Solve 4t^3+4t^2-t-1

$$4 t ^ { 3 } + 4 t ^ { 2 } - t - 1$$

Factor

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

Solution Steps

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

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

Factor out common term t+1 by using distributive property.

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

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

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

Rewrite the complete factored expression.

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

Evaluate

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