Solve 3t^3-7t^2-3t+7

$$3 t ^ { 3 } - 7 t ^ { 2 } - 3 t + 7$$

Factor

$$\left(t-1\right)\left(3t-7\right)\left(t+1\right)$$

Solution Steps

Do the grouping 3t^{3}-7t^{2}-3t+7=\left(3t^{3}-7t^{2}\right)+\left(-3t+7\right), and factor out t^{2} in the first and -1 in the second group.

$$t^{2}\left(3t-7\right)-\left(3t-7\right)$$

Factor out common term 3t-7 by using distributive property.

$$\left(3t-7\right)\left(t^{2}-1\right)$$

Consider t^{2}-1. Rewrite t^{2}-1 as t^{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).

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

Rewrite the complete factored expression.

$$\left(3t-7\right)\left(t-1\right)\left(t+1\right)$$

Evaluate

$$\left(3t-7\right)\left(t^{2}-1\right)$$