Solve 4c^3+8c^2-9c-18

$$4 c ^ { 3 } + 8 c ^ { 2 } - 9 c - 18$$

Factor

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

Solution Steps

Do the grouping 4c^{3}+8c^{2}-9c-18=\left(4c^{3}+8c^{2}\right)+\left(-9c-18\right), and factor out 4c^{2} in the first and -9 in the second group.

$$4c^{2}\left(c+2\right)-9\left(c+2\right)$$

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

$$\left(c+2\right)\left(4c^{2}-9\right)$$

Consider 4c^{2}-9. Rewrite 4c^{2}-9 as \left(2c\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).

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

Rewrite the complete factored expression.

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

Evaluate

$$\left(c+2\right)\left(4c^{2}-9\right)$$