How do you divide #(9x^4-x^2-6x-2x)/(3x-1)#?

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1 Answer
Jake M.
Aug 1, 2018

#( 9x^4 - x^2 - 6x - 2x )/(3x-1) = 3x^3 + 2/3 x - (22/3 x)/(3x-1)#

Explanation:

We ask how many times the leading term of the denominator goes into each part of the numerator and then factor that out. The leading term of the bottom is #3x#, so we start with #9x^4#
#(9x^4)/(3x) = 3x^3 #
so
#3x^3(3x-1) + 3x^3 = 9x^4 #

We can repeat this and get the following:
#( 9x^4 - x^2 - 6x - 2x )/(3x-1) = (3x^3(3x-1) + 2x^2 - 6x - 2x)/(3x-1)#

#=3x^3 + ( 2/3 x(3x - 1) + 2/3x - 6x - 2x )/(3x-1)#

#= 3x^3 + 2/3 x - (22/3 x)/(3x-1)#

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