How do you divide #( x^4-2x^3-2x^2+11x+12)/(2x-6)#?

1 Answer
Jan 7, 2016

Answer:

Long divide coefficients to find:

#(x^4-2x^3-2x^2+11x+12)/(2x-6) = 1/2x^3+1/2x^2+1/2x+7#

with remainder #27#

Explanation:

I like to long divide coefficients, but first note that #2x-6 = 2(x-3)#, so to simplify the calculation, let's divide by #x-3#, then divide the result (and any remainder) by #2# at the end:

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We find:

#(x^4-2x^3-2x^2+11x+12)/(x-3) = x^3+x^2+x+14#

with remainder #54#.

Hence (dividing by #2#):

#(x^4-2x^3-2x^2+11x+12)/(2x-6) = 1/2x^3+1/2x^2+1/2x+7#

with remainder #27#