Algebra 4

Factorising is the opposite of expanding brackets. When we factorise, we put brackets in. In the previous session we looked at the expression x(y + 1). If we expand the brackets we get xy + x. So what happens if we factorise xy + x? We go back to x(y + 1). How does this work?
If you look at the expression xy + x we can see that there is an x common to both terms. So we can take this x outside the bracket and then ask ourselves , in turn, what do we need to multiple x by to get xy (this is the first term) and x (this is the second term). The answers are y and 1 so
xy + x = x(y + 1).

Example 1 (Foundation Nov 14 Paper 2 Q18(a) )
Factorise 6x - 8
Solution: At first sight there isn`t much we can do with this expression. However, there is something that will divide into each of the two terms 6x and 8. That something is 2. So we can write 2 outside the bracket and ask what we need to multiply 2 by to get 6x (answer 3x) and 8 (answer 4). So the factored expression is 2(3x - 4).
Example 2
Factorise 2xy + 10x
Solution: Here both x and 2 will divide into each of the two terms 2xy and 10x. So we can write 2x outside the bracket and ask what we need to multiply 2x by to get 2xy (answer y) and 10x (answer 5). So the factored expression is 2x(y + 5).
Example 3
Factorise 5x + 25xy (Answer 5x(1 + 5y) )
Example 4
Factorise 4x + 8xy + 2xz (Answer 2x(2 + 4y + z) )
Example 5 (Foundation Nov 12 Paper 1 Q13(a) )
Factorise 24x + 3 (Answer 3(8x + 1) )
Example 6 (Foundation Nov 12 Paper 1 Q13(b) )
Factorise x2 - 6x (Answer x(x - 6) )
Example 7
Factorise 12x - 2xy - 4x3 (Answer 2x(6 - y - 2x2 )
Example 8
Factorise xy + y - x (Answer xy + y - x. There are no terms common to all three terms which we can factor out.)