Question

How do you divide `(9x^4-x^2-6x-2x)/(3x-1)`?

Answer 1

`( 9x^4 - x^2 - 6x - 2x )/(3x-1) = 3x^3 + 2/3 x - (22/3 x)/(3x-1)`

Explanation:

We ask how many times the leading term of the denominator goes into each part of the numerator and then factor that out. The leading term of the bottom is `3x`, so we start with `9x^4`
`(9x^4)/(3x) = 3x^3`
so
`3x^3(3x-1) + 3x^3 = 9x^4`

We can repeat this and get the following:
`( 9x^4 - x^2 - 6x - 2x )/(3x-1) = (3x^3(3x-1) + 2x^2 - 6x - 2x)/(3x-1)`

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

`= 3x^3 + 2/3 x - (22/3 x)/(3x-1)`