Question

How do you divide `( -x^3 - 6x^2+3x+4 )/(2x^2 - x )`?

Answer 1

`-1/2x - 13/4 + (-x/4 + 4)/(2x^2 - x)`

Explanation:

Given: `(-x^3 - 6x^2 + 3x += 4)/(2x^2 - x)`

Using long division:

`2x^2 - x|bar(-x^3 -6x^2+3x+4)`

What times `2x^2 = -x^3`? `" "-1/2x`

`" "-1/2x`
`2x^2 - x|bar(-x^3 -6x^2+3x+4)`
`" "ul (-x^3 + 1/2x^2)`
`" "-13/2 x^2 + 3x`

What times `2x^2 = -13/2x^2`? `" "-13/4`

`" "-1/2x -13/4`
`2x^2 - x|bar(-x^3 -6x^2+3x+4" ")`
`" "ul (-x^3 + 1/2x^2)`
`" "-13/2 x^2 + 3x`
`" "ul(-13/2x^2 + 13/4x`
`" "-1/4x + 4` This is the remainder

`(-x^3 - 6x^2 + 3x += 4)/(2x^2 - x) = -1/2x - 13/4 + (-x/4 + 4)/(2x^2 - x)`

The remainder can be simplified if desired.