How do you divide $\frac{- 2 x^{3} - 16 x^{2} + 13 x - 40}{x - 3}$ $(-2x^3-16x^2+13x-40)/(x-3)$ ?

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How do you divide $\frac{- 2 x^{3} - 16 x^{2} + 13 x - 40}{x - 3}$ $(-2x^3-16x^2+13x-40)/(x-3)$ ?

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Answer 1 · 2017-03-15T23:21:02

$- 2 x^{2} - 22 x - 53 - \frac{199}{x - 3}$ $-2x^2-22x-53-199/(x-3)$

Explanation:

$- 2 x^{3} - 16 x^{2} + 13 x - 40$ $" "-2x^3-16x^2+13x-40$
$color(red)(-2x^2)(x-3)->" "ul(-2x^3+6x^2) larr" subtract"$
$" "0-22x^2+13x-40$
$color(red)(-22x)(x-3)->" "ul(-22x^2+66x) larr" subtract"$
$" "0-53x-40$
$color(red)(-53)(x-3)-> " "ul(-53x+159) larr" subtract"$
$" "0color(red)(-199 larr" remainder")$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$color(red)(-2x^2-22x-53-199/(x-3))$

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How do you divide $\frac{- 2 x^{3} - 16 x^{2} + 13 x - 40}{x - 3}$ $(-2x^3-16x^2+13x-40)/(x-3)$ ?

1 Answer

Explanation:

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How do you divide (-2x^3-16x^2+13x-40)/(x-3) −2x3−16x2+13x−40x−3(-2x^3-16x^2+13x-40)/(x-3) ?

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Explanation:

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How do you divide $\frac{- 2 x^{3} - 16 x^{2} + 13 x - 40}{x - 3}$ $(-2x^3-16x^2+13x-40)/(x-3)$ ?