How do you divide #(x^3+7x^2-4x-1)/(3x-1) #?

Question

1373 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-05T14:34:22

#y = 1/3 x^2 + 22/9 x - 14/27 -41/27 frac{1}{3x - 1}#

Divide #x^3# by #3x#, the quotient is #1/3 x^2#

#y = 1/3 x^2 + frac{1/3 x^2 + 7x^2 - 4x - 1}{3x - 1}#

#y = 1/3 x^2 + frac{22/3 x^2 - 4x - 1}{3x - 1}#

Divide #22/3 x^2# by #3x#, the quotient is #22/9 x#

#y = 1/3 x^2 + 22/9 x + frac{22/9 x - 4x - 1}{3x - 1}#

#y = 1/3 x^2 + 22/9 x + frac{-14/9 x - 1}{3x - 1}#

Divide #14/9 x# by #3x#, the quotient is #-14/27#

#y = 1/3 x^2 + 22/9 x - 14/27 + frac{-14/27 - 1}{3x - 1}#

Degree above < degree below.