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How do you divide $\frac{x^{4} + 2 x^{2} + 2 x - 3}{2 x^{2} - x}$ $(x^4+2x^2+2x-3)/(2x^2-x)$ ?

1 Answer

Jan 21, 2016

Using polynomial long division (below)
$\frac{1}{2} x^{2} + \frac{1}{4} x + \frac{9}{8}$ $1/2x^2+1/4x+9/8$ with remainder $\frac{25 x + 24}{16 x^{2} - 8 x}$ $(25x+24)/(16x^2-8x)$

Explanation:

Remember to allow for the (zero) $x^{3}$ $x^3$ term:
enter image source here

Note that the remainder $\frac{25}{8} x + 3$ $25/8 x +3$ (over $2 x^{2} - x$ $2x^2-x$ )
can be simplified as
$XXX \frac{25 x + 24}{16 x^{2} - 8 x}$ $color(white)("XXX")(25x+24)/(16x^2-8x)$ to avoid fractions of fractions.

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