Factorising Quadratic Expressions for the TMUA

Factorising Monic Quadratics of the form $x^2 + bx + c$

A monic quadratic is one where the coefficient of $x^2$ is 1. Many such expressions can be written as a product of two linear factors with integer coefficients: $(x + p)(x + q)$. To find the values of $p$ and $q$, we look for two numbers that satisfy two conditions: their sum must be $b$ and their product must be $c$.

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

To factorise $x^2 + 17x + 72$, we need 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 work because $8 + 9 = 17$ and $8 \times 9 = 72$. Therefore, $x^2 + 17x + 72 = (x + 8)(x + 9)$.

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

For $x^2 - 1x - 72$, we need a product of $-72$ and a sum of $-1$. The numbers $-9$ and $+8$ satisfy this. Thus, $x^2 - x - 72 = (x - 9)(x + 8)$.

The Difference of Two Squares

Expressions in the form $x^2 - a^2$ are known as the difference of two squares. They always factorise into the pattern $(x + a)(x - a)$. This occurs because the middle $x$ term cancels out during expansion ($ax - ax = 0$).

Worked Example: $x^2 - 16$

Since $16 = 4^2$, we use the identity $x^2 - 4^2 = (x + 4)(x - 4)$. Alternatively, we can view this as $x^2 + 0x - 16$, seeking a product of $-16$ and a sum of 0, which leads to $+4$ and $-4$.

Worked Example: $x^2 - 3$

This identity can even be applied when the constant is not a perfect square integer. Since $3 = (\sqrt{3})^2$, we can write $x^2 - 3 = (x + \sqrt{3})(x - \sqrt{3})$.

Factorising Non Monic Quadratics of the form $ax^2 + bx + c$

When the coefficient of $x^2$ is not 1, the process is more involved. We seek to find $(px + q)(rx + s)$ such that $pr = a$, $qs = c$, and $ps + qr = b$.

Method 1: The Table Method

We can systematically test pairs of factors for $a$ and $c$. For $6x^2 + 23x + 20$, we identify $a=6, b=23, c=20$.

Factors for $a$ ($p, r$): (6, 1), (1, 6), (2, 3), (3, 2). Factors for $c$ ($q, s$): (1, 20), (20, 1), (2, 10), (10, 2), (4, 5), (5, 4).

By testing $ps + qr$, we find that the combination $p=2, r=3, q=5, s=4$ gives $(2 \times 4) + (5 \times 3) = 8 + 15 = 23$. The correct factorisation is $(2x + 5)(3x + 4)$.

Method 2: Splitting the Middle Term

Alternatively, we find two numbers that multiply to give $ac$ (the product of the $x^2$ coefficient and the constant) and add to give $b$.

For $6x^2 + 23x + 20$, we need a product of $6 \times 20 = 120$ and a sum of 23. These numbers are $+15$ and $+8$. We then split the $23x$ term: $6x^2 + 15x + 8x + 20$. Factorise the first two terms and the last two terms separately: $3x(2x + 5) + 4(2x + 5)$. Since $(2x + 5)$ is now a common factor, we have $(2x + 5)(3x + 4)$.

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

We need two numbers that multiply to $14 \times -3 = -42$ and add to $-1$. These are $-7$ and $+6$.

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

Difference of Two Squares with Coefficients

Quadratic expressions of the form $a^2x^2 - b^2$ also follow the difference of two squares identity, factorising as $(ax + b)(ax - b)$.

Worked Example: $16x^2 - 49$

Both terms are perfect squares: $16x^2 = (4x)^2$ and $49 = 7^2$. Therefore, $16x^2 - 49 = (4x + 7)(4x - 7)$.