Solve x^3/3+3+6x^2+20x-5

$$\frac { x ^ { 3 } } { 3 } + 3 + 6 x ^ { 2 } + 20 x - 5$$

Evaluate

$$\frac{x^{3}}{3}+6x^{2}+20x-2$$

Short Solution Steps

Subtract 5 from 3 to get -2.

$$\frac{x^{3}}{3}-2+6x^{2}+20x$$

To add or subtract expressions, expand them to make their denominators the same. Multiply -2+6x^{2}+20x times \frac{3}{3}.

$$\frac{x^{3}}{3}+\frac{3\left(-2+6x^{2}+20x\right)}{3}$$

Since \frac{x^{3}}{3} and \frac{3\left(-2+6x^{2}+20x\right)}{3} have the same denominator, add them by adding their numerators.

$$\frac{x^{3}+3\left(-2+6x^{2}+20x\right)}{3}$$

Do the multiplications in x^{3}+3\left(-2+6x^{2}+20x\right).

$$\frac{x^{3}-6+18x^{2}+60x}{3}$$

Factor

$$\frac{x^{3}+18x^{2}+60x-6}{3}$$

Solution Steps

Factor out \frac{1}{3}. Polynomial x^{3}+18x^{2}+60x-6 is not factored since it does not have any rational roots.

$$\frac{x^{3}-6+18x^{2}+60x}{3}$$