Exercise solution 4, page 185 - AQA GCSE MATHS HIGHER

⭐ What you need to know:

A) To factorise an expression means to rewrite it as a product of simpler expressions. For quadratic expressions, this often means writing them in the form \((x+a)(x+b)\) or \((x-a)(x-b)\). Expanding and factorising are opposite processes.

B) When a quadratic has the form \(x^2+bx+c\), you look for two numbers whose product is \(c\) and whose sum is \(b\).

For example, to factorise \(x^2+9x+20\), we look for two numbers that multiply to \(20\) and add to \(9\). These numbers are \(4\) and \(5\), so

\[ x^2+9x+20=(x+4)(x+5) \]

You can check by expanding:

\[ \begin{aligned} (x+4)(x+5) &= x^2+5x+4x+20 \\ &= x^2+9x+20 \end{aligned} \]

C) If the constant term is negative, as in \(x^2+bx-c\), the two numbers will have different signs. Their product must be negative, and their sum must match the coefficient of \(x\).

For example, to factorise \(x^2+x-12\), we need two numbers whose product is \(-12\) and whose sum is \(1\). These numbers are \(4\) and \(-3\), so

\[ x^2+x-12=(x+4)(x-3) \]

D) Some quadratics are perfect squares. These come from expanding the square of a binomial:

\[ (x+a)^2=x^2+2ax+a^2 \]

and

\[ (x-a)^2=x^2-2ax+a^2 \]

For example, since

\[ (x-7)^2=x^2-14x+49 \]

we can reverse this and say

\[ x^2-14x+49=(x-7)^2 \]

A useful clue is that the last term is a square number, and the middle term matches \(2ax\).

E) Another important pattern is the difference of two squares:

\[ a^2-b^2=(a-b)(a+b) \]

This works because

\[ \begin{aligned} (a-b)(a+b) &= a^2+ab-ab-b^2 \\ &= a^2-b^2 \end{aligned} \]

For example,

\[ y^2-64=y^2-8^2=(y-8)(y+8) \]

This pattern only works when one square is being subtracted from another square.

F) It is always a good idea to check a factorisation by expanding the brackets again. If you return to the original expression, your factorisation is correct.

For example, if you think that

\[ x^2-11x+24=(x-3)(x-8) \]

then check:

\[ \begin{aligned} (x-3)(x-8) &= x^2-8x-3x+24 \\ &= x^2-11x+24 \end{aligned} \]

So the factorisation is correct.

G) Before factorising, it helps to recognise the type of expression you are dealing with:

• If it looks like \(x^2+bx+c\), search for two numbers with the correct sum and product.

• If it looks like \(x^2-2ax+a^2\) or \(x^2+2ax+a^2\), check whether it is a perfect square.

• If it looks like one square minus another square, use \(a^2-b^2=(a-b)(a+b)\).

Recognising the pattern is often the most important step.