What is #4-x^2+x^4# divided by #x^2+x+5#?

1 Answer
George C.
Jul 27, 2018

#(4-x^2+x^4)/(x^2+x+5) = x^2-x-5+(10x+29)/(x^2+x+5)#

Explanation:

#(4-x^2+x^4)/(x^2+x+5)#

#= (x^4-x^2+4)/(x^2+x+5)#

#= (x^4+x^3+5x^2-x^3-6x^2+4)/(x^2+x+5)#

#= (x^2(x^2+x+5)-x^3-6x^2+4)/(x^2+x+5)#

#= x^2+(-x^3-6x^2+4)/(x^2+x+5)#

#= x^2+(-x^3-x^2-5x-5x^2+5x+4)/(x^2+x+5)#

#= x^2+(-x(x^2+x+5)-5x^2+5x+4)/(x^2+x+5)#

#= x^2-x+(-5x^2+5x+4)/(x^2+x+5)#

#= x^2-x+(-5x^2-5x-25+10x+29)/(x^2+x+5)#

#= x^2-x+(-5(x^2+x+5)+10x+29)/(x^2+x+5)#

#= x^2-x-5+(10x+29)/(x^2+x+5)#

Alternatively, we can find the same result by long dividing tuples representing the coefficients of the powers of #x#.

In our example, we want to divide "#1 \ 0 \ -1 \ 0 \ 4#" by "#1 \ 1 \ 5#". Note well the inclusion of #0#'s in the dividend, to represent the missing powers of #x#.
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So #x^4-x^2+4# divided by #x^2+x+5# is #x^2-x-5# with remainder #10x+29#.

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