How do you divide (9x^4-x^2-6x-2x)/(3x-1)9x4−x2−6x−2x3x−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)9x4−x2−6x−2x3x−1=3x3+23x−223x3x−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 3x3x, so we start with 9x^49x4
(9x^4)/(3x) = 3x^3 9x43x=3x3
so
3x^3(3x-1) + 3x^3 = 9x^4 3x3(3x−1)+3x3=9x4

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)9x4−x2−6x−2x3x−1=3x3(3x−1)+2x2−6x−2x3x−1

=3x^3 + ( 2/3 x(3x - 1) + 2/3x - 6x - 2x )/(3x-1)=3x3+23x(3x−1)+23x−6x−2x3x−1

= 3x^3 + 2/3 x - (22/3 x)/(3x-1)=3x3+23x−223x3x−1

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