Solve 2c^3+3c^2-2c-3

$$2 c ^ { 3 } + 3 c ^ { 2 } - 2 c - 3$$

Factor

$$\left(c-1\right)\left(c+1\right)\left(2c+3\right)$$

Solution Steps

Do the grouping 2c^{3}+3c^{2}-2c-3=\left(2c^{3}+3c^{2}\right)+\left(-2c-3\right), and factor out c^{2} in the first and -1 in the second group.

$$c^{2}\left(2c+3\right)-\left(2c+3\right)$$

Factor out common term 2c+3 by using distributive property.

$$\left(2c+3\right)\left(c^{2}-1\right)$$

Consider c^{2}-1. Rewrite c^{2}-1 as c^{2}-1^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(c-1\right)\left(c+1\right)$$

Rewrite the complete factored expression.

$$\left(c-1\right)\left(c+1\right)\left(2c+3\right)$$

Evaluate

$$\left(2c+3\right)\left(c^{2}-1\right)$$