Multiply the coefficients of the first and last term together.

$6 \times - 20 = - 120$

Find two numbers that multiply to -120 and also add to +7 (the middle term).

$\begin{array}{rcl} 15 \times - 8 & = & - 120 \\ 15 + (- 8) & = & 7 \end{array}$

Split the middle term into two terms with these coefficients.

$\begin{array}{rcl} 6 x^{2} + 7 x - 20 & = & 6 x^{2} + 15 x - 8 x - 20 \end{array}$

Find the highest common factor (HCF) of the first two terms and the last two terms.

The HCF of $6 x^{2}$ and $15 x$ is $3 x$ .

$\begin{array}{rcl} 6 x^{2} + 15 x & = & 3 x (2 x + 5) \end{array}$

The HCF of $- 8 x$ and $- 20$ is $- 4$ .

$\begin{array}{rcl} - 8 x - 20 & = & - 4 (2 x + 5) \end{array}$

So the whole expression becomes

$\begin{array}{rcl} 6 x^{2} + 15 x - 8 x - 20 & = & 3 x (2 x + 5) - 4 (2 x + 5) \end{array}$

Both terms are multiplied by $\begin{array}{rcl} 2 x + 5 \end{array}$ so the coefficients of these terms can be brought together in a bracket.

$\begin{array}{rcl} 6 x^{2} + 7 x - 20 & = & (3 x - 4) (2 x + 5) \end{array}$
The product of the first terms in both brackets is 6x² or the product of the second terms in both brackets is -20 [1]
Correct expression [1]

Expanding & Factorising Brackets (Cambridge O Level Maths)

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