Solve n^3+2n^2-9n-18

$$n ^ { 3 } + 2 n ^ { 2 } - 9 n - 18$$

Factor

$$\left(n-3\right)\left(n+2\right)\left(n+3\right)$$

Solution Steps

Do the grouping n^{3}+2n^{2}-9n-18=\left(n^{3}+2n^{2}\right)+\left(-9n-18\right), and factor out n^{2} in the first and -9 in the second group.

$$n^{2}\left(n+2\right)-9\left(n+2\right)$$

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

$$\left(n+2\right)\left(n^{2}-9\right)$$

Consider n^{2}-9. Rewrite n^{2}-9 as n^{2}-3^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(n-3\right)\left(n+3\right)$$

Rewrite the complete factored expression.

$$\left(n-3\right)\left(n+2\right)\left(n+3\right)$$

Evaluate

$$\left(n+2\right)\left(n^{2}-9\right)$$