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

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

Factor

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

Solution Steps

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

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

Factor out common term n+2 by using distributive property.

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

Consider n^{2}-4. Rewrite n^{2}-4 as n^{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).

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

Rewrite the complete factored expression.

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

Evaluate

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