# How to Factorise a Quadratic Algebraic Equation

*I am a former maths teacher and owner of DoingMaths. I love writing about maths, its applications and fun mathematical facts.*

## A Quadratic Equation

## What is a Quadratic Equation?

A quadratic equation is any equation that can be written in the form ax^{2} + bx + c = 0 where a, b and c are numbers with a ≠ 0 (if a = 0 we have a linear equation). Another way of thinking about quadratic equations is that the highest power of x is the x squared.

If we remove the = 0 from the end, we no longer have an equation, but instead have an expression of the form ax^{2} + bx + c.

In this article we are going to look at how to convert a quadratic expression from ax^{2} + bx + c into the form (px + m)(qx + n) where m, n, p and q are whole numbers. This process is called factorising as both brackets are factors of the original expression.

## Factorising a Quadratic When a = 1

When factorising a quadratic of the form x^{2} + bx + c (i.e. the coefficient of x^{2} is 1), we have a simple method which can be explained best by looking at what happens when brackets are expanded.

Let's expand (x + 2)(x + 3). By multiplying each term in the left bracket by each term in the right bracket we get x^{2} + 3x + 2x + 6 which simplifies to x^{2} + 5x + 6. We can see quite easily that the 6 in our new expression came from multiplying the 2 and 3 from the brackets, while the 5 came from adding them. This will always be the case, so when factorising x^{2} + bx + c we need to find a pair of numbers that will sum to b and multiply to make c.

__Example 1__

Factorise x^{2} + 11x + 24

We need two numbers that will add to make 11 and multiply to make 24. These must be 3 and 8 so we get,

x^{2} + 11x + 24 = (x + 3)(x + 8)

__Example 2__

Factorise x^{2} + 4x − 12

Our two numbers must add to make 4 and multiply to make -12 (notice how we have a negative this time). These must be -2 and 6 so we get,

x^{2} + 4x − 12 = (x − 2)(x + 6)

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## Factorising a Quadratic When a Does Not Equal 1

If the coefficient of x^{2} ≠ 1 things get a little trickier. Our current method no longer works as our factorised form will now be (px + m)(qx + n) with coefficients of the x's. When we expand this, we get pqx^{2} + mqx + npx + mn = pqx^{2} + (mq + np)x + mn. Our numbers m and n will no longer add to make the coefficient of x due to them being multiplied by our new numbers p and q.

There is still a simple method for factorising these however. Take the quadratic 8x^{2} + 22x + 5 for example. If we multiply the coefficient of x^{2} and the last number together, we get 8 × 5 = 40. We now look for two numbers that multiply to make this and add together to make the coefficient of x, 22. We can soon work out that these two numbers are 2 and 20. We now split our coefficient of x into these two numbers and get:

8x^{2} + 22x + 5 = 8x^{2} + 2x + 20x + 5

We can now factorise the first pair of terms: 8x^{2} + 2x = 2x(4x + 1)

and the second pair of terms: 20x + 5 = 5(4x + 1)

We now have:

8x^{2} + 22x + 5 = 2x(4x + 1) + 5(4x +1) and by factorising again we get

(2x + 5)(4x + 1)

## Factorising Quadratics

__Example 1__

Factorise 3x^{2} + 13x + 14

3 × 14 = 42

Two numbers which multiply to make 42 and add to make 13 are 6 and 7.

3x^{2} + 13x + 14 = 3x^{2} + 6x + 7x + 14

=3x(x + 2) + 7(x + 2)

=(3x + 7)(x + 2)

__Example 2__

Factorise 20x^{2} − 23x + 6

20 × 6 = 120

Two numbers which multiply to make 120 and add to make -23 are -8 and -15

20x^{2} − 23x + 6 = 20x^{2} − 8x −15x + 6

= 4x(5x − 2) − 3(5x − 2)

= (4x − 3)(5x − 2)

**© 2021 David**