Solve 5t^3-45t+3t^2-27

$$5 t ^ { 3 } - 45 t + 3 t ^ { 2 } - 27$$

Factor

$$\left(t-3\right)\left(t+3\right)\left(5t+3\right)$$

Solution Steps

Do the grouping 5t^{3}-45t+3t^{2}-27=\left(5t^{3}-45t\right)+\left(3t^{2}-27\right), and factor out 5t in the first and 3 in the second group.

$$5t\left(t^{2}-9\right)+3\left(t^{2}-9\right)$$

Factor out common term t^{2}-9 by using distributive property.

$$\left(t^{2}-9\right)\left(5t+3\right)$$

Consider t^{2}-9. Rewrite t^{2}-9 as t^{2}-3^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

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

Rewrite the complete factored expression.

$$\left(t-3\right)\left(t+3\right)\left(5t+3\right)$$

Evaluate

$$\left(5t+3\right)\left(t^{2}-9\right)$$