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

A quadratic expression can sometimes be factorised into two brackets in the form of \((x + a)(x + b)\). \(a\) 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 four ways of making +10 by multiplying two whole numbers. These are \(1 \times 10\), \(-1 \times -10\), \(2 \times 5\) 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.

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

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 \(11 \times 1\) or \(-11 \times -1\). \(11 + 1\) 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 difference of two squares means one squared term minus 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 \(x^2 + 0x -4\).

The products that make -4 are \(-1 \times 4, 1 \times -4\) or \(-2 \times 2\). Only −2 and 2 give a sum of 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 \(x^2 - 4\).