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

1 Answer
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...

enter image source here

We find:

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