4 \). This gives a difference of squares that shares a common factor with the first part.

Exam Tip: Be careful with signs. Factoring out \( -4 \) from \( -4a^2 + 4b^2 \) leaves \( -4(a^2 - b^2) \), not \( -4(a^2 + b^2) \).

Question 8. Factorise:
\( (2a - 3)^2 - 2(2a - 3)(a - 1) + (a - 1)^2 \)
Answer:
Recognizing the quadratic pattern \( x^2 - 2xy + y^2 = (x - y)^2 \):
Here, let \( x = (2a - 3) \) and \( y = (a - 1) \):
\( = \left[(2a - 3) - (a - 1)\right]^2 \)
Simplifying inside the brackets:
\( = [2a - 3 - a + 1]^2 \)
\( = (a - 2)^2 \)
In simple words: This long expression follows the standard perfect square identity \( x^2 - 2xy + y^2 \). Just group it as \( (x - y)^2 \) and simplify inside.

Exam Tip: Identifying the overall algebraic form immediately can save you from tedious expansion and regrouping.

Question 9. Factorise:
\( (a^2 - 3a)(a^2 - 3a + 7) + 10 \)
Answer:
Suppose we substitute \( a^2 - 3a = x \):
The expression becomes:
\( = x(x + 7) + 10 \)
\( = x^2 + 7x + 10 \)
Splitting the middle term to factor this quadratic trinomial:
\( = x^2 + 5x + 2x + 10 \)
\( = x(x + 5) + 2(x + 5) \)
\( = (x + 5)(x + 2) \)
Now, substitute the value of \( x \) back:
\( = (a^2 - 3a + 5)(a^2 - 3a + 2) \)
The second trinomial can be factored further by splitting the middle term:
\( = (a^2 - 3a + 5)(a^2 - 2a - a + 2) \)
\( = (a^2 - 3a + 5)[a(a - 2) - 1(a - 2)] \)
\( = (a^2 - 3a + 5)(a - 1)(a - 2) \)
In simple words: Substitute \( x \) for \( a^2 - 3a \) to make the quadratic simple to factor. Once factored, replace \( x \) with \( a^2 - 3a \) and factor the resulting quadratics if possible.

Exam Tip: Always check both factored binomial brackets at the end. In this case, only one of the quadratic terms could be factored further, while the other could not.

Question 10. Factorise:
\( (a^2 - a)(4a^2 - 4a - 5) - 6 \)
Answer:
Let us assume that \( a^2 - a = x \):
This means the expression can be rewritten as:
\( = x(4x - 5) - 6 \)
\( = 4x^2 - 5x - 6 \)
Splitting the middle term to factor:
\( = 4x^2 - 8x + 3x - 6 \)
\( = 4x(x - 2) + 3(x - 2) \)
\( = (4x + 3)(x - 2) \)
Placing the value of \( x \) back into the expression:
\( = [4(a^2 - a) + 3](a^2 - a - 2) \)
\( = (4a^2 - 4a + 3)(a^2 - a - 2) \)
Factoring the second quadratic trinomial:
\( = (4a^2 - 4a + 3)(a^2 - 2a + a - 2) \)
\( = (4a^2 - 4a + 3)[a(a - 2) + 1(a - 2)] \)
\( = (4a^2 - 4a + 3)(a - 2)(a + 1) \)
In simple words: Replace \( a^2 - a \) with a single letter \( x \). This makes it a standard quadratic to factor. After factoring, replace \( x \) and look to factor any quadratics that are left.

Exam Tip: Be on the lookout for hidden common components like \( 4a^2 - 4a \) being \( 4(a^2 - a) \) to apply substitution successfully.

Question 11. Factorise:
\( x^4 + y^4 - 3x^2y^2 \)
Answer:
Splitting the last term to find a perfect square:
\( = x^4 + y^4 - 2x^2y^2 - x^2y^2 \)
\( = (x^2 - y^2)^2 - (xy)^2 \)
Using the difference of squares identity:
\( = (x^2 - y^2 - xy)(x^2 - y^2 + xy) \)
In simple words: Rewrite \( -3x^2y^2 \) as \( -2x^2y^2 - x^2y^2 \). Now, group the first part as a perfect square and factor it using difference of squares.

Exam Tip: Modifying middle terms by adding/subtracting is a great strategy when standard quadratic grouping methods fail.

Question 12. Factorise:
\( 5a^2 - b^2 - 4ab + 7a - 7b \)
Answer:
Splitting \( 5a^2 \) into \( 4a^2 + a^2 \) to facilitate grouping:
\( = 4a^2 + a^2 - b^2 - 4ab + 7a - 7b \)
Rearranging terms:
\( = (a^2 - b^2) + 4a(a - b) + 7(a - b) \)
Applying the difference of squares identity to the first term:
\( = (a - b)(a + b) + 4a(a - b) + 7(a - b) \)
Factoring out the common binomial \( (a - b) \):
\( = (a - b)[(a + b) + 4a + 7] \)
Combining like terms inside the brackets:
\( = (a - b)[5a + b + 7] \)
In simple words: Break \( 5a^2 \) into \( 4a^2 \) and \( a^2 \). This allows you to group and factor the expression into parts that all share the common factor \( (a - b) \).

Exam Tip: Splitting terms is not just for constants; sometimes you need to split the coefficient of a squared term like \( 5a^2 \) to make factorization work.

Question 13. Factorise:
\( 12(3x - 2y)^2 - 3x + 2y - 1 \)
Answer:
Group the linear terms:
\( = 12(3x - 2y)^2 - (3x - 2y) - 1 \)
Suppose we substitute \( 3x - 2y = a \):
The expression simplifies to:
\( = 12a^2 - a - 1 \)
Splitting the middle term of the quadratic trinomial:
\( = 12a^2 - 4a + 3a - 1 \)
\( = 4a(3a - 1) + 1(3a - 1) \)
\( = (4a + 1)(3a - 1) \)
Substituting back the value of \( a \):
\( = [4(3x - 2y) + 1][3(3x - 2y) - 1] \)
\( = (12x - 8y + 1)(9x - 6y - 1) \)
In simple words: Substitute a single letter \( a \) for \( 3x - 2y \) to factor the expression as a quadratic. Replace \( a \) back at the end and simplify the brackets.

Exam Tip: Ensure you expand the coefficients into both terms inside the brackets when resubstituting \( a \); for instance, \( 4(3x - 2y) + 1 \) becomes \( 12x - 8y + 1 \).

Question 14. Factorise:
\( 4(2x - 3y)^2 - 8x + 12y - 3 \)
Answer:
Factor out \( -4 \) from the middle terms to group them:
\( = 4(2x - 3y)^2 - 4(2x - 3y) - 3 \)
Let us set \( 2x - 3y = a \):
The expression becomes:
\( = 4a^2 - 4a - 3 \)
Splitting the middle term of this quadratic:
\( = 4a^2 - 6a + 2a - 3 \)
\( = 2a(2a - 3) + 1(2a - 3) \)
\( = (2a - 3)(2a + 1) \)
Replacing \( a \) with its original expression:
\( = [2(2x - 3y) - 3][2(2x - 3y) + 1] \)
\( = (4x - 6y - 3)(4x - 6y + 1) \)
In simple words: Group the middle terms as \( -4(2x - 3y) \) and substitute a single variable \( a \). Factor the quadratic expression, then put the original terms back and expand.

Exam Tip: Be careful when factoring out the negative constant from \( -8x + 12y \). Pulling out \( -4 \) ensures you get the matching binomial \( 2x - 3y \).

Question 15. Factorise:
\( 3 - 5x + 5y - 12(x - y)^2 \)
Answer:
Group the linear terms by pulling out a negative sign:
\( = 3 - 5(x - y) - 12(x - y)^2 \)
Suppose we substitute \( x - y = a \):
This gives us:
\( = 3 - 5a - 12a^2 \)
Splitting the middle term of this quadratic expression:
\( = 3 - 9a + 4a - 12a^2 \)
\( = 3(1 - 3a) + 4a(1 - 3a) \)
\( = (3 + 4a)(1 - 3a) \)
Substituting back the value of \( a \):
\( = [3 + 4(x - y)][1 - 3(x - y)] \)
\( = (3 + 4x - 4y)(1 - 3x + 3y) \)
In simple words: Factor out \( -5 \) to get \( -5(x - y) \), then substitute \( a \) for \( x - y \). Factor the quadratic trinomial, substitute back, and simplify.

Exam Tip: Be careful with signs when substituting back. For example, \( -3(x - y) \) expands to \( -3x + 3y \) due to the double negative.

Question 16. Factorise:
\( 9x^2 + 3x - 8y - 64y^2 \)
Answer:
Rearranging the terms to group squares together:
\( = 9x^2 - 64y^2 + 3x - 8y \)
Factoring the difference of squares in the first part:
\( = \left[(3x)^2 - (8y)^2\right] + (3x - 8y) \)
\( = (3x + 8y)(3x - 8y) + (3x - 8y) \)
Pulling out the common binomial factor \( (3x - 8y) \):
\( = (3x - 8y)(3x + 8y + 1) \)
In simple words: Put the squared terms together to make a difference of squares. Factor them, then notice that the remaining terms match one of the factors, allowing you to pull it out.

Exam Tip: Remember to include the \( + 1 \) term when you factor out the common bracket \( (3x - 8y) \) from itself.

Question 17. Factorise:
\( 2\sqrt{3}x^2 + x - 5\sqrt{3} \)
Answer:
Find two numbers whose product is \( 2\sqrt{3} \times (-5\sqrt{3}) = -30 \) and whose sum is \( 1 \). These are \( 6 \) and \( -5 \).
Splitting the middle term:
\( = 2\sqrt{3}x^2 + 6x - 5x - 5\sqrt{3} \)
Factoring by grouping:
\( = 2\sqrt{3}x(x + \sqrt{3}) - 5(x + \sqrt{3}) \)
\( = (2\sqrt{3}x - 5)(x + \sqrt{3}) \)
In simple words: Split the middle term \( x \) into \( 6x - 5x \). Group the terms and pull out \( 2\sqrt{3}x \) from the first pair and \( -5 \) from the second pair.

Exam Tip: Note that \( 6 \) can be written as \( 2\sqrt{3} \times \sqrt{3} \), which is why we can factor \( 2\sqrt{3}x \) out of \( 6x \) to leave \( \sqrt{3} \).

Question 18. Factorise:
\( \frac{1}{4}(a + b)^2 - \frac{9}{16}(2a - b)^2 \)
Answer:
Factor out \( \frac{1}{4} \) from the expression:
\( = \frac{1}{4}\left[(a + b)^2 - \frac{9}{4}(2a - b)^2\right] \)
Write as a difference of squares inside the brackets:
\( = \frac{1}{4}\left[(a + b)^2 - \left(\frac{3}{2}(2a - b)\right)^2\right] \)
Applying the difference of squares identity \( x^2 - y^2 = (x + y)(x - y) \):
\( = \frac{1}{4}\left[\left(a + b + \frac{3}{2}(2a - b)\right)\left(a + b - \frac{3}{2}(2a - b)\right)\right] \)
Expanding the terms inside the brackets:
\( = \frac{1}{4}\left[\left(a + b + 3a - \frac{3b}{2}\right)\left(a + b - 3a + \frac{3b}{2}\right)\right] \)
Simplifying the terms:
\( = \frac{1}{4}\left[\left(4a - \frac{b}{2}\right)\left(\frac{5b}{2} - 2a\right)\right] \)
Taking the denominator \( 2 \) out of both brackets:
\( = \frac{1}{4}\left[\left(\frac{8a - b}{2}\right)\left(\frac{5b - 4a}{2}\right)\right] \)
\( = \frac{1}{16}(8a - b)(5b - 4a) \)
In simple words: Factor out \( \frac{1}{4} \), then use the difference of squares identity. Simplify the fractions inside each bracket, factor out the common denominators, and combine them.

Exam Tip: Factoring out fractions early prevents complex fractional arithmetic inside the main algebraic steps.

Question 19. Factorise:
\( 2(ab + cd) - a^2 - b^2 + c^2 + d^2 \)
Answer:
First, expand the expression:
\( = 2ab + 2cd - a^2 - b^2 + c^2 + d^2 \)
Group the terms into two perfect square trinomials:
\( = (c^2 + d^2 + 2cd) - (a^2 + b^2 - 2ab) \)
\( = (c + d)^2 - (a - b)^2 \)
Using the difference of squares formula:
\( = [(c + d) + (a - b)][(c + d) - (a - b)] \)
\( = (c + d + a - b)(c + d - a + b) \)
In simple words: Expand and then group the terms into two separate perfect squares: one positive and one negative. Finally, apply the difference of squares identity.

Exam Tip: Be careful when subtracting the second term \( (a-b) \). The negative sign distributes to make it \( -a + b \).

Question 20. Evaluate each of the following using identities:
(i) \( (987)^2 - (13)^2 \)
(ii) \( (67.8)^2 - (32.2)^2 \)
(iii) \( \frac{(6.7)^2 - (3.3)^2}{6.7 - 3.3} \)
(iv) \( \frac{(18.5)^2 - (6.5)^2}{18.5 + 6.5} \)
Answer:
(i) Using the identity \( a^2 - b^2 = (a + b)(a - b) \):
\( (987)^2 - (13)^2 = (987 + 13)(987 - 13) \)
\( = 1000 \times 974 \)
\( = 974000 \)

(ii) Using the identity \( a^2 - b^2 = (a + b)(a - b) \):
\( (67.8)^2 - (32.2)^2 = (67.8 + 32.2)(67.8 - 32.2) \)
\( = 100 \times 35.6 \)
\( = 3560 \)

(iii) Factoring the numerator using \( a^2 - b^2 = (a + b)(a - b) \):
\( \frac{(6.7)^2 - (3.3)^2}{6.7 - 3.3} = \frac{(6.7 + 3.3)(6.7 - 3.3)}{6.7 - 3.3} \)
Canceling the common term \( (6.7 - 3.3) \):
\( = 6.7 + 3.3 \)
\( = 10 \)

(iv) Factoring the numerator using \( a^2 - b^2 = (a + b)(a - b) \):
\( \frac{(18.5)^2 - (6.5)^2}{18.5 + 6.5} = \frac{(18.5 + 6.5)(18.5 - 6.5)}{18.5 + 6.5} \)
Canceling the common term \( (18.5 + 6.5) \):
\( = 18.5 - 6.5 \)
\( = 12 \)
In simple words: Use the difference of squares formula \( a^2 - b^2 = (a + b)(a - b) \) to simplify these numerical calculations easily without squaring large numbers.

Exam Tip: Notice how identities allow difficult calculations to simplify into easy products like multiplying by \( 1000 \) or \( 100 \). Look for these simplifications to confirm you are on the right track.