Sec 2 Maths · Expansion and factorisation of quadratic expressions

Factorising by grouping and special products questions with answers

Practice questions on factorising by grouping and special products, from the basics up to exam level, each with a full worked solution. Try them on paper first, then open the solution to check your working.

Worked example

Factorise $2hm + 8hn + km + 4kn$ completely.

Step 1 of 2

Group the terms in pairs and take out the HCF of each pair

2hm + 8hn + km + 4kn = 2h(m + 4n) + k(m + 4n)

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Questions

Question 1Grouping: four terms
Factorise $15ax - 35a + 9bx - 21b$ completely.
Show the worked solution
1. Step 1: Group the terms in pairs and take out the HCF of each pair $15ax - 35a + 9bx - 21b = 5a(3x - 7) + 3b(3x - 7)$
2. Step 2: Take out the common bracket $= (3x - 7)(5a + 3b)$
Question 2Grouping: taking out a negative factor
Factorise $xy + 5x - y - 5$ completely.
Show the worked solution
1. Step 1: Group the terms in pairs and take out the HCF of each pair (a minus HCF changes the signs in its bracket) $xy + 5x - y - 5 = x(y + 5) - 1(y + 5)$
2. Step 2: Take out the common bracket $= (y + 5)(x - 1)$
Question 3Difference of two squares
Factorise $81x^{2} - 64y^{2}$ completely.
Show the worked solution
1. Step 1: Write each term as a square $81x^{2} - 64y^{2} = (9x)^{2} - (8y)^{2}$
2. Step 2: Use a² − b² = (a + b)(a − b) $= (9x + 8y)(9x - 8y)$
Question 4Common factor first, then a² − b²
Factorise $4x^{2} - 64$ completely.
Show the worked solution
1. Step 1: Take out the HCF of the terms $4x^{2} - 64 = 4(x^{2} - 16)$
2. Step 2: Factorise the difference of two squares $= 4(x + 4)(x - 4)$
Question 5Common factor first, then a quadratic
Factorise $10x^{2} + 25xy - 125y^{2}$ completely.
Show the worked solution
1. Step 1: Take out the HCF of the terms first $10x^{2} + 25xy - 125y^{2} = 5(2x^{2} + 5xy - 25y^{2})$
2. Step 2: Factorise the quadratic in the bracket $= 5(x + 5y)(2x - 5y)$
Question 6Quadratics in two letters
Factorise $2p^{2} - pq - 10q^{2}$ completely.
Show the worked solution
1. Step 1: Find two numbers that multiply to 2 × (-10) and add to -1 $-5 \times 4 = -20$ and $-5 + 4 = -1$
2. Step 2: Split the middle term with them, group in pairs, and take out the common bracket $2p^{2} - pq - 10q^{2} = 2p^{2} - 5pq + 4pq - 10q^{2}$ $= p(2p - 5q) + 2q(2p - 5q)$ $= (p + 2q)(2p - 5q)$
Question 7Hence: factorise by substitution
Given that $2x^{2} - 9x + 10 = (x - 2)(2x - 5)$ , factorise $2(y + 4)^{2} - 9(y + 4) + 10$ completely.
Show the worked solution
1. Step 1: It is the given expression with x replaced by y + 4: do the same in the brackets $2(y + 4)^{2} - 9(y + 4) + 10 = [(y + 4) - 2][2(y + 4) - 5]$
2. Step 2: Expand inside each bracket and simplify $= (y + 2)(2y + 3)$
Question 8Grouping, then a² − b²
Factorise $p^{2} - 4pq + 4q^{2} - 4$ completely.
Show the worked solution
1. Step 1: The first three terms make a perfect square: (a − b)² = a² − 2ab + b² $p^{2} - 4pq + 4q^{2} - 4 = (p - 2q)^{2} - 2^{2}$
2. Step 2: Use a² − b² = (a + b)(a − b) $= (p - 2q + 2)(p - 2q - 2)$

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Questions

Question 1Grouping: four terms

Factorise $15ax - 35a + 9bx - 21b$ completely.

Show the worked solution

Step 1: Group the terms in pairs and take out the HCF of each pair $15ax - 35a + 9bx - 21b = 5a(3x - 7) + 3b(3x - 7)$
Step 2: Take out the common bracket $= (3x - 7)(5a + 3b)$

Question 2Grouping: taking out a negative factor

Factorise $xy + 5x - y - 5$ completely.

Show the worked solution

Step 1: Group the terms in pairs and take out the HCF of each pair (a minus HCF changes the signs in its bracket) $xy + 5x - y - 5 = x(y + 5) - 1(y + 5)$
Step 2: Take out the common bracket $= (y + 5)(x - 1)$

Question 3Difference of two squares

Factorise $81x^{2} - 64y^{2}$ completely.

Show the worked solution

Step 1: Write each term as a square $81x^{2} - 64y^{2} = (9x)^{2} - (8y)^{2}$
Step 2: Use a² − b² = (a + b)(a − b) $= (9x + 8y)(9x - 8y)$

Question 4Common factor first, then a² − b²

Factorise $4x^{2} - 64$ completely.

Show the worked solution

Step 1: Take out the HCF of the terms $4x^{2} - 64 = 4(x^{2} - 16)$
Step 2: Factorise the difference of two squares $= 4(x + 4)(x - 4)$

Question 5Common factor first, then a quadratic

Factorise $10x^{2} + 25xy - 125y^{2}$ completely.

Show the worked solution

Step 1: Take out the HCF of the terms first $10x^{2} + 25xy - 125y^{2} = 5(2x^{2} + 5xy - 25y^{2})$
Step 2: Factorise the quadratic in the bracket $= 5(x + 5y)(2x - 5y)$

Question 6Quadratics in two letters

Factorise $2p^{2} - pq - 10q^{2}$ completely.

Show the worked solution

Step 1: Find two numbers that multiply to 2 × (-10) and add to -1 $-5 \times 4 = -20$ and $-5 + 4 = -1$
Step 2: Split the middle term with them, group in pairs, and take out the common bracket $2p^{2} - pq - 10q^{2} = 2p^{2} - 5pq + 4pq - 10q^{2}$ $= p(2p - 5q) + 2q(2p - 5q)$ $= (p + 2q)(2p - 5q)$

Question 7Hence: factorise by substitution

Given that $2x^{2} - 9x + 10 = (x - 2)(2x - 5)$ , factorise $2(y + 4)^{2} - 9(y + 4) + 10$ completely.

Show the worked solution

Step 1: It is the given expression with x replaced by y + 4: do the same in the brackets $2(y + 4)^{2} - 9(y + 4) + 10 = [(y + 4) - 2][2(y + 4) - 5]$
Step 2: Expand inside each bracket and simplify $= (y + 2)(2y + 3)$

Question 8Grouping, then a² − b²

Factorise $p^{2} - 4pq + 4q^{2} - 4$ completely.

Show the worked solution

Step 1: The first three terms make a perfect square: (a − b)² = a² − 2ab + b² $p^{2} - 4pq + 4q^{2} - 4 = (p - 2q)^{2} - 2^{2}$
Step 2: Use a² − b² = (a + b)(a − b) $= (p - 2q + 2)(p - 2q - 2)$