Alternative Method for Factorising Quadratics

This method involves splitting the middle term into two parts which can then be paired with the numbers on either side and factorised, so that both brackets contain exactly the same terms.

This bracket then becomes one term; the other is found by collecting what is left and putting it into a bracket. This is really just a reversal of FOIL.

To find the correct split, the numbers must add together to make the middle number (coefficient of \(x\)) and multiply together to make the first number × last number (coefficient of \(x^2\) times the constant).

Example

Factorise:

\(x^2 + 4x + 3\)

The middle term is \(4x\). We need two numbers that:

add to give 4
multiply to give 3

Use 1 and 3, so split \(4x\) into \(3x + x\).

\(x^2 + 4x + 3\)
\(= x^2 + x + 3x + 3\)
\(= x(x + 1) + 3(x + 1)\)
\(= (x + 3)(x + 1)\)

Example

Factorise:

\(x^2 - 7x - 8\)

The middle term is \(-7x\). We need two numbers that:

add to give –7
multiply to give –8

Use –8 and 1, so split \(-7x\) into \(-8x + x\).

\(x^2 - 7x - 8\)
\(= x^2 + x - 8x - 8\)
\(= x(x + 1) - 8(x + 1)\)
\(= (x - 8)(x + 1)\)

Example

Factorise:

\(6x^2 + 23x + 10\)

We need two numbers that:

add to give 23
multiply to give \(6 \times 10 = 60\)

Use 3 and 20, so split \(23x\) into \(3x + 20x\).

\(6x^2 + 23x + 10\)
\(= 6x^2 + 3x + 20x + 10\)
\(= 3x(2x + 1) + 10(2x + 1)\)
\(= (3x + 10)(2x + 1)\)

Factorising Quadratics

Warning: Sometimes trial and error is involved.