Solve by factorisation - OCR - GCSE Maths - Question 16 - 2018 - Paper 1

To factor the expression, we look for two numbers that multiply to give the product of the coefficient of $x^2$ (which is 3) and the constant term (which is -20).

This means we need two numbers that multiply to $3 \times -20 = -60$ and add to the coefficient of $x$ which is 11. The numbers are 15 and -4.

Now we can rewrite the middle term:

$3x^2 + 15x - 4x - 20 = 0$

Next, group the terms:

$(3x^2 + 15x) + (-4x - 20) = 0$

Factoring by grouping gives:

$3x(x + 5) - 4(x + 5) = 0$

This can be factored as:

$(3x - 4)(x + 5) = 0$

Now, we set each factor to zero:

1. From $3x - 4 = 0$, we get: $3x = 4 \Rightarrow x = \frac{4}{3}$

2. From $x + 5 = 0$, we get: $x = -5$

Thus, the solutions are:

$x = \frac{4}{3} \quad \text{or} \quad x = -5$