# How do you factor the quadratic expression completely? 2x^2 - 13x + 20

Jun 10, 2018

$\left(x - 4\right) \left(2 x - 5\right)$

#### Explanation:

$2 {x}^{2} - 13 x + 20$

Factors of $20 \cdot 2$ which add up to -13 are -5 and -8

so you can replace -13 with -5 and -8 such that:

$2 {x}^{2} - 5 x - 8 x + 20$
which goes to:
$x \left(2 x - 5\right) - 4 \left(2 x - 5\right)$

Take the expressions not in the brackets together to get:
$\left(x - 4\right) \left(2 x - 5\right)$

Jun 10, 2018

$\left(x - 4\right) \left(x - \frac{5}{2}\right)$

#### Explanation:

$2 {x}^{2} - 13 x + 20$

The simplest way to do this is using the quadratic equation:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Where,
$a = 2$
$b = - 13$
$c = 20$

$x = \frac{- \left(- 13\right) \pm \sqrt{{\left(- 13\right)}^{2} - 4 \cdot 2 \cdot 20}}{2 \cdot 2}$

$x = \frac{\left(13\right) \pm \sqrt{169 - 160}}{4}$
$x = \frac{\left(13\right) \pm \sqrt{9}}{4}$
$x = 4$ and $x = \frac{5}{2}$

$x - 4 = 0$
$x - \frac{5}{2} = 0$

Therefore,

$2 {x}^{2} - 13 x + 20 = \left(x - 4\right) \left(x - \frac{5}{2}\right)$