Solve 16n^3+80n^2-9n-45

$$16 n ^ { 3 } + 80 n ^ { 2 } - 9 n - 45$$

Factor

$$\left(4n-3\right)\left(n+5\right)\left(4n+3\right)$$

Solution Steps

Do the grouping 16n^{3}+80n^{2}-9n-45=\left(16n^{3}+80n^{2}\right)+\left(-9n-45\right), and factor out 16n^{2} in the first and -9 in the second group.

$$16n^{2}\left(n+5\right)-9\left(n+5\right)$$

Factor out common term n+5 by using distributive property.

$$\left(n+5\right)\left(16n^{2}-9\right)$$

Consider 16n^{2}-9. Rewrite 16n^{2}-9 as \left(4n\right)^{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).

$$\left(4n-3\right)\left(4n+3\right)$$

Rewrite the complete factored expression.

$$\left(4n-3\right)\left(4n+3\right)\left(n+5\right)$$

Evaluate

$$\left(n+5\right)\left(16n^{2}-9\right)$$