Expansion and Factorisation
Expansion and factorisation of algebraic terms. This is a continuiation of Basic Processes of Algebra
Expansion of brackets
The expression (a + b)(c + d) means (a + b) × (c + d). What it means is that every term inside the first bracket multiplies every term inside the second bracket.

Lets start with a and multiply it with c the first term inside the second bracket:
(a + b)(c + d) = ac + bc + ad + bd 
Then b multiplies c:
(a + b)(c + d) = ac + bc + ad + bd 
Then a multiplies d:
(a + b)(c + d) = ac + bc + ad + bd 
Finally b multiplies d:
(a + b)(c + d) = ac + bc + ad + bd
Whatever order you take in expanding the brackets, as long as you make sure that every term inside the first bracket multiplies every term inside the second bracket.
Examples
Expand the following, collecting the like terms where possible.

(p + q)(r + s)
p by r, then p by s, then q by r, and finally q by s.
= pr + ps + qr + qs 
(6 + y)(5 + x)
= 30 + 6x + 5y + xy 
(a – b)(c – d)
= ac – ad – bc + bd 
(2x + 1)(3x – 1)
= 6x^{2} + – 2x + 3x – 1
2x and 3x are like terms so we simplify them
6x^{2} + x – 1 
(2n + 3)(2n + 5)
= 4n^{2} + 10n + 6n + 15
= 4n^{2} + 16n + 15 
(4m – 9)(2m – 3)
= 8m^{2} – 12m – 18m + 27
= 8m^{2} – 30m + 27 
(5x – 2)(x + 4)
= 5x^{2} + 20x – 2x – 8
= 5x^{2} + 18x – 8 
(a + 5)(3a – 2)
= 3a^{2} – 2a + 15a – 10
= 3a^{2} + 13a – 10 
(x – 8)(x + 9)
= x^{2} + 9x – 8x – 72
= x^{2} + x – 72 
(3h + 4)(h – 2)
= 3h^{2} – 6h + 4h – 8
= 3h^{2} – 2h – 8 
(b – 5)(b + 5)
= b^{2} + 5b – 5b – 25
= b^{2} – 25 
(c + 7)(c – 7)
= c^{2} + 7c – 7c – 49
= c^{2} – 49 
(3t – 2)^{2}
= (3t – 2)(3t – 2)
= 9^{2} – 6t – 6t + 4
= 9^{2} – 12t + 4 
(2x – 5y)^{2}
= (2x – 5y)(2x – 5y)
= 4x^{2} – 10xy – 10xy + 25y^{2}
= 4x^{2} – 20xy + 25y^{2} 
(5x – y)(2x – 3y)
= 10x^{2} – 15xy – 2xy + 3y^{2}
= 10x^{2} – 17xy + 3y^{2}
Factorisation
Factorising an algebraic expression simply means pulling out common factors from a expression and expressing the factors as products of each other. For example, given an expression like 2m + 8n, the highest common factor of 2m and 8n is 2, because 2m and 8n are both exactly divisible by 2.
Factorising 2m + 8n gives us 2(m + 4n).
After pulling out the common factor, we divide every term by the common factor and put the the results in the brackets.
Examples
Factorise the following:

2m + 8n
= 2(m + 4n) 
3a – 15b
= 3(a – 5b) 
10x – 5
= 5(2x – 1) 
3h – 12k
= 3(h + 4k)
usually when pulling out common factors, we usually take the sign on the first term. 
2x + 18
= 2(x – 9) 
5a – 8ab
= a(5 – 8b)
a is a common factor in this case because it exists on both 5a and 8ab 
9x + 3xz
= 3x(3 + z) 
8cm + 12dm – 16em
= 4m(2c + 3d – 4e) 
3x^{3} – 12x^{2} – 9x
= 3x(x^{2} – 4x – 3)
when dividing terms that have powers, we subtract the powers. Which means 3x^{3} divided by 3x gives us x^{2}. 
10a^{2}b^{2} – 15a^{2}b + 20ab^{2}
= 5ab(2ab – 3a + 4b) 
4a – 8b
= 4(a – 2b) 
9x + 12y
= 3(3x + 4y) 
3ab – 6ac + 3ad
= 3a(b – 2c + d) 
8px – 4qx + 8rx
= 4x(2p – q + 2r) 
3m^{3} – 2m^{2} + m
= m(3m^{2} – 2m + 1)