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

1 Answer
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.