Solving Quadratic Equations: Factorising

Where $a$, $b$, and $c$ are constant coefficients, and $a$ cannot be zero. An algebraic expression in this complete form, containing exactly three terms, is known as a quadratic trinomial. If you are given an equation with terms on both sides, such as $x^2 + 5x = 14$, you must use inverse operations to move all terms to one side. It is highly recommended to keep the $a$ value (the coefficient of $x^2$) positive, as this makes the factorisation process much easier.

The Null Product Property

Why is a zero so important in these equations? It all comes down to a mathematical rule called the Null Product Property. This rule states that if the product of two or more factors is zero, then at least one of those factors must be exactly zero.

Once a quadratic is factorised into linear factors, such as $(x - p)(x - q) = 0$, the property allows you to split the problem into two separate linear equations. You simply set $x - p = 0$ and $x - q = 0$ and solve them independently to find your two roots.

Solving Monic Quadratics ($a = 1$)

A monic quadratic is a quadratic expression where the leading coefficient $a$ is exactly 1. To solve these by factorising, you need to find two numbers that multiply to give the constant term $c$, and add to give the middle coefficient $b$.

Solve $x^2 + 2x = 15$ by factorising.

Step 1: Rearrange the equation to equal zero. $x^2 + 2x - 15 = 0$

Step 2: Find factors of $-15$ that add to $+2$. The factor pairs for $-15$ are $(1, -15)$, $(-1, 15)$, $(3, -5)$, and $(-3, 5)$. The correct pair is $-3$ and $+5$ because $-3 + 5 = +2$.

Step 3: Factorise into two linear brackets. $(x - 3)(x + 5) = 0$

Step 4: Apply the null product property to set each bracket to zero and solve. $x - 3 = 0 \Rightarrow x = 3$ $x + 5 = 0 \Rightarrow x = -5$

Solving Non-Monic Quadratics ($a > 1$)

A non-monic quadratic has a leading coefficient greater than 1. These require the "ac method", where you multiply $a$ and $c$ together first. You then use these factors for splitting the middle term, which allows you to finish by factorising by grouping.

Solve $2x^2 + 11x + 12 = 0$ by factorising.

Step 1: Identify $a, b, c$ and calculate $ac$. $a = 2$, $b = 11$, $c = 12$ $ac = 2 \times 12 = 24$

Step 2: Find factors of $24$ that add to the $b$ value ($11$). The factors of $24$ are $(1, 24)$, $(2, 12)$, $(3, 8)$, and $(4, 6)$. The pair that adds to $11$ is $+3$ and $+8$.

Step 3: Split the middle term ($11x$) using these numbers. $2x^2 + 8x + 3x + 12 = 0$

Step 4: Factorise the first two terms and the last two terms separately (grouping). $2x(x + 4) + 3(x + 4) = 0$ $(2x + 3)(x + 4) = 0$

Step 5: Apply the null product property to find the roots. $2x + 3 = 0 \Rightarrow 2x = -3 \Rightarrow x = -1.5$ $x + 4 = 0 \Rightarrow x = -4$

Special Cases: Missing Terms

If the constant $c$ is zero, your equation will look like $x^2 = 5x$. You must never divide both sides by $x$. Doing this is a critical error that deletes one of your valid answers.

Instead, rearrange it to $x^2 - 5x = 0$ and factorise out the common $x$ to get $x(x - 5) = 0$. Using the null product property, the roots are $x = 0$ or $x = 5$. Similarly, for a difference of two squares like $x^2 = 9$, factorising to $(x - 3)(x + 3) = 0$ guarantees you find both the positive and negative roots ($x = 3$ and $x = -3$).