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

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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