# 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.*

## 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 that 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)

## 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 them 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**