Question

How do you divide `(5x^3-7x^2-4x+1)/(3x-1)`?

Answer 1

Use long division.

`5/3x^2 - 16/9x - 52/27 -25/(27(3x-1))`

Explanation:

`(3x-1)` will go into `(5x^3-7x^2-4x+1)` a total of `5/3x^2` times.

`5/3x^2 (3x - 1) = (5x^3 - 5/3x^2)`

This will leave a remainder of:

`" "" "" "" ""5/3x^2`
`" "" "" "" ""----------------------------"`
`3x-1" "| 5x^3-7x^2-4x+1`
`" "" "" "-(5x^3 - 5/3x^2)`
`" "" "" "" ""--------------"`
`" "" "" "" "" "" "-16/3x^2`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

So now we're left with `-16/3x^2-4x+1`.

`(3x-1)` will go into this `-16/9x` times.

`-16/9x(3x - 1) = (-16/3x^2 + 16/9x)`

This will leave a remainder of:

`" "" "" "" ""5/3x^2 - 16/9x`
`" "" "" "" ""----------------------------"`
`3x-1" "| 5x^3-7x^2-4x+1`
`" "" "" "-(5x^3 - 5/3x^2)`
`" "" "" "" ""-----------------"`
`" "" "" "" "" "" "-16/3x^2 -4x+1`
`" "" "" "" "-(-16/3x^2+16/9x)`
`" "" "" "" "" ""-------------------------"`
`" "" "" "" "" "" "" "" "" "-52/9x`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

So now we're left with `-52/9x + 1`.

`(3x-1)` will go into this `-52/27` times.

`-52/27(3x-1) = (-52/9x + 52/27)`

This will leave a remainder of:

`" "" "" "" ""5/3x^2 - 16/9x - 52/27`
`" "" "" "" ""----------------------------"`
`3x-1" "| 5x^3-7x^2-4x+1`
`" "" "" "-(5x^3 - 5/3x^2)`
`" "" "" "" ""-----------------"`
`" "" "" "" "" "" "-16/3x^2 -4x+1`
`" "" "" "" "-(-16/3x^2+16/9x)`
`" "" "" "" "" ""-------------------------"`
`" "" "" "" "" "" "" "" "" "-52/9x + 1`
`" "" "" "" "" "" "" "-(-52/9x + 52/27)`
`" "" "" "" "" "" "" ""-------------------------"`
`color(white)"MMMMMMMMMMMMMM-"-25/27`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Since the degree of this remainder is smaller than the degree of our divisor, we will just divide it by the divisor as the last term in our answer:

`5/3x^2 - 16/9x - 52/27 -(25/27)/(3x-1)`

`5/3x^2 - 16/9x - 52/27 -25/(27(3x-1))`

Final Answer