How do you divide $(-x^3 - x^2-6x+5 )/((-x + 10 )$ ?

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Answer 1 · 2016-02-21T17:35:31

$(−x^3−x^2−6x+5)/(-x+10)$ is

$x^2+11x+116-1155/(-x+10)$

To divide $−x^3−x^2−6x+5$ by $-x+10$ , first we observe that $-x$ goes $(-x^3)/-x=x^2$ times in $-x^3$ .

Hence, $x^2(-x+10)-10x^2−x^2−6x+5$ or
(note we have added and subtracted $10x^2)$ x^2(-x+10)-11x^2−6x+5#

As $-11x^2/-x=11x$ , above can be written is

$x^2(-x+10)+11x(-x+10)-110x−6x+5$ or

$x^2(-x+10)+11x(-x+10)-116x+5$

and as $-116x/-x=116$ , above can be written as

$x^2(-x+10)+11x(-x+10)+116x(-x+10)-1160+5$ or

$x^2(-x+10)+11x(-x+10)+116x(-x+10)-1155$ or

$(x^2+11x+116)(-x+10)-1155$ or

Hence $(−x^3−x^2−6x+5)/(-x+10)$ is

$x^2+11x+116-1155/(-x+10)$

How do you divide (-x^3 - x^2-6x+5 )/((-x + 10 )(-x^3 - x^2-6x+5 )/((-x + 10 )?