How do you divide #(2x^5-3x^3+2x-12)/(-3x^2+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)#

Explanation:

#(2x^5-3x^3+2x-12)/(-3x^2+3)#

#=-1/3(2x^5-3x^3+2x-12)/(x^2-1)#

#=-1/3(2x^5-2x^3-x^3+x+x-12)/(x^2-1)#

#=-1/3((2x^3-x)(x^2-1)+x-12)/(x^2-1)#

#=-1/3(2x^3-x+(x-12)/(x^2-1))#

#=-2/3x^3+1/3x+(x-12)/(-3x^2+3)#

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