# Is it possible to factor y=2x^3-2x^2+2x-2 ? If so, what are the factors?

Apr 4, 2018

Yes, it is possible to factor this. Here is what I did:

#### Explanation:

$y = 2 {x}^{3} - 2 {x}^{2} + 2 x - 2$

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

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

The factors are $\left(2 {x}^{2} + 2\right)$ and $\left(x - 1\right)$.

Apr 4, 2018

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

#### Explanation:

$y = 2 {x}^{3} - 2 {x}^{2} + 2 x - 2$
Greatest common factor:
$y = 2 \cdot \left({x}^{3} - {x}^{2} + x - 1\right)$
Factor by grouping:
$y = 2 \cdot \left(\left({x}^{3} - {x}^{2}\right) + \left(x - 1\right)\right)$
$y = 2 \cdot \left({x}^{2} \cdot \left(x - 1\right) + 1 \cdot \left(x - 1\right)\right)$
$y = 2 \cdot \left({x}^{2} + 1\right) \cdot \left(x - 1\right)$
Thus, the factors are $2$, $\left({x}^{2} + 1\right)$, and $\left(x - 1\right)$
graph{2x^3-2x^2+2x-2 [-10, 10, -5, 5]}