Solve -x^3+2*6x-8:(x+1)

$$- x ^ { 3 } + 2 \cdot 6 x - 8 : ( x + 1 )$$

Evaluate

$$\frac{-x^{4}-x^{3}+12x^{2}+12x-8}{x+1}$$

Short Solution Steps

Multiply 2 and 6 to get 12.

$$-x^{3}+12x-\frac{8}{x+1}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply -x^{3}+12x times \frac{x+1}{x+1}.

$$\frac{\left(-x^{3}+12x\right)\left(x+1\right)}{x+1}-\frac{8}{x+1}$$

Since \frac{\left(-x^{3}+12x\right)\left(x+1\right)}{x+1} and \frac{8}{x+1} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(-x^{3}+12x\right)\left(x+1\right)-8}{x+1}$$

Do the multiplications in \left(-x^{3}+12x\right)\left(x+1\right)-8.

$$\frac{-x^{4}-x^{3}+12x^{2}+12x-8}{x+1}$$

Differentiate w.r.t. x

$$\frac{20+24x+9x^{2}-6x^{3}-3x^{4}}{\left(x+1\right)^{2}}$$