Factorising Quadratic Expressions for the ESAT

Factorising expressions of the form $x^2 + bx + c$

Many quadratic expressions of the form $x^2 + bx + c$ can be written as the product of two linear expressions with integer coefficients, formatted as $(x + p)(x + q)$. In this structure, the relationships between the coefficients are defined by $b = p + q$ and $c = pq$. This means that $p$ and $q$ are two numbers whose product equals the constant term $c$ and whose sum equals the linear coefficient $b$.

For example, to factorise $x^2 + 5x + 6$, we identify two numbers that multiply to 6 and add to 5. These numbers are 3 and 2, which gives us the factorisation $(x + 3)(x + 2)$.

Worked Example: $x^2 + 17x + 72$

To factorise $x^2 + 17x + 72$, we seek two numbers $p$ and $q$ such that $p + q = 17$ and $pq = 72$. By checking factor pairs of 72, we find that 8 and 9 satisfy both conditions because $8 + 9 = 17$ and $8 \times 9 = 72$. Therefore, $x^2 + 17x + 72 = (x + 8)(x + 9)$.

Worked Example: $x^2 - x - 72$

In the expression $x^2 - x - 72$, the linear coefficient is -1 and the constant term is -72. We need two numbers whose product is -72 and whose sum is -1. The numbers are -9 and +8. Thus, $x^2 - x - 72 = (x - 9)(x + 8)$.

The Difference of Two Squares: $x^2 - a^2$

Quadratic expressions consisting of one squared term subtracted from another are known as the difference of two squares. These always factorise according to the pattern $x^2 - a^2 = (x + a)(x - a)$.

Worked Example: $x^2 - 16$

There are two ways to approach this factorisation:

1. Treat it as a standard quadratic: $x^2 + 0x - 16$. We need two numbers with a product of -16 and a sum of 0. These are +4 and -4, leading to $(x + 4)(x - 4)$.
2. Use the difference of two squares identity: Since $16 = 4^2$, we immediately write $(x + 4)(x - 4)$.

Worked Example: $x^2 - 3$

Even if the constant is not a perfect square, we can still use this method by using square roots. Because $3 = (\sqrt{3})^2$, we can factorise the expression as $x^2 - 3 = (x + \sqrt{3})(x - \sqrt{3})$.

Factorising expressions of the form $ax^2 + bx + c$

When the coefficient of $x^2$ is not 1, the expression can be expressed as $(px + q)(rx + s)$. Here, the coefficients must satisfy $a = pr$, $c = qs$, and $b = ps + rq$.

Worked Example: $6x^2 + 23x + 20$

Method 1: Systematic Trial

If $6x^2 + 23x + 20 = (px + q)(rx + s)$, we expand the right side to get $prx^2 + (ps + qr)x + qs$. We must find $p, r$ such that $p \times r = 6$ and $q, s$ such that $q \times s = 20$, while ensuring $ps + qr = 23$. Because all terms are positive, we only need to consider positive factors.

Pairs for $pr=6$ include (6,1), (1,6), (2,3), and (3,2). Pairs for $qs=20$ include (1,20), (20,1), (2,10), (10,2), (4,5), and (5,4). By testing combinations, we find that when $p=2, r=3, q=5$, and $s=4$, the middle term is $2(4) + 5(3) = 8 + 15 = 23$. The correct factorisation is $(2x + 5)(3x + 4)$.

Method 2: Splitting the Middle Term

First, multiply the coefficient of $x^2$ by the constant term: $6 \times 20 = 120$. Now find two numbers that multiply to 120 and add to the linear coefficient, 23.

The numbers are +15 and +8. We split $23x$ into $15x + 8x$ and factorise by grouping:

$6x^2 + 15x + 8x + 20 = 3x(2x + 5) + 4(2x + 5)$

Since $(2x + 5)$ is a common factor, we get $(2x + 5)(3x + 4)$.

Worked Example: $14x^2 - x - 3$

Using the splitting method, we multiply $14 \times -3 = -42$. We need two numbers that multiply to -42 and add to -1.

These numbers are -7 and +6. Split the middle term and factorise:

$14x^2 + 6x - 7x - 3 = 2x(7x + 3) - 1(7x + 3)$

Note that when factorising the second pair, the negative sign outside the bracket is treated as -1. This results in $(7x + 3)(2x - 1)$.

The Difference of Two Squares: $a^2x^2 - b^2$

This pattern extends to cases where the $x^2$ term has a coefficient. The general rule is $a^2x^2 - b^2 = (ax + b)(ax - b)$.

Worked Example: $16x^2 - 49$

We identify the square roots of both terms: $\sqrt{16x^2} = 4x$ and $\sqrt{49} = 7$. Using the identity, we get $16x^2 - 49 = (4x + 7)(4x - 7)$.