6.12 Factorising
The distributive law says that for any numbers $A,B$A,B, and $C$C, $A\left(B+C\right)=AB+AC$A(B+C)=AB+AC. We saw how to use this rule to expand algebraic terms, but we can also use the rule in reverse.
The reverse of expanding algebraic expressions is called factorising. Factorising an algebraic expression means writing the expression with any common factors between the terms taken outside of the brackets.
Worked Examples
Example 1
Fully factorise $12x+20y$12x+20y.
Think: Before we can factorise the expression, we should look for the factors of each term. $12x$12x has the factors $12$12 and $x$x and $20y$20y has the factors $20$20 and $y$y.
We should check for any common factors in the coefficients. $12=4\times3$12=4×3 and $20=4\times5$20=4×5, so the highest common factor of the coefficients is $4$4.
Do:
| $12x+20y$12x+20y | $=$= | $4\times3x+4\times5y$4×3x+4×5y |
Write the coefficients as products of HCF. |
| $=$= | $4\left(3x+5y\right)$4(3x+5y) |
Factorise using the distributive law, $AB+AC=A\left(B+C\right)$AB+AC=A(B+C). Here, $A=4,B=3x,$A=4,B=3x, and $C=5y$C=5y. |
Reflect: It is helpful to find the highest common factor of the coefficients before we factorise the expression. Otherwise we might end up having to factorise a second time. For example, if we factorised $2$2 instead, we would have $2\left(6x+10y\right)$2(6x+10y). Since $6$6 and $10$10 have a common factor of $2$2 we would still need to factorise this again.
Example 2
Fully factorise $-3xz-5yz$−3xz−5yz.
Think: In this case, there is no highest common factor of the coefficients. However, we can still factorise this expression.
First, notice that both terms have a pronumeral factor of $z$z. This means that we can factorise $z$z.
Also, both terms are negative. Being negative is the same as having a factor of $-1$−1. So we can also factorise $-1$−1.
Do:
| $-3xz-5yz$−3xz−5yz | $=$= | $-1\times3xz+\left(-1\right)\times5yz$−1×3xz+(−1)×5yz |
Write the terms as products of $-1$−1. Note that the sign of the terms changes. |
| $=$= | $-1\times z\times3x+\left(-1\right)\times z\times5y$−1×z×3x+(−1)×z×5y |
Write the terms as products of $z$z. |
|
| $=$= | $-1\times z\left(3x+5y\right)$−1×z(3x+5y) |
Factorise using the law, $AB+AC=A\left(B+C\right)$AB+AC=A(B+C). Here, $A=-1\times z,B=3x,$A=−1×z,B=3x, and $C=5y$C=5y. |
|
| $=$= | $-z\left(3x+5y\right)$−z(3x+5y) |
Simplify the factor $-1\times z$−1×z. |
Reflect: It is important to be careful about factorising $-1$−1, because the sign of each term will change.
We can use the reverse of distributive law to factorise an algebraic expression like so.
$AB+AC=A\left(B+C\right)$AB+AC=A(B+C)
This means means writing the expression with any common factors between the terms taken outside of the brackets. Factorising is the reverse of expanding.
Practice questions
Question 1
Factorise $3w+15$3w+15.
Question 2
Factorise $40c-35$40c−35.
Question 3
Factorise $10g-35h+15$10g−35h+15.