Factorising quadratics

In a quadratic expression using \(x\) as the variable, the highest power of \(x\) is \(x^2\).

A can sometimes be factorised into two brackets in the form of \((x + a)(x + b)\) where \(a\) and \(b\) can be any number, positive, negative or zero. \(a\) and \(b\) can be found by using a and method.

Expanding the brackets \((x + 2)(x + 3)\) gives \(x^2 + 3x + 2x + 6\)

Which simplifies to: \(x^2 + 5x + 6\)

Factorising is the reverse process of expanding brackets, so factorising \(x^2 + 5x + 6\) gives \((x + 2)(x + 3)\).

Example 1

Factorise \(x^2 + 7x + 10\).

To factorise this expression, find two numbers that have a product of +10 and a sum of +7.

There are two of ways of making +10 by multiplying two whole numbers. These are \(1 \times 10\) and \(2 \times 5\). Only the combination of 2 and 5 will also give a sum of +7, so the two numbers are 2 and 5.

\(x^2 + 7x + 10 = (x + 2)(x + 5)\)

To check this answer is correct, expand out the brackets.

\((x + 2)(x + 5) = x \times x + x \times 5 + 2 \times x + 2 \times 5\)

\(= x^2 + 5x + 2x + 10 = x^2 + 7x + 10\)

This is the original expression, so \((x + 2)(x + 5)\) is the correct factorisation.

Example 2

\(y^2 - 12y + 11\)

As there is a negative sign in the quadratic, it is helpful to remember the sign rules - ie two same signs make a positive and two different signs make a negative.

To factorise this quadratic, find two numbers that have a product of +11 and a sum of -12. 11 is a number, so the only way of multiplying two numbers to make 11 is \(11 \times 1\).

\(11 + 1\) will add up to +12, not -12, so the numbers must be -11 and -1 as \(-11 \times -1 = +11\).

\(= (y - 11)(y - 1)\)

To check this answer is correct, expand out the brackets:

\((y - 11)(y - 1) = (y \times y) + (y \times -1) + (-11 \times y) + (-11 \times -1)\)

\(= y^2 - y - 11y + 11\) which simplifies to \(y^2 - 12y + 11\).

This is the original expression, so \((y - 11)(y - 1)\) is the correct factorisation.

The difference of two squares

The of two numbers is found by subtracting the smaller number from the larger number. The difference of two squares means one squared term subtracted from another squared term. For example, \(x^2 - 9\) would be the difference of two squares as \(x^2\) is a squared term (\(x\) has been multiplied by itself) and 9 is a square number (\(3 \times 3\)).

The difference of two squares can be factorised into brackets using the method above for factorising quadratics.

Example

Factorise \(x^2 - 4\).

Quadratics can be factorised into the form \((x+a)(x+b)\).

\(x^2 - 4\) can be written as \(x^2 + 0x - 4\).

To factorise this quadratic, find two numbers that have a product of -4 and a sum of 0. The that make -4 are either \(-1 \times 4\), \(1 \times -4\), or \(-2 \times 2\). The factor pair that has a product of -4 but a sum of 0 is −2 and 2 because \(-2 \times 2 = -4\) and \(-2 + 2 = 0\).

Therefore, the two numbers to go into the brackets are -2 and 2.

This gives \((x - 2)(x + 2)\).

To check this answer, expand the brackets to see if the answer gives the original expression. If it does, then this is a correct factorisation.

\((x - 2)(x + 2) = (x \times x) + (x \times 2) + (-2 \times x) + (-2 \times 2)\)

\(= x^2 + 2x - 2x - 4\). Collecting the like terms gives the original expression of \(x^2 - 4\).

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Algebraic expressions - AQAFactorising quadratics