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iGCSE (2021 Edition)

13.01 Factorising binomial products (Extended)

Lesson

We saw how to use the distributive law to expand binomial products. We can also use it to factorise binomial products. In order to find what the factors are, we factorise the terms in pairs.

Worked examples

Example 1

Factorise $xy+8x-3y-24$xy+8x3y24

Think: We can use the rule $\left(A+B\right)\left(C+D\right)=AC+AD+BC+BD$(A+B)(C+D)=AC+AD+BC+BD. First we will factorise $xy+8x$xy+8x and $-3y-24$3y24 separately. Then we can factorise the whole expression.

Do:

$xy+8x-3y-24$xy+8x3y24 $=$= $x\left(y+8\right)-3\left(y+8\right)$x(y+8)3(y+8)

Factorising $x$x from $xy+8x$xy+8x

and $3$3 from $-3y-24$3y24

  $=$= $\left(x-3\right)\left(y+8\right)$(x3)(y+8)

Since $\left(y+8\right)$(y+8) is a factor

of both terms we can factorise it

Reflect: We can check that this is the answer by expanding it. Note that sometimes we have to change the order of the terms in order to factorise them this way. For example, if we rearranged the terms to be $xy-3y+8x-24$xy3y+8x24 we would get the same answer.

We can also use the special cases of expansion that we learned previously to factorise.

Example 2

Factorise $x^2-6x+9$x26x+9

Think: Notice that $-6x=2\times\left(-3x\right)$6x=2×(3x). Therefore, $x^2-6x+9$x26x+9 is the result of expanding a perfect square.

Do: Using the rule $\left(A+B\right)^2=A^2+2AB+B^2$(A+B)2=A2+2AB+B2, we can see that $A=x$A=x and $B=-3$B=3. Therefore,

$x^2-6x+9=\left(x-3\right)^2$x26x+9=(x3)2

Reflect: Once again, we can check the answer by expanding it. It's always worth checking if an expression fits the pattern $A^2+2AB+B^2$A2+2AB+B2, because then we can use this special rule. . We call this perfect square factorisation.

Example 3

Factorise $x^2-9$x29

Think: Notice that $9=3^2$9=32. Therefore, $x^2-9$x29 is the difference of two squares.

Do: Using the rule $\left(A+B\right)\left(A-B\right)=A^2-B^2$(A+B)(AB)=A2B2, we can see that $A=x$A=x and $B=3$B=3. Therefore,

$x^2-9=\left(x+3\right)\left(x-3\right)$x29=(x+3)(x3)

Reflect: Once again, we can check the answer by expanding it. It's always worth checking if an expression fits the pattern $A^2-B^2$A2B2, because then we can use this special rule. . We call this difference of two squares factorisation. Also notice that it would not make a difference if we said that $B=-3$B=3.

Summary

We can factorise the product of two binomial expressions using the rule $\left(A+B\right)\left(C+D\right)=AC+AD+BC+BD$(A+B)(C+D)=AC+AD+BC+BD.

There are two special cases of factorising these expressions:

  • $\left(A+B\right)^2=A^2+2AB+B^2$(A+B)2=A2+2AB+B2 (called a perfect square)
  • $\left(A+B\right)\left(A-B\right)=A^2-B^2$(A+B)(AB)=A2B2 (called a difference of two squares)

Practice questions

Question 1

Fully factorise: $5\left(a+b\right)+v\left(a+b\right)$5(a+b)+v(a+b)

Question 2

Factorise the following expression by grouping in pairs:

$8x+xz-16y-2yz$8x+xz16y2yz

Question 3

Factorise: $n^2-25$n225

Outcomes

0607E2.8C

Factorisation of four term expressions.

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