How do you divide (-x^4+12x^3-5x^2-6x-2)/(x^2-4) −x4+12x3−5x2−6x−2x2−4?

1 Answer
George C.
May 1, 2016

(-x^4+12x^3-5x^2-6x-2)/(x^2-4) = -x^2+12x-9−x4+12x3−5x2−6x−2x2−4=−x2+12x−9 with remainder 42x-3842x−38

Explanation:

I like to just long divide the coefficients, not forgetting to include 00's for any missing powers of xx. In this example that means the missing xx 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−x4+12x3−5x2−6x−2x2−4=−x2+12x−9 with remainder 42x-3842x−38

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