How do you divide (2x^5-3x^3+2x-12)/(-3x^2+3)2x5−3x3+2x−12−3x2+3?

1 Answer
George C.
Jul 29, 2018

(2x^5-3x^3+2x-12)/(-3x^2+3)=-2/3x^3+1/3x+(x-12)/(-3x^2+3)2x5−3x3+2x−12−3x2+3=−23x3+13x+x−12−3x2+3

Explanation:

(2x^5-3x^3+2x-12)/(-3x^2+3)2x5−3x3+2x−12−3x2+3

=-1/3(2x^5-3x^3+2x-12)/(x^2-1)=−132x5−3x3+2x−12x2−1

=-1/3(2x^5-2x^3-x^3+x+x-12)/(x^2-1)=−132x5−2x3−x3+x+x−12x2−1

=-1/3((2x^3-x)(x^2-1)+x-12)/(x^2-1)=−13(2x3−x)(x2−1)+x−12x2−1

=-1/3(2x^3-x+(x-12)/(x^2-1))=−13(2x3−x+x−12x2−1)

=-2/3x^3+1/3x+(x-12)/(-3x^2+3)=−23x3+13x+x−12−3x2+3

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