Method 1
Following the "hence" (using what was done before), let $a = x^{2} + 4$ and $b = x^{2} - 2$

$a^{2} - b^{2}$

[1]

Use the "difference of two squares" to factorise

$(a + b) (a - b)$

Replace $a$ with $x^{2} + 4$ in brackets and $b$ with $x^{2} - 2$ in brackets

$((x^{2} + 4) + (x^{2} - 2)) ((x^{2} + 4) - (x^{2} - 2))$

Expand inside the brackets

$(x^{2} + 4 + x^{2} - 2) (x^{2} + 4 - x^{2} + 2)$

[1]

Collect "like" terms

$(2 x^{2} + 2) (6)$

Multiply out the brackets (it may help to write the 6 at the front)

$12 x^{2} + 12$ [1]

Method 2
Following the "or otherwise", write out the squared brackets as the product of two brackets

$(x^{2} + 4) (x^{2} + 4) - (x^{2} - 2) (x^{2} - 2)$

Multiply out the two sets of brackets and collect like terms (keeping the results in their own separate brackets)

$(x^{4} + 4 x^{2} + 4 x^{2} + 16) - (x^{4} - 2 x^{2} - 2 x^{2} + 4) = (x^{4} + 8 x^{2} + 16) - (x^{4} - 4 x^{2} + 4)$

[1]

Remove the brackets (by multiplying all terms by -1 in the second bracket)

$x^{4} + 8 x^{2} + 16 - x^{4} + 4 x^{2} - 4$

[1]

Collect like terms

$12 x^{2} + 12$ [1]

Factorising (AQA GCSE Maths)

Exam Questions