Factorisation
Class 8 Maths · Factorisation · 34 questions with answers
Most children who are stuck on factorisation are not unable to factorise - they are guessing which method to try. There are four, and they go in a fixed order: common factor, identity, grouping, splitting the middle term. A child working down that list rarely gets stuck; a child starting anywhere gets stuck constantly.
- Takes out the highest common factor before anything else
- Recognises a difference of squares and a perfect square trinomial
- Factorises a four-term expression by grouping
- Splits the middle term of a quadratic, including when the leading coefficient is not 1
- Chooses the method in the right order rather than by trial
Exercise 1 of 3 · 10 questions
Factorise by taking out the common factor.
Write the answer · Warm-up
Here’s one done for you
12a²b - 18ab² = 6ab(2a - 3b). Two things to check every time: the number factor is the highest common one (6, not 2 or 3), and the letters are taken to the lowest power that appears in both (ab, not a²b). Taking out a partial factor is not wrong exactly, but it leaves work behind and loses marks that were already earned.
- 1)6x + 9
- 2)12a²b - 18ab²
- 3)5x² + 10x
- 4)7pq + 14p
- 5)4m³ - 8m²
- 6)8x + 20
- 7)15a²b - 25ab²
- 8)6x² + 9x
- 9)3m³ - 12m²
- 10)9pq + 12p
Answers
- 1) 3(2x + 3).
- 2) 6ab(2a - 3b). The common factor is 6ab - take the highest one, not just any one.
- 3) 5x(x + 2).
- 4) 7p(q + 2).
- 5) 4m²(m - 2).
- 6) 4(2x + 5)
- 7) 5ab(3a - 5b)
- 8) 3x(2x + 3)
- 9) 3m²(m - 4)
- 10) 3p(3q + 4)
Exercise 2 of 3 · 12 questions
Factorise using an identity, or by grouping.
Write the answer · Guided
Here’s one done for you
ax + ay + bx + by = (a + b)(x + y). Grouping works when there are four terms and no factor common to all of them. Take a from the first pair and b from the second: a(x + y) + b(x + y). Both now share (x + y), so it comes out. If the second bracket does not match the first, try pairing the terms differently before deciding grouping will not work.
- 1)x² - 16
- 2)9a² - 25
- 3)x² + 10x + 25
- 4)4y² - 12y + 9
- 5)ax + ay + bx + by
- 6)x² + 3x + 2x + 6
- 7)x² - 49
- 8)16a² - 81
- 9)x² + 14x + 49
- 10)9y² - 24y + 16
- 11)px + py + qx + qy
- 12)x² + 5x + 3x + 15
Answers
- 1) (x + 4)(x - 4). Difference of squares.
- 2) (3a + 5)(3a - 5). Since 9a² = (3a)².
- 3) (x + 5)². A perfect square, because 25 = 5² and 10x = 2 × x × 5.
- 4) (2y - 3)². Since 4y² = (2y)², 9 = 3² and 12y = 2 × 2y × 3.
- 5) (a + b)(x + y). Group as a(x + y) + b(x + y).
- 6) (x + 3)(x + 2). Group as x(x + 3) + 2(x + 3).
- 7) (x + 7)(x - 7)
- 8) (4a + 9)(4a - 9)
- 9) (x + 7)²
- 10) (3y - 4)²
- 11) (x + y)(p + q)
- 12) (x + 3)(x + 5)
Exercise 3 of 3 · 12 questions
Factorise by splitting the middle term.
Write the answer · Practice
Here’s one done for you
2x² + 7x + 3 = (2x + 1)(x + 3). When the number in front of x² is not 1, the two numbers must multiply to the first coefficient times the last - here 2 × 3 = 6 - and add to the middle one, 7. That gives 6 and 1, so write 2x² + 6x + x + 3 and group: 2x(x + 3) + 1(x + 3). Children who learn the shortcut for x² + bx + c and are never shown this extension get stuck on every question where the leading coefficient is not 1.
- 1)x² + 7x + 12
- 2)x² - 5x + 6
- 3)x² + x - 12
- 4)x² - 2x - 15
- 5)2x² + 7x + 3
- 6)3x² - 10x + 8
- 7)x² + 11x + 30
- 8)x² - x - 20
- 9)x² - 9x + 20
- 10)x² + 2x - 24
- 11)2x² + 9x + 4
- 12)3x² + 11x + 6
Answers
- 1) (x + 3)(x + 4). Two numbers multiplying to 12 and adding to 7.
- 2) (x - 2)(x - 3). Product 6, sum -5, so both are negative.
- 3) (x + 4)(x - 3). Product -12, sum +1.
- 4) (x - 5)(x + 3). Product -15, sum -2.
- 5) (2x + 1)(x + 3). Product 2 × 3 = 6, sum 7, so split as 6x + x: 2x² + 6x + x + 3.
- 6) (3x - 4)(x - 2). Product 3 × 8 = 24, sum -10, so split as -6x - 4x.
- 7) (x + 5)(x + 6)
- 8) (x - 5)(x + 4)
- 9) (x - 4)(x - 5)
- 10) (x + 6)(x - 4)
- 11) (2x + 1)(x + 4)
- 12) (3x + 2)(x + 3)
While your child works
- Ask 'which method, and why that one?' before any factorising happens. Common factor, identity, grouping, middle term - in that order.
- Check the answer by expanding it back. It takes fifteen seconds and catches every error.
- When the number in front of x² is not 1, the product to aim for is first coefficient times last. That extension is the one most often never taught.