How do you divide (2x^5-3x^3+2x-12)/(-3x^2+3)2x53x3+2x123x2+3?

1 Answer
Jul 29, 2018

(2x^5-3x^3+2x-12)/(-3x^2+3)=-2/3x^3+1/3x+(x-12)/(-3x^2+3)2x53x3+2x123x2+3=23x3+13x+x123x2+3

Explanation:

(2x^5-3x^3+2x-12)/(-3x^2+3)2x53x3+2x123x2+3

=-1/3(2x^5-3x^3+2x-12)/(x^2-1)=132x53x3+2x12x21

=-1/3(2x^5-2x^3-x^3+x+x-12)/(x^2-1)=132x52x3x3+x+x12x21

=-1/3((2x^3-x)(x^2-1)+x-12)/(x^2-1)=13(2x3x)(x21)+x12x21

=-1/3(2x^3-x+(x-12)/(x^2-1))=13(2x3x+x12x21)

=-2/3x^3+1/3x+(x-12)/(-3x^2+3)=23x3+13x+x123x2+3