How do you simplify (y^2 + 3y - 4)/( 7y^3 + 7y^2 - 14y )y2+3y−47y3+7y2−14y?

1 Answer
Konstantinos Michailidis
Oct 2, 2015

See explanation

Explanation:

It is

(y^2 + 3y - 4)/(7y^3 + 7y^2 - 14y)=(y^2-1+3y-3)/(7y^3-7y+7y^2-7y)= ((y-1)*(y+1+3))/(7y(y^2-1)+7y(y-1))=((y-1)(y+4))/(7y(y-1)(y+1+1))= (y+4)/(7y(y+2))y2+3y−47y3+7y2−14y=y2−1+3y−37y3−7y+7y2−7y=(y−1)⋅(y+1+3)7y(y2−1)+7y(y−1)=(y−1)(y+4)7y(y−1)(y+1+1)=y+47y(y+2)

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