Question

How do you divide `(4x^3-5x^2-4x-12)/(3x-4)`?

Answer 1

`4/3 x^2 + 1/9x - 32/27 " Remainder: " (-196)/(27(3x-4))`

Explanation:

This is best solved using long division. The question is, what do I have to multiply by `3x-4` to get `4x^3-5x^2-4x-12`?

`" "4/3 x^2 + 1/9x - 32/27 " Remainder: " (-196)/(27(3x-4))`
`3x-4 |bar(4x^3-5x^2-4x-12)`
`" "-ul((4x^3-16/3 x^2)) downarrow " " downarrow`
`" "1/3x^2-4x`
`" "-ul((1/3x^2-4/9x))`
`" "-32/9x-12`
`" "-ul((-32/9x+128/27))`
`" "-196/27`

Answer 2

`color(magenta)(1.3x^2+0.1x-1.2` and remainder of `color(magenta)(-6`

Explanation:

`(4x^3-5x^2-4x-12)/(3x-4)`

`color(white)(................)color(magenta)(1.3x^2+0.1x-1.2`
`color(white)(a)3x-4` `|` `overline(4x^3-5x^2-4x-12)`
`color(white)(..............)ul(4x^3-5.3x^2)`
`color(white)(........................)0.3x^2-4x`
`color(white)(........................)ul(0.3x^2-0.4x)`
`color(white)(...............................)-3.6x-12`
`color(white)(................................)ul(-3.6x+4.8)`
`color(white)(...........................................)color(magenta)(-6`

`color(magenta)((4x^3-5x^2-4x-12)/(3x-4)=1.3x^2+0.1x-1.2`and remainder of `color(magenta)(-6`