Solve 4x^3+8x^2-x-2

$$4 x ^ { 3 } + 8 x ^ { 2 } - x - 2$$

Factor

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

Solution Steps

Do the grouping 4x^{3}+8x^{2}-x-2=\left(4x^{3}+8x^{2}\right)+\left(-x-2\right), and factor out 4x^{2} in the first and -1 in the second group.

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

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

$$\left(x+2\right)\left(4x^{2}-1\right)$$

Consider 4x^{2}-1. Rewrite 4x^{2}-1 as \left(2x\right)^{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).

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

Rewrite the complete factored expression.

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

Evaluate

$$\left(x+2\right)\left(4x^{2}-1\right)$$