Solve 2v^3-4v^2+4v-8

$$2 v ^ { 3 } - 4 v ^ { 2 } + 4 v - 8$$

Factor

$$2\left(v-2\right)\left(v^{2}+2\right)$$

Solution Steps

Factor out 2.

$$2\left(v^{3}-2v^{2}+2v-4\right)$$

Consider v^{3}-2v^{2}+2v-4. Do the grouping v^{3}-2v^{2}+2v-4=\left(v^{3}-2v^{2}\right)+\left(2v-4\right), and factor out v^{2} in the first and 2 in the second group.

$$v^{2}\left(v-2\right)+2\left(v-2\right)$$

Factor out common term v-2 by using distributive property.

$$\left(v-2\right)\left(v^{2}+2\right)$$

Rewrite the complete factored expression. Polynomial v^{2}+2 is not factored since it does not have any rational roots.

$$2\left(v-2\right)\left(v^{2}+2\right)$$

Evaluate

$$2\left(v-2\right)\left(v^{2}+2\right)$$