# How do you factor 4x^2-8x-12+6x?

Oct 30, 2016

$4 {x}^{2} - 8 x - 12 + 6 x = 4 {x}^{2} - 2 x - 12 = \left(4 x + 6\right) \left(x - 2\right)$,
[ or $2 \left(2 x + 3\right) \left(x - 2\right)$ ]

#### Explanation:

We want to factorise $4 {x}^{2} - 8 x - 12 + 6 x = 4 {x}^{2} - 2 x - 12$

We look for two numbers which add to give the coefficient of $x$; so we seek two numbers which add to give $- 2$.

However, as the coefficient of ${x}^{2}$ is not $1$ then instead of looking for two numbers which multiply to give the constant term $- 12$ we must look for two numbers which multiply to give $- 48$, (that is, the coefficient of ${x}^{2}$ multiplied by the constant term, ie 4 × −12.

a × b = -48
a + b = -2

By inspection (or trial and error) we can find two numbers a=-8 and b=6

So we have,
$4 {x}^{2} - 2 x - 12 = 4 {x}^{2} - 8 x + 6 x - 12$
$\therefore 4 {x}^{2} - 2 x - 12 = 4 x \left(x - 2\right) + 6 x - 12$ (by factorising the first two terms)
$\therefore 4 {x}^{2} - 2 x - 12 = 4 x \left(x - 2\right) + 6 \left(x - 2\right)$ (collecting common terms)
$\therefore 4 {x}^{2} - 2 x - 12 = \left(4 x + 6\right) \left(x - 2\right)$ (by factorising the last two terms)