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

A quadratic expression 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 product and sum 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 \((x + 2)(x + 3)\).

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

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.

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

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

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

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

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

To factorise this quadratic, find two numbers that have a product of +11 and a sum of -12. 11 is a prime 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 \(-11 \times -1 = +11\).

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

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

\(= y^2 - y - 11y + 11\) 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 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 (\(3 \times 3\)).

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 factor pairs 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 \(-2 + 2 = 0\).

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

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

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