Solve x^4+2x^2+1div(x-4)

$${ x }^{ 4 } +2 { x }^{ 2 } +1 \div (x-4)$$

Evaluate

$$\frac{x^{5}-4x^{4}+2x^{3}-8x^{2}+1}{x-4}$$

Short Solution Steps

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

$$\frac{\left(x^{4}+2x^{2}\right)\left(x-4\right)}{x-4}+\frac{1}{x-4}$$

Since \frac{\left(x^{4}+2x^{2}\right)\left(x-4\right)}{x-4} and \frac{1}{x-4} have the same denominator, add them by adding their numerators.

$$\frac{\left(x^{4}+2x^{2}\right)\left(x-4\right)+1}{x-4}$$

Do the multiplications in \left(x^{4}+2x^{2}\right)\left(x-4\right)+1.

$$\frac{x^{5}-4x^{4}+2x^{3}-8x^{2}+1}{x-4}$$

Differentiate w.r.t. x

$$\frac{4x^{5}-32x^{4}+68x^{3}-32x^{2}+64x-1}{\left(x-4\right)^{2}}$$