Solve x^2-4/1+4/x-4

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

Evaluate

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

Short Solution Steps

Anything divided by one gives itself.

$$x^{2}-4+\frac{4}{x-4}$$

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

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

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

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

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

$$\frac{x^{3}-4x^{2}-4x+16+4}{x-4}$$

Combine like terms in x^{3}-4x^{2}-4x+16+4.

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

Differentiate w.r.t. x

$$\frac{2\left(x^{3}-8x^{2}+16x-2\right)}{\left(x-4\right)^{2}}$$