How do you divide #(-x^4+12x^3-5x^2-6x-2)/(x^2-4) #?

1 Answer
George C.
May 1, 2016

#(-x^4+12x^3-5x^2-6x-2)/(x^2-4) = -x^2+12x-9# with remainder #42x-38#

Explanation:

I like to just long divide the coefficients, not forgetting to include #0#'s for any missing powers of #x#. In this example that means the missing #x# term in the divisor...

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We find:

#(-x^4+12x^3-5x^2-6x-2)/(x^2-4) = -x^2+12x-9# with remainder #42x-38#

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