Solve 2x^2+8/x+40x

$$2 x ^ { 2 } + \frac { 8 } { x } + 40 x$$

Evaluate

$$2x^{2}+40x+\frac{8}{x}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 2x^{2}+40x times \frac{x}{x}.

$$\frac{\left(2x^{2}+40x\right)x}{x}+\frac{8}{x}$$

Since \frac{\left(2x^{2}+40x\right)x}{x} and \frac{8}{x} have the same denominator, add them by adding their numerators.

$$\frac{\left(2x^{2}+40x\right)x+8}{x}$$

Do the multiplications in \left(2x^{2}+40x\right)x+8.

$$\frac{2x^{3}+40x^{2}+8}{x}$$

Factor

$$\frac{2\left(x^{3}+20x^{2}+4\right)}{x}$$