17.02 Factorising binomial products (Extended)

Factorise $xy+8x-3y-24$xy+8x−3y−24

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 $-3y-24$−3y−24 separately. Then we can factorise the whole expression.

Do:

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 $xy-3y+8x-24$xy−3y+8x−24 we would get the same answer.

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

Factorise $x^2-6x+9$x2−6x+9

Think: Notice that $-6x=2\times\left(-3x\right)$−6x=2×(−3x). Therefore, $x^2-6x+9$x2−6x+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,

$x^2-6x+9=\left(x-3\right)^2$x2−6x+9=(x−3)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.

Factorise $x^2-9$x2−9

Think: Notice that $9=3^2$9=32. Therefore, $x^2-9$x2−9 is the difference of two squares.

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

$x^2-9=\left(x+3\right)\left(x-3\right)$x2−9=(x+3)(x−3)

Reflect: Once again, we can check the answer by expanding it. It's always worth checking if an expression fits the pattern $A^2-B^2$A2−B2, 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.