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.

3x + 9b + 5ax + 15ab= 3(x + 3b) + 5a(x + 3b)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
= (x + 3b)(3 + 5a)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.

2ce  2cf + de  df= 2c(e – f) + d(e – f)
= (e – f)(2c + d) 
4u + 4v + vx + ux= 4(u + v) + x(v + u)
= (u + v)(4 + x)(u + v) and (v + u) are identical despite their order
 mn  3my  3nx + 9xy= m(n – 3y) – 3x(n – 3y)
= (n – 3y)(m – 3x)we took 3x as a common factor of the second pair of terms, thereby getting 3y when we divided 9xy by 3x.
 6a^{2}  3a + 4a  2= 3a(2a – 1) + 2(2a – 1)
= (2a – 1)(3a + 2)
 cd  ce  d^{2} + de= c(d – e) – d(d – e)
= (d – e)(c – d)
 12eg  4eh  6fg + 2hf= 4e(3g – h) – 2f(3g – h)
= (3g – h)(4e – 2f)
= (4e – 2f)(3g – h)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.
= 2(2e – f)(3g – h)
 4ab + 6bn  2a  3n= 2b(2a + 3n) – 1(2a + 3n)
= (2a + 3n)(2b – 1)sometimes, if there are no common factors, 1 and 1 can be used as common factors
 3ax  2a  6bx + 4b= a(3x – 2) – 2b(3x – 2)
= (3x – 2)(a – 2b)
 ax  a + x  1= a(x – 1) + 1(x – 1)
= (x – 1)(a + 1)
 ac + ad  bc  bd= a(c + d) – b(c + d)
= (c + d)(a – b)
 4a  7b + 28bx  16ax= 1(4a – 7b) + 4x(7b – 4a)
= 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.
= (4a – 7b)(1 – 4x)now, (4a – 7b) is identical to (7b + 4a)
 p + q + 5ap + 5aq= 1(p + q) + 5a(p + q)
= (p + q)(1 + 5a)
 2c^{2} + 8cm  3cm + 12m^{2}= 2c(c + 4m) – 3m(c + 4m)
= (c + 4m)(2c – 3m)
 x^{2} + 2x  5x  10= x(x + 2) – 5(x + 2)
= (x + 2)(x – 5)
 2pr  sq + 2qr  ps= 2pr + 2qr – ps – sqsometimes if it appears as if there is no common factor, you regroup the terms
= 2r(p + q) – s(p + q)
= (p + q)(2r – s)
 y^{2}  5y + 4y  20= y(y – 5) + 4(y – 5)
= (y – 5)(y + 4)
 6ac + bd  3bc  2ad= 6ac – 3bc + bd – 2ad
= 3c(2a – b) + d(b – 2a)
= 3c(2a – b) – d(b + 2a)
= (2a – b)(3c – d)
 2ab  10cd  5bc + 4ad= 2ab + 4ad – 5bc – 10cd
= 2a(b + 2d) – 5c(b + 2d)
= (b + 2d)(2a – 5c)
 6xy  2z  4y + 3xz= 6xy – 4y + 3xz – 2z
= 2y(3x – 2) + z(3x – 2)
= (3x – 2)(2y + z)
Factorising quadratic expressions
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
first we multiply the end terms, x^{2} and 14 in this case to get 14x^{2}
Then we list the factors of 14x^{2} and find among them, 2 factors whose sum is 9x, the middle term.
14x^{2} =
 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.
= x^{2} + 9x + 14
= x^{2} + 2x + 7x + 14
= x(x + 2) + 7(x + 2)
= (x + 2)(x + 7)