Frank Brothers Solutions for ICSE Class 9 Mathematics Chapter 5 Factorisation

Exercise 5.1

Question 1A. Factorise: \(4x^2y^3 - 6x^3y^2 - 12xy^2\)
Answer: The largest common factor here is \(2xy^2\). We divide each term by \(2xy^2\): \[ \frac{4x^2y^3}{2xy^2} - \frac{6x^3y^2}{2xy^2} - \frac{12xy^2}{2xy^2} \]
\( \implies 2xy - 3x^2 - 6 \)
This gives us:
\( 4x^2y^3 - 6x^3y^2 - 12xy^2 = 2xy^2(2xy - 3x^2 - 6) \)
In simple words: Find the largest term that divides into all three parts, which is \(2xy^2\). Pull it outside of the parentheses and write the remaining terms inside.
Exam Tip: Always double-check your factoring by distributing the common factor back into the parentheses to see if you get the original expression.

Question 1B. Factorise: \(5a(x^2 - y^2) + 35b(x^2 - y^2)\)
Answer: The shared factor for both terms is \(5(x^2 - y^2)\). Dividing both parts by \(5(x^2 - y^2)\) gives: \[ \frac{5a(x^2 - y^2)}{5(x^2 - y^2)} + \frac{35b(x^2 - y^2)}{5(x^2 - y^2)} \]
\( \implies a + 7b \)
This gives:
\( 5a(x^2 - y^2) + 35b(x^2 - y^2) = 5(x^2 - y^2)(a + 7b) \)
In simple words: Take out the common bracket \((x^2 - y^2)\) along with the number 5. Then write what is left over in another bracket.
Exam Tip: Remember that coefficients like 5 and 35 have a common numerical factor of 5 that must be factored out too.

Question 1C. Factorise: \(2x^5y + 8x^3y^2 - 12x^2y^3\)
Answer: The common factor in this expression is \(2x^2y\). We divide each term by \(2x^2y\): \[ \frac{2x^5y}{2x^2y} + \frac{8x^3y^2}{2x^2y} - \frac{12x^2y^3}{2x^2y} \]
\( \implies x^3 + 4xy - 6y^2 \)
This gives us:
\( 2x^5y + 8x^3y^2 - 12x^2y^3 = 2x^2y(x^3 + 4xy - 6y^2) \)
In simple words: Find the highest power of each variable and the largest number that divides all parts, which is \(2x^2y\), and pull it out.
Exam Tip: When looking for common variables, always take the lowest power of each variable present in all the terms.

Question 1D. Factorise: \(12a^3 + 15a^2b - 21ab^2\)
Answer: The common factor here is \(3a\). We divide each term by \(3a\): \[ \frac{12a^3}{3a} + \frac{15a^2b}{3a} - \frac{21ab^2}{3a} \]
\( \implies 4a^2 + 5ab - 7b^2 \)
This gives us:
\( 12a^3 + 15a^2b - 21ab^2 = 3a(4a^2 + 5ab - 7b^2) \)
In simple words: Look at the numbers 12, 15, and 21. They can all be divided by 3. The letter 'a' is also in all terms, so we pull \(3a\) outside.
Exam Tip: Ensure that no other common factor is left inside the parentheses after you factor out.

Question 1E. Factorise: \(24m^4n^6 + 56m^6n^4 - 72m^2n^2\)
Answer: The common factor here is \(8m^2n^2\). We divide each term by \(8m^2n^2\): \[ \frac{24m^4n^6}{8m^2n^2} + \frac{56m^6n^4}{8m^2n^2} - \frac{72m^2n^2}{8m^2n^2} \]
\( \implies 3m^2n^4 + 7m^4n^2 - 9 \)
This gives us:
\( 24m^4n^6 + 56m^6n^4 - 72m^2n^2 = 8m^2n^2(3m^2n^4 + 7m^4n^2 - 9) \)
In simple words: Find the largest number that divides 24, 56, and 72, which is 8. Then pull out the lowest powers of \(m\) and \(n\), which is \(m^2n^2\).
Exam Tip: Be careful with exponent rules; when dividing terms with powers, subtract the powers: \(m^4 / m^2 = m^{2}\).

Question 1F. Factorise: \((a - b)^2 - 2(a - b)\)
Answer: The common factor here is \((a - b)\). We divide each part by \((a - b)\): \[ \frac{(a - b)^2}{(a - b)} - \frac{2(a - b)}{(a - b)} \]
\( \implies a - b - 2 \)
This gives us:
\( (a - b)^2 - 2(a - b) = (a - b)(a - b - 2) \)
In simple words: Treat the bracket \((a - b)\) like a single unit. Since it is present in both terms, take one out.
Exam Tip: When a binomial is squared, factoring it out leaves one copy of that binomial inside the remaining group.

Question 1G. Factorise: \(2a(p^2 + q^2) + 4b(p^2 + q^2)\)
Answer: The common factor here is \(2(p^2 + q^2)\). We divide each term by \(2(p^2 + q^2)\): \[ \frac{2a(p^2 + q^2)}{2(p^2 + q^2)} + \frac{4b(p^2 + q^2)}{2(p^2 + q^2)} \]
\( \implies a + 2b \)
This gives us:
\( 2a(p^2 + q^2) + 4b(p^2 + q^2) = 2(p^2 + q^2)(a + 2b) \)
In simple words: Both terms share the bracket \((p^2 + q^2)\) and the numbers can both be divided by 2. Pulling \(2(p^2 + q^2)\) out leaves \(a + 2b\).
Exam Tip: Always check if the coefficients outside the brackets share any common factors that should also be taken out.

Question 1H. Factorise: \(81(p + q)^2 - 9p - 9q\)
Answer: First, group and factor the last two terms:
\( 81(p + q)^2 - 9(p + q) \)
Now, the common factor is \(9(p + q)\). We divide each part by \(9(p + q)\): \[ \frac{81(p + q)^2}{9(p + q)} - \frac{9(p + q)}{9(p + q)} \]
\( \implies 9(p + q) - 1 \)
This gives us:
\( 81(p + q)^2 - 9p - 9q = 9(p + q)[9(p + q) - 1] \)
In simple words: Group the last two terms to create a common bracket \((p + q)\). Then pull out the common factor \(9(p + q)\).
Exam Tip: Watch out for negative signs when grouping. Factoring out \(-9\) from \(-9p - 9q\) changes the signs inside the bracket to positive, giving \(-9(p + q)\).

Question 1I. Factorise: \((mx + ny)^2 + (nx - my)^2\)
Answer: First, expand the squared terms:
\( = m^2x^2 + n^2y^2 + 2mnxy + n^2x^2 + m^2y^2 - 2mnxy \)
Cancel the middle terms and group what is left:
\( = m^2x^2 + n^2x^2 + m^2y^2 + n^2y^2 \)
Factor each pair:
\( = x^2(m^2 + n^2) + y^2(m^2 + n^2) \)
Now, the common factor is \((m^2 + n^2)\). We divide each part by \((m^2 + n^2)\): \[ \frac{x^2(m^2 + n^2)}{m^2 + n^2} + \frac{y^2(m^2 + n^2)}{m^2 + n^2} \]
\( \implies x^2 + y^2 \)
This gives us:
\( (mx + ny)^2 + (nx - my)^2 = (m^2 + n^2)(x^2 + y^2) \)
In simple words: First expand both parts using algebra rules. Clean up the terms, group them together, and factor them in pairs.
Exam Tip: Use the identity \((A \pm B)^2 = A^2 \pm 2AB + B^2\) carefully during expansion to avoid sign errors.

Question 1J. Factorise: \(36(x + y)^3 - 54(x + y)^2\)
Answer: The common factor here is \(18(x + y)^2\). We divide each term by \(18(x + y)^2\): \[ \frac{36(x + y)^3}{18(x + y)^2} - \frac{54(x + y)^2}{18(x + y)^2} \]
\( \implies 2(x + y) - 3 \)
This gives us:
\( 36(x + y)^3 - 54(x + y)^2 = 18(x + y)^2[2(x + y) - 3] \)
In simple words: Look at 36 and 54, the largest number dividing both is 18. The shared bracket is \((x + y)\), and its smallest power is 2. Pull them out.
Exam Tip: Don't forget that after factoring out \((x + y)^2\) from the first term, one copy of \((x + y)\) still remains.

Question 1K. Factorise: \(p(p^2 + q^2 - r^2) + q(r^2 - q^2 - p^2) - r(p^2 + q^2 - r^2)\)
Answer: First, change the sign of the middle term to make the brackets match:
\( = p(p^2 + q^2 - r^2) - q(p^2 + q^2 - r^2) - r(p^2 + q^2 - r^2) \)
Now we divide each term by this factor \((p^2 + q^2 - r^2)\): \[ \frac{p(p^2 + q^2 - r^2)}{p^2 + q^2 - r^2} - \frac{q(p^2 + q^2 - r^2)}{p^2 + q^2 - r^2} - \frac{r(p^2 + q^2 - r^2)}{p^2 + q^2 - r^2} \]
\( \implies p - q - r \)
This gives us:
\( p(p^2 + q^2 - r^2) + q(r^2 - q^2 - p^2) - r(p^2 + q^2 - r^2) = (p^2 + q^2 - r^2)(p - q - r) \)
In simple words: Flip the signs inside the middle bracket by changing the sign outside it from plus to minus. Now that all brackets are the same, pull them out.
Exam Tip: Factoring a negative sign out of a bracket reverses the sign of every single term inside that bracket.

Question 2A. Factorise: \(15xy - 9x - 25y + 15\)
Answer: Group the terms in pairs and factor out a minus sign:
\( = (15xy - 9x) - (25y - 15) \)
Factor out the common parts from each pair:
\( = 3x(5y - 3) - 5(5y - 3) \)
Take out the common bracket:
\( = (5y - 3)(3x - 5) \)
In simple words: Group the first two terms together and the last two terms together. Factor out what is common in each pair to find a shared bracket.
Exam Tip: Ensure the signs inside the brackets match exactly. If they differ by a sign, factor out a negative value.

Question 2B. Factorise: \(15x^2 + 7y - 3x - 35xy\)
Answer: Rearrange the terms to group common letters:
\( = 15x^2 - 3x - 35xy + 7y \)
Group them in pairs:
\( = (15x^2 - 3x) - (35xy - 7y) \)
Factor each pair:
\( = 3x(5x - 1) - 7y(5x - 1) \)
Take out the common bracket:
\( = (5x - 1)(3x - 7y) \)
In simple words: Rearrange the terms so that parts with common letters are together. Then group and factor them out in pairs.
Exam Tip: Rearranging terms is often necessary before grouping to make common factors apparent.

Question 2C. Factorise: \(9 + 3xy + x^2y + 3x\)
Answer: First, rearrange the terms:
\( = 9 + 3xy + 3x + x^2y \)
Group them in pairs:
\( = (9 + 3xy) + (3x + x^2y) \)
Factor each pair:
\( = 3(3 + xy) + x(3 + xy) \)
Take out the common bracket:
\( = (3 + xy)(3 + x) \)
In simple words: Rearrange and split the four terms into two groups. Take 3 out of the first group and \(x\) out of the second group.
Exam Tip: Always double-check your rearranged groups to ensure they lead to a common binomial factor.

Question 2D. Factorise: \(8(2a + b)^2 - 8a - 4b\)
Answer: Group and factor the last two terms:
\( = 8(2a + b)^2 - (8a + 4b) \)
\( = 8(2a + b)^2 - 4(2a + b) \)
Take out the common factor:
\( = 4(2a + b)[2(2a + b) - 1] \)
Simplify the terms inside the bracket:
\( = 4(2a + b)[4a + 2b - 1] \)
In simple words: Factor \(-4\) from the last two terms to create the same bracket \((2a+b)\). Then take out \(4(2a+b)\).
Exam Tip: Expand the remaining terms inside the square bracket fully to get the final simplified factor.

Question 2E. Factorise: \(x(x - 4) - x + 4\)
Answer: Factor out \(-1\) from the last two terms:
\( = x(x - 4) - 1(x - 4) \)
Take out the common bracket:
\( = (x - 4)(x - 1) \)
In simple words: Change \(-x + 4\) into \(-1(x - 4)\). This gives you two parts with the same bracket, so you can pull it out.
Exam Tip: Remember that writing \(-x + 4\) as \(-1(x - 4)\) is a helpful trick to find a hidden factor of 1.

Question 2F. Factorise: \(2m^3 - 5n^2 - 5m^2n + 2mn\)
Answer: Rearrange the terms to group them:
\( = 2m^3 + 2mn - 5m^2n - 5n^2 \)
Group them in pairs:
\( = (2m^3 + 2mn) - (5m^2n + 5n^2) \)
Factor each pair:
\( = 2m(m^2 + n) - 5n(m^2 + n) \)
Take out the common bracket:
\( = (m^2 + n)(2m - 5n) \)
In simple words: Rearrange the terms, group them in pairs, factor each pair, and then pull out the common bracket \((m^2+n)\).
Exam Tip: Be careful with signs when factoring out negative terms like \(-5n\); it changes the sign inside the bracket to positive.

Question 2G. Factorise: \(x^3y^3 - 8x^2y^2 + 15xy\)
Answer: Split the middle term:
\( = x^3y^3 - 3x^2y^2 - 5x^2y^2 + 15xy \)
Group them in pairs:
\( = (x^3y^3 - 3x^2y^2) - (5x^2y^2 - 15xy) \)
Factor each pair:
\( = x^2y^2(xy - 3) - 5xy(xy - 3) \)
Take out the common bracket:
\( = (xy - 3)(x^2y^2 - 5xy) \)
Factor out \(xy\) from the second bracket:
\( = xy(xy - 3)(xy - 5) \)
In simple words: Split the middle term to make four terms, group them in pairs to factor out, and then pull out the extra \(xy\) from the final bracket.
Exam Tip: Always check if the final factors can be simplified or factored further. Here, \(x^2y^2 - 5xy\) has a common factor of \(xy\).

Question 2H. Factorise: \(9x^3 + 6x^2y^2 - 4y^3 - 6xy\)
Answer: First, rearrange the terms:
\( = 9x^3 + 6x^2y^2 - 6xy - 4y^3 \)
Group them in pairs:
\( = (9x^3 + 6x^2y^2) - (6xy + 4y^3) \)
Factor each pair:
\( = 3x^2(3x + 2y^2) - 2y(3x + 2y^2) \)
Take out the common bracket:
\( = (3x + 2y^2)(3x^2 - 2y) \)
In simple words: Group the terms and pull out what is common in each pair, making sure the brackets inside are exactly the same.
Exam Tip: Always write the grouped expressions clearly with correct signs before factoring out.

Question 2I. Factorise: \(3ax^2 - 5bx^2 + 9az^2 + 6ay^2 - 10by^2 - 15bz^2\)
Answer: First, group the \(a\) terms and the \(b\) terms:
\( = 3ax^2 + 6ay^2 + 9az^2 - 5bx^2 - 10by^2 - 15bz^2 \)
Group them into two parts:
\( = (3ax^2 + 6ay^2 + 9az^2) - (5bx^2 + 10by^2 + 15bz^2) \)
Factor each part:
\( = 3a(x^2 + 2y^2 + 3z^2) - 5b(x^2 + 2y^2 + 3z^2) \)
Take out the common bracket:
\( = (x^2 + 2y^2 + 3z^2)(3a - 5b) \)
In simple words: Group all the parts with \(a\) together and all the parts with \(b\) together. Factor out \(3a\) and \(5b\), then pull out the matching large bracket.
Exam Tip: For expressions with six terms, try to group them into two sets of three terms each.

Question 2J. Factorise: \(8x^3 - 24x^2y + 54xy^2 - 162y^3\)
Answer: Group them in pairs:
\( = (8x^3 - 24x^2y) + (54xy^2 - 162y^3) \)
Factor each pair:
\( = 8x^2(x - 3y) + 54y^2(x - 3y) \)
Take out the common bracket:
\( = (x - 3y)(8x^2 + 54y^2) \)
(We can also factor out \(2\) from the second bracket to write it as \(2(x - 3y)(4x^2 + 27y^2)\)).
In simple words: Group in pairs, pull out the common bracket \((x-3y)\), then pull out the number 2 if needed.
Exam Tip: Always check if the final brackets have any remaining common numerical factors to factor them out completely.

Question 2K. Factorise: \(2a + b + 3c - d + (2a + b)^3 + (2a + b)^2(3c - d)\)
Answer: Group the terms into two parts:
\( = (2a + b + 3c - d) + [(2a + b)^3 + (2a + b)^2(3c - d)] \)
Factor out the square bracket from the second part:
\( = 1(2a + b + 3c - d) + (2a + b)^2(2a + b + 3c - d) \)
Take out the common bracket:
\( = (2a + b + 3c - d)[1 + (2a + b)^2] \)
In simple words: Group the first four terms together, and the next two together. Factor out the bracket from the second group, then pull out the giant shared bracket.
Exam Tip: Placing a '1' in front of a bracket helps keep track of the remaining terms when factoring out the entire expression.

Question 2L. Factorise: \(xy(a^2 + 1) + a(x^2 + y^2)\)
Answer: First, expand the brackets:
\( = a^2xy + xy + ax^2 + ay^2 \)
Rearrange and group them in pairs:
\( = (a^2xy + ax^2) + (ay^2 + xy) \)
Factor each pair:
\( = ax(ay + x) + y(ay + x) \)
Take out the common bracket:
\( = (ay + x)(ax + y) \)
In simple words: Multiply out the brackets first, rearrange the terms, group them in pairs, and factor them out.
Exam Tip: Expansion is often the key first step when terms are locked inside brackets with no obvious common factors.

Question 2M. Factorise: \(p^2x^2 + (px^2 + 1)x + p\)
Answer: First, expand the middle term:
\( = p^2x^2 + px^3 + x + p \)
Rearrange and group in pairs:
\( = (p^2x^2 + px^3) + (x + p) \)
Factor each pair:
\( = px^2(p + x) + 1(p + x) \)
Take out the common bracket:
\( = (p + x)(px^2 + 1) \)
In simple words: Multiply the \(x\) into the middle bracket first, then group the terms in pairs and factor them.
Exam Tip: Be systematic when multiplying variables with powers, e.g., \(px^2 \times x = px^3\).

Question 2N. Factorise: \(x^2 - (p + q)x + pq\)
Answer: First, expand the terms:
\( = x^2 - px - qx + pq \)
Group in pairs and watch the negative signs:
\( = (x^2 - px) - (qx - pq) \)
Factor each pair:
\( = x(x - p) - q(x - p) \)
Take out the common bracket:
\( = (x - p)(x - q) \)
In simple words: Distribute the \(x\) into the bracket, split into two pairs, and pull out the common factor \((x-p)\).
Exam Tip: Pay close attention to the negative sign before the second group, which changes \(+pq\) to \(-pq\) inside the factored group.

Question 2O. Factorise: \(p^2 + \frac{1}{p^2} - 2 - 5p + \frac{5}{p}\)
Answer: Group the terms into two parts:
\( = \left(p^2 + \frac{1}{p^2} - 2\right) - \left(5p - \frac{5}{p}\right) \)
The first part is a perfect square:
\( = \left(p^2 + \left(\frac{1}{p}\right)^2 - 2 \times p \times \frac{1}{p}\right) - 5\left(p - \frac{1}{p}\right) \)
\( = \left(p - \frac{1}{p}\right)^2 - 5\left(p - \frac{1}{p}\right) \)
Take out the common bracket:
\( = \left(p - \frac{1}{p}\right)\left(p - \frac{1}{p} - 5\right) \)
In simple words: The first three terms form a perfect square formula. Write them as a square bracket, factor out 5 from the rest, and then take out the common bracket.
Exam Tip: Recognise the perfect square pattern \(a^2 + \frac{1}{a^2} - 2 = \left(a - \frac{1}{a}\right)^2\), which is very common in algebraic identity questions.

Question 2P. Factorise: \(x + y + m(x + y)\)
Answer: Group the first two terms:
\( = (x + y) + m(x + y) \)
Take out the common bracket:
\( = (x + y)(1 + m) \)
In simple words: Treat \(x + y\) as a single unit, which is multiplied by 1. Since it is shared, pull it out.
Exam Tip: Remember that any term without an explicit coefficient has an implicit coefficient of 1.

Question 2Q. Factorise: \(\frac{1}{25x^2} + 16x^2 + \frac{8}{5} - 12x - \frac{3}{5x}\)
Answer: Group the terms into two parts:
\( = \left(\frac{1}{25x^2} + 16x^2 + \frac{8}{5}\right) - \left(12x + \frac{3}{5x}\right) \)
The first part is a perfect square:
\( = \left(\left(\frac{1}{5x}\right)^2 + (4x)^2 + 2 \times \frac{1}{5x} \times 4x\right) - 3\left(4x + \frac{1}{5x}\right) \)
\( = \left(\frac{1}{5x} + 4x\right)^2 - 3\left(\frac{1}{5x} + 4x\right) \)
Take out the common bracket:
\( = \left(\frac{1}{5x} + 4x\right)\left(\frac{1}{5x} + 4x - 3\right) \)
In simple words: The first three terms form a perfect square. Factor out 3 from the remaining terms, and pull out the common bracket.
Exam Tip: Identify the middle term \(\frac{8}{5}\) as the product \(2 \times \frac{1}{5x} \times 4x\) to confirm it forms a perfect square.

Question 2R. Factorise: \(2p(a^2 - 2b^2) - 14p + (a^2 - 2b^2)^2 - 7(a^2 - 2b^2)\)
Answer: Rearrange the terms:
\( = 2p(a^2 - 2b^2) + (a^2 - 2b^2)^2 - 14p - 7(a^2 - 2b^2) \)
Group them in pairs:
\( = [2p(a^2 - 2b^2) + (a^2 - 2b^2)^2] - [14p + 7(a^2 - 2b^2)] \)
Factor each pair:
\( = (a^2 - 2b^2)(2p + a^2 - 2b^2) - 7(2p + a^2 - 2b^2) \)
Take out the common bracket:
\( = (2p + a^2 - 2b^2)(a^2 - 2b^2 - 7) \)
In simple words: Rearrange so similar groups are next to each other. Factor out common parts from each group to get a matching binomial, then factor that out.
Exam Tip: When grouping, keep large polynomial terms intact as single units to simplify the factoring process.

Exercise 5.2

Question 1A. Factorise: \(x^2 + 6x + 8\)
Answer: Split the middle term:
\( = x^2 + 4x + 2x + 8 \)
Now group the terms in pairs and factor them:
\( = x(x + 4) + 2(x + 4) \)
Factoring out the common bracket \((x + 4)\) gives:
\( = (x + 4)(x + 2) \)
In simple words: Find two numbers that multiply to 8 and add up to 6. These are 4 and 2. Split the middle term and factor.
Exam Tip: Always check that the two split factors multiply to the constant term and add up to the coefficient of the middle term.

Question 1B. Factorise: \(x^2 - 11x + 24\)
Answer: Split the middle term:
\( = x^2 - 8x - 3x + 24 \)
Grouping the terms into pairs gives:
\( = x(x - 8) - 3(x - 8) \)
Factoring out the common bracket \((x - 8)\) yields:
\( = (x - 8)(x - 3) \)
In simple words: We need two numbers that add up to \(-11\) and multiply to 24. These are \(-8\) and \(-3\).
Exam Tip: Since the product is positive (\(24\)) and the sum is negative (\(-11\)), both split factors must be negative.

Question 1C. Factorise: \(x^2 + 5x - 6\)
Answer: Split the middle term:
\( = x^2 + 6x - x - 6 \)
Grouping in pairs gives:
\( = x(x + 6) - 1(x + 6) \)
Factoring out \((x + 6)\) gives:
\( = (x + 6)(x - 1) \)
In simple words: Find two numbers that multiply to \(-6\) and add up to 5. These are 6 and \(-1\).
Exam Tip: When the constant term is negative, the two numbers must have opposite signs.

Question 1D. Factorise: \(p^2 - 12p - 64\)
Answer: Split the middle term:
\( = p^2 - 16p + 4p - 64 \)
Grouping into pairs gives:
\( = p(p - 16) + 4(p - 16) \)
Factoring out the binomial \((p - 16)\) yields:
\( = (p - 16)(p + 4) \)
In simple words: Look for two numbers that multiply to \(-64\) and add up to \(-12\). These are \(-16\) and 4.
Exam Tip: Ensure the larger number has the same sign as the middle term when the constant term is negative.

Question 1E. Factorise: \(y^2 - 2y - 24\)
Answer: Split the middle term:
\( = y^2 - 6y + 4y - 24 \)
Grouping and factoring:
\( = y(y - 6) + 4(y - 6) \)
Factoring out \((y - 6)\) yields:
\( = (y - 6)(y + 4) \)
In simple words: Find two numbers that multiply to \(-24\) and add up to \(-2\). These are \(-6\) and 4.
Exam Tip: Double-check your basic arithmetic to avoid simple sign errors during splitting.

Question 1F. Factorise: \(3x^2 + 19x - 14\)
Answer: Split the middle term:
\( = 3x^2 + 21x - 2x - 14 \)
Grouping the terms into pairs gives:
\( = 3x(x + 7) - 2(x + 7) \)
Factoring out the common bracket \((x + 7)\) yields:
\( = (x + 7)(3x - 2) \)
In simple words: We need two numbers that multiply to \(-42\) (which is \(3 \times -14\)) and add up to 19. These are 21 and \(-2\).
Exam Tip: When the coefficient of the quadratic term is greater than 1, multiply it by the constant term to find the target product.

Question 1G. Factorise: \(15a^2 - 14a - 16\)
Answer: Split the middle term:
\( = 15a^2 - 24a + 10a - 16 \)
Grouping the terms in pairs:
\( = 3a(5a - 8) + 2(5a - 8) \)
Factoring out the binomial \((5a - 8)\) gives:
\( = (5a - 8)(3a + 2) \)
In simple words: Find two numbers that multiply to \(-240\) and add up to \(-14\). These are \(-24\) and 10. Group and factor.
Exam Tip: Finding factors of large numbers like 240 can be done easily by listing its factor pairs or using prime factorisation.

Question 1H. Factorise: \(12 + x - 6x^2\)
Answer: Split the middle term:
\( = 12 + 9x - 8x - 6x^2 \)
Grouping into pairs gives:
\( = 3(4 + 3x) - 2x(4 + 3x) \)
Factoring out the common bracket \((4 + 3x)\) yields:
\( = (4 + 3x)(3 - 2x) \)
In simple words: We need two numbers that multiply to \(-72\) (from \(12 \times -6\)) and add up to 1. These are 9 and \(-8\).
Exam Tip: Do not be confused if the quadratic term is written last; the rules of splitting the middle term remain exactly the same.

Question 1I. Factorise: \(7x^2 + 40x - 12\)
Answer: Split the middle term:
\( = 7x^2 + 42x - 2x - 12 \)
Grouping the terms into pairs gives:
\( = 7x(x + 6) - 2(x + 6) \)
Factoring out the binomial \((x + 6)\) yields:
\( = (x + 6)(7x - 2) \)
In simple words: Find two numbers that multiply to \(-84\) and add up to 40. These are 42 and \(-2\).
Exam Tip: Always write the intermediate grouping step clearly to show the examiner how you factored the expression.

Question 2A. Factorise: \(5x^2 - 17xy + 6y^2\)
Answer: Split the middle term:
\( = 5x^2 - 15xy - 2xy + 6y^2 \)
Grouping the terms into pairs:
\( = 5x(x - 3y) - 2y(x - 3y) \)
Factoring out the shared bracket \((x - 3y)\) yields:
\( = (x - 3y)(5x - 2y) \)
In simple words: Look for two numbers that multiply to 30 and add up to \(-17\). These are \(-15\) and \(-2\).
Exam Tip: For terms containing two variables like \(x\) and \(y\), write the split middle terms with both variables (i.e., \(xy\)).

Question 2B. Factorise: \(9x^2 - 22xy + 8y^2\)
Answer: Split the middle term:
\( = 9x^2 - 18xy - 4xy + 8y^2 \)
Grouping the terms into pairs:
\( = 9x(x - 2y) - 4y(x - 2y) \)
Factoring out the shared binomial \((x - 2y)\) gives:
\( = (x - 2y)(9x - 4y) \)
In simple words: Find two numbers that multiply to 72 and add up to \(-22\). These are \(-18\) and \(-4\).
Exam Tip: Be careful when factoring out negative terms like \(-4y\) to ensure the signs inside the bracket are correct.

Question 2C. Factorise: \(2x^3 + 5x^2y - 12xy^2\)
Answer: Split the middle term:
\( = 2x^3 + 8x^2y - 3x^2y - 12xy^2 \)
Grouping the terms into pairs gives:
\( = 2x^2(x + 4y) - 3xy(x + 4y) \)
Factoring out the common bracket \((x + 4y)\) yields:
\( = (x + 4y)(2x^2 - 3xy) \)
We can factor out \(x\) from the second bracket:
\( = x(x + 4y)(2x - 3y) \)
In simple words: Split the middle term first. Group and factor, then notice that \(2x^2 - 3xy\) has a common \(x\) that must be pulled outside.
Exam Tip: Always examine your final expression to ensure every term is fully factorised.

Question 2D. Factorise: \(x^2y^2 + 15xy - 16\)
Answer: Split the middle term:
\( = x^2y^2 + 16xy - xy - 16 \)
Grouping the terms into pairs:
\( = xy(xy + 16) - 1(xy + 16) \)
Factoring out the common binomial \((xy + 16)\) gives:
\( = (xy + 16)(xy - 1) \)
In simple words: We need two numbers that multiply to \(-16\) and add up to 15. These are 16 and \(-1\).
Exam Tip: Treating composite variables like \(xy\) as a single variable makes the quadratic trinomial easy to solve.

Question 2E. Factorise: \((2p + q)^2 - 10p - 5q - 6\)
Answer: First, group and factor the middle terms:
\( = (2p + q)^2 - 5(2p + q) - 6 \)
Split the middle term:
\( = (2p + q)^2 - 6(2p + q) + 1(2p + q) - 6 \)
Grouping and factoring:
\( = (2p + q)[(2p + q) - 6] + 1[(2p + q) - 6] \)
Factoring out the common polynomial bracket yields:
\( = (2p + q - 6)(2p + q + 1) \)
In simple words: Group the middle terms to make a bracket \((2p+q)\). Then treat it like a simple variable, split the middle term, and factor.
Exam Tip: Substituting a temporary letter like \(u = 2p + q\) can make the steps of splitting the middle term clearer.

Question 2F. Factorise: \(y^2 + 3y + 2 + by + 2b\)
Answer: First, expand the middle term:
\( = y^2 + y + 2y + 2 + by + 2b \)
Rearrange and group in pairs:
\( = y^2 + y + by + 2y + 2 + 2b \)
Group them into two parts:
\( = (y^2 + y + by) + (2y + 2 + 2b) \)
Factor each pair:
\( = y(y + 1 + b) + 2(y + 1 + b) \)
Take out the common bracket:
\( = (y + 1 + b)(y + 2) \)
In simple words: Split \(3y\) into \(y\) and \(2y\), rearrange the six terms, and factor them in groups of three.
Exam Tip: Grouping six terms into two trinomials is a common strategy when a shared trinomial factor can be found.

Question 2H. Factorise: \(6\sqrt{3}x^2 - 19x + 5\sqrt{3}\)
Answer: Split the middle term:
\( = 6\sqrt{3}x^2 - 10x - 9x + 5\sqrt{3} \)
Grouping the terms into pairs:
\( = 2x(3\sqrt{3}x - 5) - \sqrt{3}(3\sqrt{3}x - 5) \)
Factoring out the common bracket \((3\sqrt{3}x - 5)\) gives:
\( = (3\sqrt{3}x - 5)(2x - \sqrt{3}) \)
In simple words: Multiply the coefficients of the first and last terms to get 90. Split \(-19\) into \(-10\) and \(-9\), then factor. Remember that \(9 = 3\sqrt{3} \times \sqrt{3}\).
Exam Tip: When dealing with surds, remember that \(3\sqrt{3} \times \sqrt{3} = 9\). This is key to factoring the second group.

Question 2I. Factorise: \(2\sqrt{5}x^2 - 7x - 3\sqrt{5}\)
Answer: Split the middle term:
\( = 2\sqrt{5}x^2 - 10x + 3x - 3\sqrt{5} \)
Group them in pairs:
\( = 2\sqrt{5}x(x - \sqrt{5}) + 3(x - \sqrt{5}) \)
Take out the common bracket:
\( = (x - \sqrt{5})(2\sqrt{5}x + 3) \)
In simple words: Find two numbers that multiply to \(-30\) and add up to \(-7\). These are \(-10\) and 3. Pull out the shared brackets.
Exam Tip: Factor surds carefully by writing integers as products of roots, such as \(10 = 2\sqrt{5} \times \sqrt{5}\).

Question 3A. Factorise: \(5(3x + y)^2 + 6(3x + y) - 8\)
Answer: Split the middle term:
\( = 5(3x + y)^2 + 10(3x + y) - 4(3x + y) - 8 \)
Group them in pairs:
\( = 5(3x + y)(3x + y + 2) - 4(3x + y + 2) \)
Take out the common bracket:
\( = (3x + y + 2)[5(3x + y) - 4] \)
(This can also be written as \((3x + y + 2)(15x + 5y - 4)\)).
In simple words: Treat the bracket \((3x+y)\) as a single letter. Split the middle term and factor.
Exam Tip: Always perform the final distribution inside the square bracket to fully simplify the factored expression.

Question 3B. Factorise: \(5 - 4(a - b) - 12(a - b)^2\)
Answer: Split the middle term:
\( = 5 - 10(a - b) + 6(a - b) - 12(a - b)^2 \)
Group them in pairs:
\( = 5[1 - 2(a - b)] + 6(a - b)[1 - 2(a - b)] \)
Take out the common bracket:
\( = [5 + 6(a - b)][1 - 2(a - b)] \)
Expand the terms:
\( = (5 + 6a - 6b)(1 - 2a + 2b) \)
In simple words: Treat \((a-b)\) like a single term. Split the middle term into \(-10\) and \(6\) of that term, group, factor, and then expand the brackets.
Exam Tip: Pay attention to distributing negative signs correctly, such as \(-2(a - b) = -2a + 2b\).

Question 3(C). Factorise: \( (3a - 2b)^2 + 3(3a - 2b) - 10 \)
Answer:
\( (3a - 2b)^2 + 3(3a - 2b) - 10 \)
\( = (3a - 2b)^2 + 5(3a - 2b) - 2(3a - 2b) - 10 \)
\( = (3a - 2b)(3a - 2b + 5) - 2(3a - 2b + 5) \)
\( = (3a - 2b + 5)(3a - 2b - 2) \)

In simple words: Split the middle term \( 3(3a - 2b) \) into \( 5(3a - 2b) - 2(3a - 2b) \) since \( 5 \times -2 = -10 \). Then factor out the common binomial.

Exam Tip: Treating a repeating binomial expression as a single term simplifies the process of splitting the middle term.

Question 3(D). Factorise: \( (a^2 - 2a)^2 - 23(a^2 - 2a) + 120 \)
Answer:
\( (a^2 - 2a)^2 - 23(a^2 - 2a) + 120 \)
\( = (a^2 - 2a)^2 - 15(a^2 - 2a) - 8(a^2 - 2a) + 120 \)
\( = (a^2 - 2a)(a^2 - 2a - 15) - 8(a^2 - 2a - 15) \)
\( = (a^2 - 2a - 15)(a^2 - 2a - 8) \)
\( = (a^2 - 5a + 3a - 15)(a^2 - 4a + 2a - 8) \)
\( = [a(a - 5) + 3(a - 5)][a(a - 4) + 2(a - 4)] \)
\( = [(a - 5)(a + 3)][(a - 4)(a + 2)] \)
\( = (a - 5)(a + 3)(a - 4)(a + 2) \)
\( = (a + 2)(a + 3)(a - 4)(a - 5) \)

In simple words: First, split the middle term using the shared term \( a^2 - 2a \). Once you obtain two quadratic expressions, split the middle terms for each of them to find the final four linear factors.

Exam Tip: Be careful when factoring the secondary quadratics; ensure you split the middle terms correctly with proper signs.

Question 3(E). Factorise: \( (x + 4)^2 - 5xy - 20y - 6y^2 \)
Answer:
\( (x + 4)^2 - 5xy - 20y - 6y^2 \)
\( = (x + 4)^2 - 5y(x + 4) - 6y^2 \)
\( = (x + 4)^2 - 6y(x + 4) + y(x + 4) - 6y^2 \)
\( = (x + 4)(x + 4 - 6y) + y(x + 4 - 6y) \)
\( = (x + 4 - 6y)(x + y + 4) \)
\( = (x - 6y + 4)(x + y + 4) \)

In simple words: Group the middle two terms by pulling out \( -5y \) to reveal the common term \( x + 4 \). Then split the middle term with respect to \( y \) and factorise.

Exam Tip: Grouping terms to find a common binomial factor makes complex-looking algebraic expressions much easier to solve.

Question 3(F). Factorise: \( 7(x - 2)^2 - 13(x - 2) - 2 \)
Answer:
\( 7(x - 2)^2 - 13(x - 2) - 2 \)
\( = 7(x - 2)^2 - 14(x - 2) + (x - 2) - 2 \)
\( = 7(x - 2)(x - 2 - 2) + 1(x - 2 - 2) \)
\( = 7(x - 2)(x - 4) + 1(x - 4) \)
\( = (x - 4)[7(x - 2) + 1] \)
\( = (x - 4)(7x - 14 + 1) \)
\( = (x - 4)(7x - 13) \)

In simple words: Split the middle term \( -13(x - 2) \) into \( -14(x - 2) \) and \( +1(x - 2) \). Then factor out the common binomials and simplify the brackets.

Exam Tip: Don't forget to expand the inner bracket \( 7(x - 2) \) completely in the final step to simplify the terms.

Question 3(G). Factorise: \( 12 - (y + y^2)(8 - y - y^2) \)
Answer:
\( 12 - (y + y^2)(8 - y - y^2) \)
Let \( y + y^2 = a \).
The expression becomes:
\( 12 - a(8 - a) \)
\( = 12 - 8a + a^2 \)
\( = 12 - 6a - 2a + a^2 \)
\( = 6(2 - a) - a(2 - a) \)
\( = (2 - a)(6 - a) \)
Substituting back \( a = y + y^2 \):
\( = [2 - (y + y^2)][6 - (y + y^2)] \)
\( = (2 - y - y^2)(6 - y - y^2) \)
Splitting the middle terms of both quadratic expressions:
\( = (2 - 2y + y - y^2)(6 - 3y + 2y - y^2) \)
\( = [2(1 - y) + y(1 - y)][3(2 - y) + y(2 - y)] \)
\( = [(1 - y)(2 + y)][(2 - y)(3 + y)] \)
\( = (1 - y)(2 + y)(2 - y)(3 + y) \)
\( = (y - 1)(y + 2)(y - 2)(y + 3) \)

In simple words: Introduce a temporary variable \( a \) to simplify the expression into a quadratic form. Factorise it, and then plug back \( y + y^2 \) to solve the resulting quadratics separately.

Exam Tip: Be mindful of signs when pulling out negative factors like \( (1 - y) \) and \( (2 - y) \) to write the final form as \( (y - 1)(y - 2) \).

Question 3(H). Factorise: \( (p^2 + p)^2 - 8(p^2 + p) + 12 \)
Answer:
\( (p^2 + p)^2 - 8(p^2 + p) + 12 \)
\( = (p^2 + p)^2 - 6(p^2 + p) - 2(p^2 + p) + 12 \)
\( = (p^2 + p)(p^2 + p - 6) - 2(p^2 + p - 6) \)
\( = (p^2 + p - 6)(p^2 + p - 2) \)
\( = (p^2 + 3p - 2p - 6)(p^2 + 2p - p - 2) \)
\( = [p(p + 3) - 2(p + 3)][p(p + 2) - 1(p + 2)] \)
\( = [(p + 3)(p - 2)][(p + 2)(p - 1)] \)
\( = (p + 3)(p - 2)(p + 2)(p - 1) \)

In simple words: This problem is solved by splitting the middle terms twice. First, split the main expression, and then split the resulting quadratics into simple linear terms.

Exam Tip: Factoring completely means reducing the polynomial to its lowest possible degree factors.

Question 4(A). Factorise: \( (y^2 - 3y)(y^2 - 3y + 7) + 10 \)
Answer:
\( (y^2 - 3y)(y^2 - 3y + 7) + 10 \)
Let \( y^2 - 3y = a \).
The expression becomes:
\( a(a + 7) + 10 \)
\( = a^2 + 7a + 10 \)
\( = a^2 + 5a + 2a + 10 \)
\( = a(a + 5) + 2(a + 5) \)
\( = (a + 5)(a + 2) \)
Substituting \( a = y^2 - 3y \) back:
\( = (y^2 - 3y + 5)(y^2 - 3y + 2) \)
Now split the middle term for the factorable quadratic part:
\( = (y^2 - 3y + 5)(y^2 - 2y - y + 2) \)
\( = (y^2 - 3y + 5)[y(y - 2) - 1(y - 2)] \)
\( = (y^2 - 3y + 5)[(y - 2)(y - 1)] \)
\( = (y - 1)(y - 2)(y^2 - 3y + 5) \)

In simple words: Substitute \( a \) for the repeated expression \( y^2 - 3y \). Solve the basic quadratic, substitute back, and factorise any parts that can still be split further.

Exam Tip: Not all quadratic factors can be split using real integers. For example, \( y^2 - 3y + 5 \) cannot be factorised further, so leave it as is.

Question 4(B). Factorise: \( (t^2 - t)(4t^2 - 4t - 5) - 6 \)
Answer:
\( (t^2 - t)(4t^2 - 4t - 5) - 6 \)
\( = (t^2 - t)[4(t^2 - t) - 5] - 6 \)
Let \( t^2 - t = a \).
The expression becomes:
\( a[4a - 5] - 6 \)
\( = 4a^2 - 5a - 6 \)
\( = 4a^2 - 8a + 3a - 6 \)
\( = 4a(a - 2) + 3(a - 2) \)
\( = (a - 2)(4a + 3) \)
Substituting \( a = t^2 - t \) back:
\( = (t^2 - t - 2)[4(t^2 - t) + 3] \)
\( = (t^2 - 2t + t - 2)(4t^2 - 4t + 3) \)
\( = [t(t - 2) + 1(t - 2)](4t^2 - 4t + 3) \)
\( = (t - 2)(t + 1)(4t^2 - 4t + 3) \)
\( = (t + 1)(t - 2)(4t^2 - 4t + 3) \)

In simple words: Rewrite \( 4t^2 - 4t \) as \( 4(t^2 - t) \). Use substitution to simplify, factorise the resulting expression, and then substitute back to finish factorising.

Exam Tip: Notice that \( 4t^2 - 4t + 3 \) has no real integer roots, so it cannot be factorised further. Do not waste time trying to split its middle term.

Question 4(C). Factorise: \( 12(2x - 3y)^2 - (2x - 3y) - 1 \)
Answer:
\( 12(2x - 3y)^2 - (2x - 3y) - 1 \)
Let \( 2x - 3y = a \).
The expression becomes:
\( 12a^2 - a - 1 \)
\( = 12a^2 - 4a + 3a - 1 \)
\( = 4a(3a - 1) + 1(3a - 1) \)
\( = (3a - 1)(4a + 1) \)
Substituting \( a = 2x - 3y \) back:
\( = [3(2x - 3y) - 1][4(2x - 3y) + 1] \)
\( = (6x - 9y - 1)(8x - 12y + 1) \)

In simple words: Replace the common bracket term with \( a \), factorise the quadratic trinomial, and then substitute the original bracket back.

Exam Tip: Make sure to distribute the coefficients (3 and 4) to both terms inside the bracket when expanding in the final step.

Question 4(D). Factorise: \( 6 - 5x + 5y + (x - y)^2 \)
Answer:
\( 6 - 5x + 5y + (x - y)^2 \)
\( = 6 - 5(x - y) + (x - y)^2 \)
\( = 6 - 3(x - y) - 2(x - y) + (x - y)^2 \)
\( = 3[2 - (x - y)] - (x - y)[2 - (x - y)] \)
\( = [2 - (x - y)][3 - (x - y)] \)
\( = (2 - x + y)(3 - x + y) \)

In simple words: Group \( -5x + 5y \) as \( -5(x - y) \) to form a quadratic expression in terms of \( (x - y) \). Then factorise it by splitting the middle term.

Exam Tip: Pay close attention to signs when removing the brackets, e.g., \( - (x - y) = -x + y \).

Question 4(E). Factorise: \( 2x^2 + \frac{x}{6} - 1 \)
Answer:
\( 2x^2 + \frac{x}{6} - 1 \)
Take \( \frac{1}{6} \) common out of the expression:
\( = \frac{1}{6}(12x^2 + x - 6) \)
\( = \frac{1}{6}(12x^2 + 9x - 8x - 6) \)
\( = \frac{1}{6}[3x(4x + 3) - 2(4x + 3)] \)
\( = \frac{1}{6}(4x + 3)(3x - 2) \)

In simple words: Pull out the fraction \( \frac{1}{6} \) first to make all terms integers. Then factorise the simple quadratic trinomial inside the bracket.

Exam Tip: Factoring out a common fraction simplifies the coefficients, making the splitting method much easier to execute.

Question 4(F). Factorise: \( p^4 + 23p^2q^2 + 90q^4 \)
Answer:
\( p^4 + 23p^2q^2 + 90q^4 \)
\( = p^4 + 18p^2q^2 + 5p^2q^2 + 90q^4 \)
\( = p^2(p^2 + 18q^2) + 5q^2(p^2 + 18q^2) \)
\( = (p^2 + 18q^2)(p^2 + 5q^2) \)

In simple words: Split the middle term \( 23p^2q^2 \) into \( 18p^2q^2 \) and \( 5p^2q^2 \), since \( 18 \times 5 = 90 \). Then group and factorise.

Exam Tip: This equation acts like a regular quadratic trinomial with \( p^2 \) as the variable. Treat it similarly for a straightforward split.

Question 4(G). Factorise: \( 2a^3 + 5a^2b - 12ab^2 \)
Answer:
\( 2a^3 + 5a^2b - 12ab^2 \)
\( = 2a^3 + 8a^2b - 3a^2b - 12ab^2 \)
\( = 2a^2(a + 4b) - 3ab(a + 4b) \)
\( = (a + 4b)(2a^2 - 3ab) \)
\( = a(a + 4b)(2a - 3b) \)

In simple words: Split the middle term \( 5a^2b \) into \( 8a^2b - 3a^2b \). After grouping and factorising, pull out the common factor \( a \) from the remaining term.

Exam Tip: Alternatively, you can pull out the common factor \( a \) at the very first step, making the remaining quadratic trinomial simpler to factorise.

Exercise 5.3

Question 1(A). Factorise: \( x^2 - 16 \)
Answer:
\( x^2 - 16 \)
\( = x^2 - 4^2 \)
\( = (x - 4)(x + 4) \)

In simple words: This expression uses the difference of squares identity, \( a^2 - b^2 = (a - b)(a + b) \), where \( a = x \) and \( b = 4 \).

Exam Tip: Identify perfect square numbers like 16, 25, 36 to quickly apply the difference of squares identity.

Question 1(B). Factorise: \( 64x^2 - 121y^2 \)
Answer:
\( 64x^2 - 121y^2 \)
\( = (8x)^2 - (11y)^2 \)
\( = (8x - 11y)(8x + 11y) \)

In simple words: Write both terms as perfect squares and apply the identity \( a^2 - b^2 = (a - b)(a + b) \).

Exam Tip: Always verify that both coefficients are perfect squares before rewriting them as a single squared term.

Question 1(C). Factorise: \( 441 - 81y^2 \)
Answer:
\( 441 - 81y^2 \)
\( = (21)^2 - (9y)^2 \)
\( = (21 - 9y)(21 + 9y) \)
Take out the common factor \( 3 \) from both factors:
\( = 3(7 - 3y) \cdot 3(7 + 3y) \)
\( = 9(7 - 3y)(7 + 3y) \)

In simple words: Use the difference of squares on \( 21^2 - (9y)^2 \), then factor out \( 3 \) from each bracket to simplify the result.

Exam Tip: Alternatively, you can factor out the greatest common divisor \( 9 \) first to get \( 9(49 - y^2) \) and then factorise.

Question 1(D). Factorise: \( x^6 - 196 \)
Answer:
\( x^6 - 196 \)
\( = (x^3)^2 - 14^2 \)
\( = (x^3 - 14)(x^3 + 14) \)

In simple words: Express \( x^6 \) as \( (x^3)^2 \) and \( 196 \) as \( 14^2 \) to apply the difference of squares rule.

Exam Tip: Remember that for exponents, \( (x^a)^b = x^{ab} \). So, \( x^6 \) is the square of \( x^3 \).

Question 1(E). Factorise: \( 625 - b^2 \)
Answer:
\( 625 - b^2 \)
\( = 25^2 - b^2 \)
\( = (25 - b)(25 + b) \)

In simple words: This can be factorised easily by identifying \( 625 \) as the square of \( 25 \) and using the difference of squares identity.

Exam Tip: Memorizing squares up to 30 helps in recognizing numbers like 625 instantly.

Question 1(F). Factorise: \( m^2 - \frac{1}{9} n^2 \)
Answer:
\( m^2 - \frac{1}{9} n^2 \)
\( = m^2 - \left(\frac{1}{3} n\right)^2 \)
\( = \left(m - \frac{1}{3} n\right)\left(m + \frac{1}{3} n\right) \)

In simple words: Write the second term as the square of \( \frac{1}{3}n \) to apply the standard difference of squares formula.

Exam Tip: Fractional squares like \( \frac{1}{9}n^2 \) are treated exactly like regular squares; just square the numerator and denominator separately.

Question 1(G). Factorise: \( 8xy^2 - 18x^3 \)
Answer:
\( 8xy^2 - 18x^3 \)
Take the common factor \( 2x \) out first:
\( = 2x(4y^2 - 9x^2) \)
\( = 2x[(2y)^2 - (3x)^2] \)
\( = 2x(2y - 3x)(2y + 3x) \)

In simple words: First factor out the common term \( 2x \). This leaves a difference of perfect squares inside, which can be easily factorised.

Exam Tip: Always look for any common factors to pull out first before applying specific factorization identities.

Question 1(H). Factorise: \( 16a^4 - 81b^4 \)
Answer:
\( 16a^4 - 81b^4 \)
\( = (4a^2)^2 - (9b^2)^2 \)
\( = (4a^2 - 9b^2)(4a^2 + 9b^2) \)
Now factorise the first term further:
\( = [(2a)^2 - (3b)^2](4a^2 + 9b^2) \)
\( = (2a - 3b)(2a + 3b)(4a^2 + 9b^2) \)

In simple words: Apply the difference of squares formula twice. The first application yields a term that can be split yet again into simpler binomials.

Exam Tip: Never stop after one round of factoring if one of the resulting terms can be factorised further.

Question 1(I). Factorise: \( a(a - 1) - b(b - 1) \)
Answer:
\( a(a - 1) - b(b - 1) \)
\( = a^2 - a - b^2 + b \)
\( = a^2 - b^2 - a + b \)
\( = (a^2 - b^2) - (a - b) \)
\( = (a - b)(a + b) - (a - b) \)
\( = (a - b)(a + b - 1) \)

In simple words: First expand the terms and group them to make \( a^2 - b^2 \) and \( a - b \). Factor the difference of squares, then factor out the common bracket \( a - b \).

Exam Tip: Pay close attention to the sign when grouping \( -a + b \) as \( -(a - b) \).

Question 1(J). Factorise: \( (x + y)^2 - 1 \)
Answer:
\( (x + y)^2 - 1 \)
\( = (x + y)^2 - 1^2 \)
\( = (x + y + 1)(x + y - 1) \)

In simple words: Treat \( x + y \) as one single term. Use the difference of squares identity with \( a = x + y \) and \( b = 1 \).

Exam Tip: The constant term 1 can always be written as \( 1^2 \) to match the difference of squares pattern.

Question 1(K). Factorise: \( x^2 + y^2 - z^2 - 2xy \)
Answer:
\( x^2 + y^2 - z^2 - 2xy \)
Rearranging the terms:
\( = x^2 + y^2 - 2xy - z^2 \)
\( = (x^2 + y^2 - 2xy) - z^2 \)
\( = (x - y)^2 - z^2 \)
\( = (x - y - z)(x - y + z) \)

In simple words: Rearrange the terms to group the perfect square trinomial \( x^2 - 2xy + y^2 \) into \( (x - y)^2 \). Then apply the difference of squares.

Exam Tip: Look out for groups of three terms that can form a perfect square binomial, like \( x^2 - 2xy + y^2 \).

Question 1(L). Factorise: \( (x - 2y)^2 - z^2 \)
Answer:
\( (x - 2y)^2 - z^2 \)
\( = (x - 2y)^2 - z^2 \)
\( = (x - 2y - z)(x - 2y + z) \)

In simple words: Directly use the difference of squares rule by treating the binomial \( x - 2y \) as the first term.

Exam Tip: There is no need to expand \( (x - 2y)^2 \) first; doing so would make the expression harder to factorise.

Question 2(A). Factorise: \( 9(a - b)^2 - (a + b)^2 \)
Answer:
\( 9(a - b)^2 - (a + b)^2 \)
\( = [3(a - b)]^2 - (a + b)^2 \)
Using the difference of squares identity:
\( = [3(a - b) - (a + b)][3(a - b) + (a + b)] \)
\( = (3a - 3b - a - b)(3a - 3b + a + b) \)
\( = (2a - 4b)(4a - 2b) \)
Factoring out 2 from each bracket:
\( = 2(a - 2b) \cdot 2(2a - b) \)
\( = 4(a - 2b)(2a - b) \)

In simple words: Rewrite the term \( 9(a - b)^2 \) as the square of \( 3(a - b) \). Apply the difference of squares formula, combine like terms inside, and factor out any common numbers.

Exam Tip: Don't forget to extract numerical common factors like 2 from both parentheses at the end to completely simplify the answer.

Question 2(B). Factorise: \( 25(x - y)^2 - 49(c - d)^2 \)
Answer:
\( 25(x - y)^2 - 49(c - d)^2 \)
\( = [5(x - y)]^2 - [7(c - d)]^2 \)
\( = [5(x - y) - 7(c - d)][5(x - y) + 7(c - d)] \)
\( = (5x - 5y - 7c + 7d)(5x - 5y + 7c - 7d) \)

In simple words: Express both terms as perfect squares and apply the difference of squares identity. Simplify by opening the brackets inside.

Exam Tip: Distribute the signs carefully when expanding \( -7(c - d) \) to get \( -7c + 7d \).

Question 2(C). Factorise: \( (2a - b)^2 - 9(3c - d)^2 \)
Answer:
\( (2a - b)^2 - 9(3c - d)^2 \)
\( = (2a - b)^2 - [3(3c - d)]^2 \)
\( = [(2a - b) - 3(3c - d)][(2a - b) + 3(3c - d)] \)
\( = (2a - b - 9c + 3d)(2a - b + 9c - 3d) \)

In simple words: Treat \( (2a - b) \) and \( 3(3c - d) \) as your squares, apply the difference of squares identity, and carefully multiply the constant through the brackets.

Exam Tip: Be cautious about sign changes when distributing a negative number over a binomial.

Question 2(D). Factorise: \( b^2 - 2bc + c^2 - a^2 \)
Answer:
\( b^2 - 2bc + c^2 - a^2 \)
\( = (b^2 - 2bc + c^2) - a^2 \)
\( = (b - c)^2 - a^2 \)
\( = (b - c - a)(b - c + a) \)

In simple words: Group the first three terms to form the perfect square \( (b - c)^2 \). Then apply the difference of squares formula.

Exam Tip: Recognizing standard expansions like \( b^2 - 2bc + c^2 = (b - c)^2 \) is key to grouping terms correctly.

Question 2(E). Factorise: \( x^2 + \frac{1}{x^2} - 2 \)
Answer:
\( x^2 + \frac{1}{x^2} - 2 \)
\( = x^2 + \frac{1}{x^2} - 2 \cdot x \cdot \frac{1}{x} \)
\( = \left(x - \frac{1}{x}\right)^2 \)
\( = \left(x - \frac{1}{x}\right)\left(x - \frac{1}{x}\right) \)

In simple words: Rewrite the expression by introducing \( x \times \frac{1}{x} \) at the end to make it a perfect square trinomial, \( \left(x - \frac{1}{x}\right)^2 \).

Exam Tip: Remember that \( 2 \) is equivalent to \( 2 \cdot x \cdot \frac{1}{x} \) in these types of expressions.

Question 2(F). Factorise: \( (x^2 + y^2 - z^2)^2 - 4x^2y^2 \)
Answer:
\( (x^2 + y^2 - z^2)^2 - 4x^2y^2 \)
\( = (x^2 + y^2 - z^2)^2 - (2xy)^2 \)
Using the difference of squares:
\( = (x^2 + y^2 - z^2 - 2xy)(x^2 + y^2 - z^2 + 2xy) \)
\( = [(x^2 + y^2 - 2xy) - z^2][(x^2 + y^2 + 2xy) - z^2] \)
\( = [(x - y)^2 - z^2][(x + y)^2 - z^2] \)
Applying the difference of squares to both brackets:
\( = (x - y - z)(x - y + z)(x + y - z)(x + y + z) \)

In simple words: Use the difference of squares first to get two brackets. Group the trinomial parts inside each bracket to form perfect squares, then apply the difference of squares again.

Exam Tip: This problem requires multiple applications of the difference of squares identity. Keep factoring until no squares remain.

Question 2(G). Factorise: \( a^2 + b^2 - c^2 - d^2 + 2ab - 2cd \)
Answer:
\( a^2 + b^2 - c^2 - d^2 + 2ab - 2cd \)
\( = (a^2 + b^2 + 2ab) - (c^2 + d^2 + 2cd) \)
\( = (a + b)^2 - (c + d)^2 \)
\( = (a + b + c + d)(a + b - c - d) \)

In simple words: Group the terms with \( a, b \) together and those with \( c, d \) together to form two perfect squares, then apply the difference of squares.

Exam Tip: Be careful with the negative sign when grouping \( -c^2 - d^2 - 2cd \) as \( -(c^2 + d^2 + 2cd) \).

Question 2(H). Factorise: \( 4xy - x^2 - 4y^2 + z^2 \)
Answer:
\( 4xy - x^2 - 4y^2 + z^2 \)
\( = z^2 - x^2 - 4y^2 + 4xy \)
\( = z^2 - (x^2 + 4y^2 - 4xy) \)
\( = z^2 - (x - 2y)^2 \)
\( = [z - (x - 2y)][z + (x - 2y)] \)
\( = (z - x + 2y)(z + x - 2y) \)

In simple words: Rearrange the terms to place \( z^2 \) first, then group the remaining terms into a perfect square binomial \( (x - 2y)^2 \) and apply the difference of squares.

Exam Tip: Keep the parenthesis intact when applying the difference of squares first, to avoid simple sign errors in the final steps.

Question 2(I). Factorise: \( 4x^2 - 12ax - y^2 - z^2 - 2yz + 9a^2 \)
Answer:
\( 4x^2 - 12ax - y^2 - z^2 - 2yz + 9a^2 \)
\( = (4x^2 - 12ax + 9a^2) - (y^2 + z^2 + 2yz) \)
\( = (2x - 3a)^2 - (y + z)^2 \)
\( = [(2x - 3a) + (y + z)][(2x - 3a) - (y + z)] \)
\( = (2x - 3a + y + z)(2x - 3a - y - z) \)

In simple words: Group the terms to find two perfect square binomials. Factorise both, and apply the difference of squares identity to finish.

Exam Tip: Spot terms like \( 4x^2 \) and \( 9a^2 \) together with their cross term \( -12ax \) to form the square \( (2x - 3a)^2 \).

Question 2(J). Factorise: \( (x + y)^3 - x - y \)
Answer:
\( (x + y)^3 - x - y \)
\( = (x + y)(x + y)^2 - (x + y) \)
\( = (x + y)[(x + y)^2 - 1] \)
\( = (x + y)[(x + y + 1)(x + y - 1)] \)
\( = (x + y)(x + y + 1)(x + y - 1) \)

In simple words: Take out the common factor \( (x + y) \) first. The remaining part is a difference of squares which can be factored further.

Exam Tip: Be careful not to miss factoring the remaining quadratic difference \( (x + y)^2 - 1 \).

Question 2(K). Factorise: \( y^4 + y^2 + 1 \)
Answer:
\( y^4 + y^2 + 1 \)
\( = y^4 + 2y^2 + 1 - y^2 \)
\( = (y^2 + 1)^2 - y^2 \)
\( = (y^2 + 1 + y)(y^2 + 1 - y) \)
\( = (y^2 + y + 1)(y^2 - y + 1) \)

In simple words: Add and subtract \( y^2 \) to complete the square, forming the expression \( (y^2 + 1)^2 - y^2 \), and then apply the difference of squares identity.

Exam Tip: This is a standard math trick for quartic expressions: add and subtract a term to create a difference of squares.

Question 2(L). Factorise: \( (a^2 - b^2)(c^2 - d^2) - 4abcd \)
Answer:
\( (a^2 - b^2)(c^2 - d^2) - 4abcd \)
Expand the product of binomials:
\( = a^2c^2 - a^2d^2 - b^2c^2 + b^2d^2 - 4abcd \)
Split and rearrange the terms:
\( = a^2c^2 + b^2d^2 - 2abcd - a^2d^2 - b^2c^2 - 2abcd \)
\( = (a^2c^2 + b^2d^2 - 2abcd) - (a^2d^2 + b^2c^2 + 2abcd) \)
Write as perfect square binomials:
\( = (ac - bd)^2 - (ad + bc)^2 \)
Apply the difference of squares identity:
\( = [(ac - bd) + (ad + bc)][(ac - bd) - (ad + bc)] \)
\( = (ac - bd + ad + bc)(ac - bd - ad - bc) \)

In simple words: Expand the expression first. Split the \( -4abcd \) term into two \( -2abcd \) parts, group them with appropriate terms to form two perfect squares, and factorise using the difference of squares.

Exam Tip: Be careful when grouping the negative terms; pulling out the negative sign changes the signs inside the parenthesis.

Question 3(A). Simplify: \( (x^2 - 2x + 3)(x^2 + 2x + 3) \)
Answer:
\( (x^2 - 2x + 3)(x^2 + 2x + 3) \)
Rearranging the terms:
\( = [x^2 + 3 - 2x][x^2 + 3 + 2x] \)
Grouping the terms:
\( = [(x^2 + 3) - 2x][(x^2 + 3) + 2x] \)
Using the identity \( (A - B)(A + B) = A^2 - B^2 \):
\( = (x^2 + 3)^2 - (2x)^2 \)
\( = (x^2 + 3)^2 - 4x^2 \)

In simple words: Group the terms \( x^2 + 3 \) together as a single term. This lets you use the difference of squares identity to simplify the expression.

Exam Tip: Rearranging terms to match the form \( (A - B)(A + B) \) makes multiplication much cleaner and faster.

Question 3(B). Simplify: \( (x^2 - 2x + 3)(x^2 - 2x - 3) \)
Answer:
\( (x^2 - 2x + 3)(x^2 - 2x - 3) \)
Grouping \( (x^2 - 2x) \):
\( = [(x^2 - 2x) + 3][(x^2 - 2x) - 3] \)
\( = (x^2 - 2x)^2 - 3^2 \)
\( = (x^2 - 2x)^2 - 9 \)

In simple words: Group the \( x^2 - 2x \) terms together as \( A \) and \( 3 \) as \( B \). Then use the difference of squares formula to write the simplified expression.

Exam Tip: Recognize when part of an expression is repeated across both factors to group it and use standard identities.

Question 3(C). Simplify: \( (x^2 + 2x - 3)(x^2 - 2x + 3) \)
Answer:
\( (x^2 + 2x - 3)(x^2 - 2x + 3) \)
\( = [x^2 + (2x - 3)][x^2 - (2x - 3)] \)
\( = (x^2)^2 - (2x - 3)^2 \)
\( = x^4 - (2x - 3)^2 \)

In simple words: Group \( 2x - 3 \) and its negative to write the terms in the form \( (A + B)(A - B) \), then apply the identity to simplify.

Exam Tip: Be careful with the signs when grouping \( -2x + 3 \) as \( -(2x - 3) \).

Question 4(A). Factorise: \( y^2 + \frac{1}{4y^2} + 1 - 6y - \frac{3}{y} \)
Answer:
\( y^2 + \frac{1}{4y^2} + 1 - 6y - \frac{3}{y} \)
\( = \left(y^2 + \frac{1}{4y^2} + 1\right) - \left(6y + \frac{3}{y}\right) \)
\( = \left(y + \frac{1}{2y}\right)^2 - 6\left(y + \frac{1}{2y}\right) \)
\( = \left(y + \frac{1}{2y}\right)\left(y + \frac{1}{2y} - 6\right) \)

In simple words: Group the first three terms as they form the perfect square of \( y + \frac{1}{2y} \). Then group and factor out \( -6 \) from the last two terms to find the common binomial.

Exam Tip: The constant 1 serves as the perfect \( 2 \cdot a \cdot b \) term for the binomial square since \( 2 \cdot y \cdot \frac{1}{2y} = 1 \).

Question 4(B). Factorise: \( 4a^2 + \frac{1}{4a^2} - 2 - 6a + \frac{3}{2a} \)
Answer:
\( 4a^2 + \frac{1}{4a^2} - 2 - 6a + \frac{3}{2a} \)
\( = \left(4a^2 + \frac{1}{4a^2} - 2\right) - \left(6a - \frac{3}{2a}\right) \)
\( = \left(2a - \frac{1}{2a}\right)^2 - 3\left(2a - \frac{1}{2a}\right) \)
\( = \left(2a - \frac{1}{2a}\right)\left(2a - \frac{1}{2a} - 3\right) \)

In simple words: Group the first three terms as the perfect square \( \left(2a - \frac{1}{2a}\right)^2 \). Factor out \( -3 \) from the remaining terms, and take the common binomial out.

Exam Tip: Remember that \( 2a \cdot \frac{1}{2a} = 1 \), so \( -2 \) serves as the middle term for the expansion of \( \left(2a - \frac{1}{2a}\right)^2 \).

Question 4(C). Factorise: \( x^4 + y^4 - 6x^2y^2 \)
Answer:
\( x^4 + y^4 - 6x^2y^2 \)
\( = (x^2)^2 + (y^2)^2 - 2x^2y^2 - 4x^2y^2 \)
\( = [(x^2)^2 + (y^2)^2 - 2x^2y^2] - (4x^2y^2) \)
\( = (x^2 - y^2)^2 - (2xy)^2 \)
\( = (x^2 - y^2 - 2xy)(x^2 - y^2 + 2xy) \)

In simple words: Split the \( -6x^2y^2 \) term into \( -2x^2y^2 \) and \( -4x^2y^2 \) to complete the square. This gives a difference of squares that is easy to factorise.

Exam Tip: Splitting coefficients to reveal a perfect square trinomial is a powerful method for quartic expressions.

Question 4(D). Factorise: \( 4x^4 + 25y^4 + 19x^2y^2 \)
Answer:
\( 4x^4 + 25y^4 + 19x^2y^2 \)
\( = 4x^4 + 25y^4 + 20x^2y^2 - x^2y^2 \)
\( = (2x^2)^2 + (5y^2)^2 + 2(2x^2)(5y^2) - x^2y^2 \)
\( = [(2x^2)^2 + (5y^2)^2 + 2(2x^2)(5y^2)] - x^2y^2 \)
\( = [2x^2 + 5y^2]^2 - (xy)^2 \)
\( = (2x^2 + 5y^2 - xy)(2x^2 + 5y^2 + xy) \)

In simple words: Add and subtract \( x^2y^2 \) to rewrite \( 19x^2y^2 \) as \( 20x^2y^2 - x^2y^2 \). This completes a perfect square trinomial, allowing you to use the difference of squares identity.

Exam Tip: Always look to add/subtract small values to make terms perfect squares, which is very common in quartic expressions.

Question 4(E). Factorise: \( p^2 + \frac{1}{p^2} - 3 \)
Answer:
\( p^2 + \frac{1}{p^2} - 3 \)
\( = p^2 + \frac{1}{p^2} - 2 - 1 \)
\( = \left(p^2 + \frac{1}{p^2} - 2 \cdot p \cdot \frac{1}{p}\right) - 1 \)
\( = \left(p - \frac{1}{p}\right)^2 - 1^2 \)
\( = \left(p - \frac{1}{p} + 1\right)\left(p - \frac{1}{p} - 1\right) \)

In simple words: Split \( -3 \) into \( -2 - 1 \) to form a perfect square trinomial \( (p - \frac{1}{p})^2 \), and then use the difference of squares with \( 1^2 \).

Exam Tip: Splitting constants is a clever way to introduce perfect squares for trinomial expressions.

Question 4(F). Factorise: \( 5x^2 - y^2 - 4xy + 3x - 3y \)
Answer:
\( 5x^2 - y^2 - 4xy + 3x - 3y \)
\( = x^2 + 4x^2 - y^2 - 4xy + 3x - 3y \)
\( = (x^2 - y^2) + (4x^2 - 4xy) + (3x - 3y) \)
\( = (x - y)(x + y) + 4x(x - y) + 3(x - y) \)
\( = (x - y)[(x + y) + 4x + 3] \)
\( = (x - y)(5x + y + 3) \)

In simple words: Split \( 5x^2 \) into \( x^2 + 4x^2 \) and group the terms into three groups to find the common binomial factor \( x - y \).

Exam Tip: Splitting terms to group them into pairs that share a common binomial factor is a very useful technique for expressions with five or more terms.