Solve [-y^3/3-2y^2-4y]_0^2

$$[ - \frac { y ^ { 3 } } { 3 } - 2 y ^ { 2 } - 4 y ] _ { 0 } ^ { 2 }$$

Evaluate

$$-\frac{y^{3}}{3}-2y^{2}-4y$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply -2y^{2}-4y times \frac{3}{3}.

$$-\frac{y^{3}}{3}+\frac{3\left(-2y^{2}-4y\right)}{3}$$

Since -\frac{y^{3}}{3} and \frac{3\left(-2y^{2}-4y\right)}{3} have the same denominator, add them by adding their numerators.

$$\frac{-y^{3}+3\left(-2y^{2}-4y\right)}{3}$$

Do the multiplications in -y^{3}+3\left(-2y^{2}-4y\right).

$$\frac{-y^{3}-6y^{2}-12y}{3}$$

Factor

$$\frac{y\left(-y^{2}-6y-12\right)}{3}$$

Solution Steps

Factor out \frac{1}{3}.

$$\frac{-y^{3}-6y^{2}-12y}{3}$$

Consider -y^{3}-6y^{2}-12y. Factor out y.

$$y\left(-y^{2}-6y-12\right)$$

Rewrite the complete factored expression. Polynomial -y^{2}-6y-12 is not factored since it does not have any rational roots.

$$\frac{y\left(-y^{2}-6y-12\right)}{3}$$