# How do you multiply (x^4-x^3+x^2-x)/(2x^3+2x^2+x+1)div (x^3-4x^2+x-4)/(2x^3-8x^2+x-4)?

Sep 5, 2016

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

#### Explanation:

Factorise each expression first. Each has 4 terms - use grouping.

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

Group into pairs.
$\frac{\left({x}^{4} - {x}^{3}\right) + \left({x}^{2} - x\right)}{\left(2 {x}^{3} + 2 {x}^{2}\right) + \left(x + 1\right)} \div \frac{\left({x}^{3} - 4 {x}^{2}\right) + \left(x - 4\right)}{\left(2 {x}^{3} - 8 {x}^{2}\right) + \left(x - 4\right)}$

Take out the common factors
(x^3(x-1)+x(x-1))/(2x^2(x+1)+(x+1))div color(red)((x^2(x-4)+(x-4)))/(color(blue)((2x^2(x-4)+(x-4))

change $\div$ to $\times$ and invert the fraction.
Take out the common brackets and factors

$\frac{x \left(x - 1\right) \left({x}^{2} + 1\right)}{\left(x + 1\right) \left(2 {x}^{2} + 1\right)} \times \frac{\textcolor{b l u e}{\left(x - 4\right) \left(2 {x}^{2} + 1\right)}}{\textcolor{red}{\left(x - 4\right) \left({x}^{2} + 1\right)}}$l

Cancel like factors

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

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

Sep 5, 2016

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

$= \frac{x \left(x - 1\right)}{x + 1}$

$= x - 2 + \frac{2}{x + 1}$

with exclusion: $x \ne 4$

#### Explanation:

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

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

$= \frac{x \left(\left({x}^{3} - {x}^{2}\right) + \left(x - 1\right)\right)}{\left(2 {x}^{3} + 2 {x}^{2}\right) + \left(x + 1\right)} \times \frac{\left(2 {x}^{3} - 8 {x}^{2}\right) + \left(x - 4\right)}{\left({x}^{3} - 4 {x}^{2}\right) + \left(x - 4\right)}$

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

=(xcolor(red)(cancel(color(black)((x^2+1))))(x-1))/(color(orange)(cancel(color(black)((2x^2+1))))(x+1)) xx (color(orange)(cancel(color(black)((2x^2+1))))color(purple)(cancel(color(black)((x-4)))))/(color(red)(cancel(color(black)((x^2+1))))color(purple)(cancel(color(black)((x-4)))

$= \textcolor{b l u e}{\frac{x \left(x - 1\right)}{x + 1}}$

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

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

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

$= \textcolor{b l u e}{x - 2 + \frac{2}{x + 1}}$

with exclusion: $x \ne 4$

This value needs to be excluded because when $x = 4$ the left hand side boils down to $\frac{0}{0}$ which is undefined, while the right hand side is $\frac{12}{5}$ which is defined.

If $x$ is allowed to take Complex values, then there are also exclusions:

$x \ne i \text{ }$ and $\text{ } x \ne \frac{\sqrt{2}}{2} i$

These values would result in ${x}^{2} + 1 = 0$ or $2 {x}^{2} + 1 = 0$ respectively, making the left hand side expression $\frac{0}{0}$ undefined.