1.09 Factorisation with common factors
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.
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 a product of the highest common factor. |
| $=$= | $4\left(3x+5y\right)$4(3x+5y) |
Use the distributive law, $A\left(B+C\right)=AB+AC$A(B+C)=AB+AC. 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 when we do this. |
| $=$= | $-1\times z\times3x+\left(-1\right)\times z\times5y$−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) |
Use the distributive law, $A\left(B+C\right)=AB+AC$A(B+C)=AB+AC. 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.
Example 3
Fully factorise $5x^2y^2-7xy^2z$5x2y2−7xy2z
Think: We can see that $y^2$y2 is a factor of both terms. We also have $x^2$x2 in one term and $x$x in the other. Since $x^2=x\times x$x2=x×x, we can factorise $x$x from both terms using index laws.
Do:
| $5x^2y^2-7xy^2z$5x2y2−7xy2z | $=$= | $xy^2\times5x-xy^2\times7z$xy2×5x−xy2×7z |
Write the terms as products of $xy^2$xy2. Note that we can take $x$x out of $x^2$x2 but then we decrease the index by $1$1. |
| $=$= | $xy^2\left(5x-7z\right)$xy2(5x−7z) |
Use the distributive law, $A\left(B+C\right)=AB+AC$A(B+C)=AB+AC. Here, $A=xy^2\times z,B=5x,$A=xy2×z,B=5x, and $C=7z$C=7z. |
Reflect: In general, when each term has powers of the same variable, we can factorise the power with the lowest index. In this case we subtract this index from all the indices.
We can use the distributive law to factorise an algebraic expression like $AB+AC$AB+AC. 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: $y^2+4y$y2+4y
Question 2
Factorise the following expression:
$5k^2t+40k^3t^3$5k2t+40k3t3
Question 3
Factorise the following expression:
$2yz-10xy+12xy^2z$2yz−10xy+12xy2z