## Factorisation by grouping

Usually when we are given an expression that contains 4 terms like
3x + 9b + 5ax + 15ab and there is no common factor throughout we factorise it by grouping terms that have common factors together.

Examples

Factorise the following completely.

1. 3x + 9b + 5ax + 15ab
3 is the common factor of 3x and 9, whilst 5a is the common factor of 5ax and 15ab, so we factorise them in pairs as follows
= 3(x + 3b) + 5a(x + 3b)
at this stage, make sure the terms in the two brackets are identical. Which means (x + 3b) is now the common factor. Take (x + 3b) and then divide it into 3(x + 3b) to get 3, then divide (x + 3b) into 5a(x + 3b) too to get 5a. Finally put 3 and 5a into one bracket as follows.
= (x + 3b)(3 + 5a)
2. 2ce - 2cf + de - df
= 2c(e – f) + d(e – f)
= (e – f)(2c + d)
3. 4u + 4v + vx + ux
= 4(u + v) + x(v + u)
(u + v) and (v + u) are identical despite their order
= (u + v)(4 + x)
4. mn - 3my - 3nx + 9xy
= m(n – 3y) – 3x(n – 3y)
we took -3x as a common factor of the second pair of terms, thereby getting -3y when we divided 9xy by -3x.
= (n – 3y)(m – 3x)
5. 6a2 - 3a + 4a - 2
= 3a(2a – 1) + 2(2a – 1)
= (2a – 1)(3a + 2)
6. cd - ce - d2 + de
= c(d – e) – d(d – e)
= (d – e)(c – d)
7. 12eg - 4eh - 6fg + 2hf
= 4e(3g – h) – 2f(3g – h)
= (3g – h)(4e – 2f)
however, at this stage we still haven’t factorised completely because (4e – 2f) still has a common factor 2. Let’s then rearange and bring (4e – 2f) upfront.
= (4e – 2f)(3g – h)
= 2(2e – f)(3g – h)
8. 4ab + 6bn - 2a - 3n
= 2b(2a + 3n) – 1(2a + 3n)
sometimes, if there are no common factors, 1 and -1 can be used as common factors
= (2a + 3n)(2b – 1)
9. 3ax - 2a - 6bx + 4b
= a(3x – 2) – 2b(3x – 2)
= (3x – 2)(a – 2b)
10. ax - a + x - 1
= a(x – 1) + 1(x – 1)
= (x – 1)(a + 1)
11. ac + ad - bc - bd
= a(c + d) – b(c + d)
= (c + d)(a – b)
12. 4a - 7b + 28bx - 16ax
= 1(4a – 7b) + 4x(7b – 4a)
you cant factorise properly unless those 2 brackets are identical. However, to make the 2 brackets identical we factor out -4x instead of 4x from the second pair.
= 1(4a – 7b) – 4x(-7b + 4a)
now, (4a – 7b) is identical to (-7b + 4a)
= (4a – 7b)(1 – 4x)
13. p + q + 5ap + 5aq
= 1(p + q) + 5a(p + q)
= (p + q)(1 + 5a)
14. 2c2 + 8cm - 3cm + 12m2
= 2c(c + 4m) – 3m(c + 4m)
= (c + 4m)(2c – 3m)
15. x2 + 2x - 5x - 10
= x(x + 2) – 5(x + 2)
= (x + 2)(x – 5)
16. 2pr - sq + 2qr - ps
sometimes if it appears as if there is no common factor, you regroup the terms
= 2pr + 2qr – ps – sq
= 2r(p + q) – s(p + q)
= (p + q)(2r – s)
17. y2 - 5y + 4y - 20
= y(y – 5) + 4(y – 5)
= (y – 5)(y + 4)
18. 6ac + bd - 3bc - 2ad
= 6ac – 3bc + bd – 2ad
= 3c(2a – b) + d(b – 2a)
= 3c(2a – b) – d(-b + 2a)
= (2a – b)(3c – d)
19. 2ab - 10cd - 5bc + 4ad
= 2ab + 4ad – 5bc – 10cd
= 2a(b + 2d) – 5c(b + 2d)
= (b + 2d)(2a – 5c)
20. 6xy - 2z - 4y + 3xz
= 6xy – 4y + 3xz – 2z
= 2y(3x – 2) + z(3x – 2)
= (3x – 2)(2y + z)

The method of factorisation by grouping can be extended even to quadratic expressions. In the previous exercise there were some quadratic expressions we factorised which were however in different form from what we are going to look at right now.

Example

Factorise the following completely

x2 + 9x + 14
This expression has 3 terms and we need 4 terms to factorise it by grouping pairs. So we split the middle term into 2 terms as follows.

first we multiply the end terms, x2 and 14 in this case to get 14x2

Then we list the factors of 14x2 and find among them, 2 factors whose sum is 9x, the middle term.

14x2 =

• x × 14x [add these and you get 15x.]
• 2x × 7x [add these and you get 9x.]

Therefore the factors we need are 2x and 7x. Go back to the oroginal expression and replace 9x with these 2 factors.

= x2 + 9x + 14
= x2 + 2x + 7x + 14

then factorise by grouping

= x(x + 2) + 7(x + 2)
= (x + 2)(x + 7)