Solve frac{x-16x^{-1}}{5}*(2-2x^-1-2x^2/x^-1+4x^-2)

$$\frac { x - 16 x ^ { - 1 } } { 5 } \cdot ( 2 - \frac { 2 x ^ { - 1 } - 2 x ^ { 2 } } { x ^ { - 1 } + 4 x ^ { - 2 } } )$$

Evaluate

$$\frac{2\left(x-4\right)\left(x^{4}+4\right)}{5x}$$

Short Solution Steps

Use the distributive property to multiply \frac{x-16x^{-1}}{5} by 2-\frac{2x^{-1}-2x^{2}}{x^{-1}+4x^{-2}}.

$$2\times \frac{x-16x^{-1}}{5}+\frac{x-16x^{-1}}{5}\left(-\frac{2x^{-1}-2x^{2}}{x^{-1}+4x^{-2}}\right)$$

Express 2\times \frac{x-16x^{-1}}{5} as a single fraction.

$$\frac{2\left(x-16x^{-1}\right)}{5}+\frac{x-16x^{-1}}{5}\left(-\frac{2x^{-1}-2x^{2}}{x^{-1}+4x^{-2}}\right)$$

Multiply \frac{x-16x^{-1}}{5} times -\frac{2x^{-1}-2x^{2}}{x^{-1}+4x^{-2}} by multiplying numerator times numerator and denominator times denominator.

$$\frac{2\left(x-16x^{-1}\right)}{5}+\frac{-\left(x-16x^{-1}\right)\left(2x^{-1}-2x^{2}\right)}{5\left(x^{-1}+4x^{-2}\right)}$$

Factor 5\left(x^{-1}+4x^{-2}\right).

$$\frac{2\left(x-16x^{-1}\right)}{5}+\frac{-\left(x-16x^{-1}\right)\left(2x^{-1}-2x^{2}\right)}{5\times \left(\frac{1}{x}\right)^{2}\left(x+4\right)}$$

Use the distributive property to multiply 2 by x-16x^{-1}.

$$\frac{2x-32x^{-1}}{5}+\frac{-\left(x-16x^{-1}\right)\left(2x^{-1}-2x^{2}\right)}{5\left(x^{-1}+4x^{-2}\right)}$$

Factor the expressions that are not already factored in \frac{-\left(x-16x^{-1}\right)\left(2x^{-1}-2x^{2}\right)}{5\left(x^{-1}+4x^{-2}\right)}.

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

Cancel out \left(\frac{1}{x}\right)^{2}\left(x+4\right) in both numerator and denominator.

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

Since \frac{2x-32x^{-1}}{5} and \frac{-2\left(x-4\right)\left(x-1\right)\left(-x^{2}-x-1\right)}{5} have the same denominator, add them by adding their numerators.

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

Do the multiplications in 2x-32x^{-1}-2\left(x-4\right)\left(x-1\right)\left(-x^{2}-x-1\right).

$$\frac{2x-32\times \frac{1}{x}+2x^{4}-2x-8x^{3}+8}{5}$$

Combine like terms in 2x-32\times \frac{1}{x}+2x^{4}-2x-8x^{3}+8.

$$\frac{-32\times \frac{1}{x}+2x^{4}-8x^{3}+8}{5}$$

Expand

$$\frac{2x^{4}}{5}-\frac{8x^{3}}{5}+\frac{8}{5}-\frac{32}{5x}$$

Short Solution Steps

Use the distributive property to multiply \frac{x-16x^{-1}}{5} by 2-\frac{2x^{-1}-2x^{2}}{x^{-1}+4x^{-2}}.

$$2\times \frac{x-16x^{-1}}{5}+\frac{x-16x^{-1}}{5}\left(-\frac{2x^{-1}-2x^{2}}{x^{-1}+4x^{-2}}\right)$$

Express 2\times \frac{x-16x^{-1}}{5} as a single fraction.

$$\frac{2\left(x-16x^{-1}\right)}{5}+\frac{x-16x^{-1}}{5}\left(-\frac{2x^{-1}-2x^{2}}{x^{-1}+4x^{-2}}\right)$$

Multiply \frac{x-16x^{-1}}{5} times -\frac{2x^{-1}-2x^{2}}{x^{-1}+4x^{-2}} by multiplying numerator times numerator and denominator times denominator.

$$\frac{2\left(x-16x^{-1}\right)}{5}+\frac{-\left(x-16x^{-1}\right)\left(2x^{-1}-2x^{2}\right)}{5\left(x^{-1}+4x^{-2}\right)}$$

Factor 5\left(x^{-1}+4x^{-2}\right).

$$\frac{2\left(x-16x^{-1}\right)}{5}+\frac{-\left(x-16x^{-1}\right)\left(2x^{-1}-2x^{2}\right)}{5\times \left(\frac{1}{x}\right)^{2}\left(x+4\right)}$$

Use the distributive property to multiply 2 by x-16x^{-1}.

$$\frac{2x-32x^{-1}}{5}+\frac{-\left(x-16x^{-1}\right)\left(2x^{-1}-2x^{2}\right)}{5\left(x^{-1}+4x^{-2}\right)}$$

Factor the expressions that are not already factored in \frac{-\left(x-16x^{-1}\right)\left(2x^{-1}-2x^{2}\right)}{5\left(x^{-1}+4x^{-2}\right)}.

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

Cancel out \left(\frac{1}{x}\right)^{2}\left(x+4\right) in both numerator and denominator.

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

Since \frac{2x-32x^{-1}}{5} and \frac{-2\left(x-4\right)\left(x-1\right)\left(-x^{2}-x-1\right)}{5} have the same denominator, add them by adding their numerators.

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

Do the multiplications in 2x-32x^{-1}-2\left(x-4\right)\left(x-1\right)\left(-x^{2}-x-1\right).

$$\frac{2x-32\times \frac{1}{x}+2x^{4}-2x-8x^{3}+8}{5}$$

Combine like terms in 2x-32\times \frac{1}{x}+2x^{4}-2x-8x^{3}+8.

$$\frac{-32\times \frac{1}{x}+2x^{4}-8x^{3}+8}{5}$$