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

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

Factor

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

Solution Steps

Do the grouping v^{3}-2v^{2}-4v+8=\left(v^{3}-2v^{2}\right)+\left(-4v+8\right), and factor out v^{2} in the first and -4 in the second group.

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

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

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

Consider v^{2}-4. Rewrite v^{2}-4 as v^{2}-2^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(v-2\right)\left(v+2\right)$$

Rewrite the complete factored expression.

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

Evaluate

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