Factorising is effectively the opposite of expanding brackets.

We are given an expression and we have to “put brackets back into it”, by spotting common factors.

So if we have 2x + 6 + 10y we could spot that each of our terms has 2 as a factor and rewrite it as the equal expression 2(x+3+5y)

Or if we have 6x² +3x we can spot that x is a common factor and 3 is a common factor, so overall 3x is a common factor, and we can rewrite the expression as the equal expression 3x(2x+1).

Exercise

Let’s complete the following exercise (exercise 12 from the core textbook):

The answers are below:

Two stage factorisation

Sometimes we can’t find a factor that is a common factor of all of our terms, but only one of some of our terms. In this situation we can do our factorisation in stages:

Example

• Factorise:
  • xy + bx + ay + ab
  • ax + 3a + 2x + 6
  • 6ac-3bc+2ad-bd
  • 2ay – 2ab + xy – bx
  • x²y² + bx² + ay² + ab

Exercise

Factorise the following expressions:

Question 1	Question 2	Question 3
(a.) ax + ay + bx + by	(a) hs + ht + ks + kt	(a) 2ax + 6ay + bx + 3by
(b.) ay + az + by + bz	(b) xs – xt + ys – yt	(b) 2ax + 2ay + bx + by
(c.) xb + xc + yb + yc	(c) ax – ay – bx + by	(c) 2mh – 2mk + nh – nk
(d.) xh + xk + yh + yk	(d) xs – xt – ys + yt	(d) 2mh + 3mk – 2nh – 3nk
(e.) xm + xn + my + ny	(e) as – ay – xs + xy	(e) 6ax + 2bx + 3ay + by
(f.) ah – ak + bh – bk	(f) hx – hy – bx + by	(f) 2ax – 2ay – bx + by
(g.) ax – ay + bx – by	(g) am – bm – an + bn	(g) x2a + x2b + ya + yb
(h.) am – bm + an – bn	(h) xk – xm – kz + mz	(h) ms + 2mt2 – ns – 2nt2

Answers

The answers are below:

Question 1	Question 2	Question 3
(a) (a+b)(x+y)	(a) (h+k)(s+t)	(a) (2a+b)(x+3y)
(b) (a+b)(y+z)	(b) (x+y)(s-t)	(b) (2a+b)(x+y)
(c) (x+y)(b+c)	(c) (a-b)(x-y)	(c) (2m+n)(h-k)
(d) (x+y)(h+k)	(d) (x-y)(s-t)	(d) (m-n)(2h+3k)
(e) (x+y)(m+n)	(e) (a-x)(s-y)	(e) (2x+y)(3a+b)
(f) (a+b)(h-k)	(f) (h-b)(x-y)	(f) (2a-b)(x-y)
(g) (a+b)(x-y)	(g) (m-n)(a-b)	(g) (x2+y)(a+b)
(h) (m+n)(a-b)	(h) (x-z)(k-m)	(h) (m-n)(s+2t2)