Question

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

Answer 1

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

Explanation:

One method involves separating out multiples of the denominator from the numerator, starting with the highest degree term. This is equivalent to polynomial long division.

We find:

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

`=((x^4+3x^3-x^2)-3x^3-8x^2-2)/(x^2+3x-1)`

`=(x^2(x^2+3x-1)-(3x^3+8x^2+2))/(x^2+3x-1)`

`=x^2-(3x^3+8x^2+2)/(x^2+3x-1)`

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

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

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

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

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