Question

How do you divide `( -x^4 + 3x^3 + 9x^2 +4x)/(2x^2-3x)`?

Answer 1

Trigonometry long division gives quotient of `(-4x^2 +6x +9)/8` and remainder of `59/8`

Explanation:

This can be simplified a bit by first taking out common factors. After that you need to do the trigonometry version of long division.
`(x(-x^3 +3x^2 +9x +4))/(x(2x - 3)) = (-x^3 +3x^2 +9x +4)/(2x -3)`
Divide the first element by the first element of the divisor:
`-x^3/(2x) = **-x^2/2**`

Multiply this result by the divisor:
`-x^2/2 *(2x - 3) = -x^3+3x^2/2`

Then subtract this from the dividend:
`(-x^3 +3x^2 +9x +4) - (-x^3+3x^2/2) = 3x^2/2 +9x +4`
Repeat this process with the remaining dividend:
`(3x^2/2) / (2x) = **3x/4**`

`(3x/4)*(2x-3) = (6x^2 -9x)/4`
`((3x^2)/2 +9x +4) - ((6x^2 - 9x)/4) = (9x)/4 +4`

And repeat again
`((9x)/4)/(2x) = **9/8**`

`(9/8)*(2x-3) = (9x)/4 -27/8`

`((9x)/4 +4) - ((9x)/4 -27/8) = **59/8**`

The quotient is then the sum of the factors in bold and the remainder is `59/8`
`-x^2/2 + 3x/4 + 9/8 = (-4x^2 +6x +9)/8`