Solve 40t^5+5t^2=

$$40 t ^ { 5 } + 5 t ^ { 2 } =$$

Factor

$$5\left(2t+1\right)t^{2}\left(4t^{2}-2t+1\right)$$

Solution Steps

Factor out 5.

$$5\left(8t^{5}+t^{2}\right)$$

Consider 8t^{5}+t^{2}. Factor out t^{2}.

$$t^{2}\left(8t^{3}+1\right)$$

Consider 8t^{3}+1. Rewrite 8t^{3}+1 as \left(2t\right)^{3}+1^{3}. The sum of cubes can be factored using the rule: a^{3}+b^{3}=\left(a+b\right)\left(a^{2}-ab+b^{2}\right).

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

Rewrite the complete factored expression. Polynomial 4t^{2}-2t+1 is not factored since it does not have any rational roots.

$$5t^{2}\left(2t+1\right)\left(4t^{2}-2t+1\right)$$

Evaluate

$$40t^{5}+5t^{2}$$