Solve 2x^3-3x^2-39x+20/x-1

$$2 { x }^{ 3 } -3 { x }^{ 2 } -39x+ \frac{ 20 }{ x-1 }$$

Evaluate

$$\frac{2x^{4}-5x^{3}-36x^{2}+39x+20}{x-1}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 2x^{3}-3x^{2}-39x times \frac{x-1}{x-1}.

$$\frac{\left(2x^{3}-3x^{2}-39x\right)\left(x-1\right)}{x-1}+\frac{20}{x-1}$$

Since \frac{\left(2x^{3}-3x^{2}-39x\right)\left(x-1\right)}{x-1} and \frac{20}{x-1} have the same denominator, add them by adding their numerators.

$$\frac{\left(2x^{3}-3x^{2}-39x\right)\left(x-1\right)+20}{x-1}$$

Do the multiplications in \left(2x^{3}-3x^{2}-39x\right)\left(x-1\right)+20.

$$\frac{2x^{4}-2x^{3}-3x^{3}+3x^{2}-39x^{2}+39x+20}{x-1}$$

Combine like terms in 2x^{4}-2x^{3}-3x^{3}+3x^{2}-39x^{2}+39x+20.

$$\frac{2x^{4}-5x^{3}-36x^{2}+39x+20}{x-1}$$

Differentiate w.r.t. x

$$\frac{6x^{4}-18x^{3}-21x^{2}+72x-59}{\left(x-1\right)^{2}}$$