How do you divide (x^4-9x^2-2) / (x^2+3x-1) x4−9x2−2x2+3x−1?
1 Answer
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)x4−9x2−2x2+3x−1
=((x^4+3x^3-x^2)-3x^3-8x^2-2)/(x^2+3x-1)=(x4+3x3−x2)−3x3−8x2−2x2+3x−1
=(x^2(x^2+3x-1)-(3x^3+8x^2+2))/(x^2+3x-1)=x2(x2+3x−1)−(3x3+8x2+2)x2+3x−1
=x^2-(3x^3+8x^2+2)/(x^2+3x-1)=x2−3x3+8x2+2x2+3x−1
=x^2-((3x^3+9x^2-3x)-x^2+3x+2)/(x^2+3x-1)=x2−(3x3+9x2−3x)−x2+3x+2x2+3x−1
=x^2-(3x(x^2+3x-1)-(x^2-3x-2))/(x^2+3x-1)=x2−3x(x2+3x−1)−(x2−3x−2)x2+3x−1
=x^2-3x+(x^2-3x-2)/(x^2+3x-1)=x2−3x+x2−3x−2x2+3x−1
=x^2-3x+((x^2+3x-1)-6x-1)/(x^2+3x-1)=x2−3x+(x2+3x−1)−6x−1x2+3x−1
=x^2-3x+1-(6x+1)/(x^2+3x-1)=x2−3x+1−6x+1x2+3x−1
