How do you factor 2x^2-21x-11?

Oct 30, 2016

$2 {x}^{2} - 21 x - 11 = \left(x - 11\right) \left(2 x + 1\right)$

Explanation:

We want to factorise $2 {x}^{2} - 21 x - 11$

We look for two numbers which add to give the coefficient of $x$; so we seek two numbers which add to give $- 21$.

However, as the coefficient of ${x}^{2}$ is not $1$ then instead of looking for two numbers which multiply to give −11 we must look for two numbers which multiply to give −22, (that is, the coefficient of ${x}^{2}$ multiplied by the constant term, 2 × −11.

a × b = -22
a + b = -21

By inspection (or trial ad error) we can find two numbers a=-22 and b=1

So we have,
$2 {x}^{2} - 21 x - 11 = 2 {x}^{2} - 22 x + x - 11$
$\therefore 2 {x}^{2} - 21 x - 11 = 2 x \left(x - 11\right) + x - 11$ (by factorising the first two terms)
$\therefore 2 {x}^{2} - 21 x - 11 = 2 x \left(x - 11\right) + \left(x - 11\right)$ (collecting common terms)
$\therefore 2 {x}^{2} - 21 x - 11 = \left(x - 11\right) \left(2 x + 1\right)$ (by factorising the last two terms)

We can check this by multiplying out:
$\left(x - 11\right) \left(2 x + 1\right) = 2 {x}^{2} + x - 22 x - 11 = 2 {x}^{2} - 21 x - 11$ QED