Solve b^3+2b^2-b-2

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

Factor

$$\left(b-1\right)\left(b+1\right)\left(b+2\right)$$

Solution Steps

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

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

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

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

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

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

Rewrite the complete factored expression.

$$\left(b-1\right)\left(b+1\right)\left(b+2\right)$$

Evaluate

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