Question

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)`?

Answer 1

`(x(x-1))/(x+1)`

Explanation:

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

`(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)`

Group into pairs.
`((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))`

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 `xx` and invert the fraction.
Take out the common brackets and factors

`(x(x-1)(x^2+1))/((x+1)(2x^2 +1)) xx (color(blue)((x-4)(2x^2+1)))/color(red)((x-4)(x^2+1))`l

Cancel like factors

`(x(x-1)cancel(x^2+1))/((x+1)cancel(2x^2 +1)) xx ((cancel(x-4)cancel(2x^2+1)))/(cancel(x-4)cancel(x^2+1))`l

`(x(x-1))/(x+1)`

Answer 2

`(x^4-x^3+x^2-x)/(2x^3+2x^2+x+1) -: (x^3-4x^2+x-4)/(2x^3-8x^2+x-4)`

`=(x(x-1))/(x+1)`

`=x-2+2/(x+1)`

with exclusion: `x != 4`

Explanation:

`(x^4-x^3+x^2-x)/(2x^3+2x^2+x+1) -: (x^3-4x^2+x-4)/(2x^3-8x^2+x-4)`

`=(x^4-x^3+x^2-x)/(2x^3+2x^2+x+1) xx (2x^3-8x^2+x-4)/(x^3-4x^2+x-4)`

`=(x((x^3-x^2)+(x-1)))/((2x^3+2x^2)+(x+1)) xx ((2x^3-8x^2)+(x-4))/((x^3-4x^2)+(x-4))`

`=(x(x^2(x-1)+1(x-1)))/(2x^2(x+1)+1(x+1)) xx (2x^2(x-4)+1(x-4))/(x^2(x-4)+1(x-4))`

`=(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)))`

`=color(blue)((x(x-1))/(x+1))`

`=(x^2+x-2x-2+2)/(x+1)`

`=(x(x+1)-2(x+1)+2)/(x+1)`

`=((x-2)(x+1)+2)/(x+1)`

`=color(blue)(x-2+2/(x+1))`

with exclusion: `x != 4`

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

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

`x != i" "` and `" "x != sqrt(2)/2i`

These values would result in `x^2+1 = 0` or `2x^2+1 = 0` respectively, making the left hand side expression `0/0` undefined.