# Question #b3042

Jan 10, 2017

(x + 1)(2x^2 - 5x + 5)

#### Explanation:

$f \left(x\right) = 2 {x}^{3} - 3 {x}^{2} + 5$.
When x = - 1 --> f(-1) = 0, then one factor is (x + 1)
After division we get:
$f \left(x\right) = \left(x + 1\right) \left(2 {x}^{2} - 5 x + 5\right)$
The quadratic function $\left(2 {x}^{2} - 5 x + 5\right)$ can't be factored because its D < 0
$D = {b}^{2} - 4 a c = 25 - 40 = - 15 < 0$
Factored form:
$f \left(x\right) = \left(x + 1\right) \left(2 {x}^{2} - 5 x + 5\right)$