Solve n^3+2n^2-36n-72

$$n ^ { 3 } + 2 n ^ { 2 } - 36 n - 72$$

Factor

$$\left(n-6\right)\left(n+2\right)\left(n+6\right)$$

Solution Steps

Do the grouping n^{3}+2n^{2}-36n-72=\left(n^{3}+2n^{2}\right)+\left(-36n-72\right), and factor out n^{2} in the first and -36 in the second group.

$$n^{2}\left(n+2\right)-36\left(n+2\right)$$

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

$$\left(n+2\right)\left(n^{2}-36\right)$$

Consider n^{2}-36. Rewrite n^{2}-36 as n^{2}-6^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(n-6\right)\left(n+6\right)$$

Rewrite the complete factored expression.

$$\left(n-6\right)\left(n+2\right)\left(n+6\right)$$

Evaluate

$$\left(n+2\right)\left(n^{2}-36\right)$$