How do you divide #(x^4 - 8x^3 + 4x^2 + 12x)/(x^2-2x+2)#?

1 Answer
George C.
May 1, 2016

#(x^4-8x^3+4x^2+12x)/(x^2-2x+2) = (x^2-6x-10)# with remainder #4x+20#.

Explanation:

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

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The process is similar to long division of numbers.

The result #1, -6, -10# means #x^2-6x-10# and remainder #4, 20# means #4x+20#

So:

#x^4-8x^3+4x^2+12x = (x^2-2x+2)(x^2-6x-10)+(4x+20)#

Or if you prefer:

#(x^4-8x^3+4x^2+12x)/(x^2-2x+2) = (x^2-6x-10)+(4x+20)/(x^2-2x+2)#

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