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

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

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Answer 1 · 2016-07-28T15:02:06

$\Rightarrow x - 1 - \frac{2 x - 6}{x^{2} + 2}$ $=> x-1- (2x-6)/(x^2+2)$

Explanation:

Write as: $\frac{x^{3} - x^{2} + 0 x + 4}{x^{2} + 0 x + 2}$ $(x^3-x^2+0x+4)/(x^2+0x+2)$

I have added the place holder of $0 x$ $0x$ to make the calculation formatting easier. It has no value.

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$x^{3} - x^{2} + 0 x + 4$ $" "x^3-x^2+0x+4$
$color(magenta)(x)(x^2+0x+2) ->ul(x^3+0x^2+2x) larr" Subtract"$
$" "0-x^2-2x+4$
$color(magenta)(-1)(x^2+0x+2)->" " ul(-x^2-0x-2 ) larr" Subtract"$
$" "color(magenta)(0-2x+6 larr" Remainder")$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$(x^3-x^2+0x+4)/(x^2+0x+2)" " =" " x-1+ (-2x+6)/(x^2+2)$

$=> x-1- (2x-6)/(x^2+2)$

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

1 Answer

Explanation:

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