How do you divide $(-x^4+6x^3+8x+12)/(x^2-3x+9)$ ?

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Answer 1 · 2016-05-17T12:59:55

Dividing $-x^4+6x^3+8x+12$ by $x^2-3x+9$ ,

the quotient is $-x^2+3x+18$ and remainder is $35x-150$

To divide $-x^4+6x^3+8x+12$ by $x^2-3x+9$ ,

as $-x^4/x^2=-x^2$ .

Now $-x^2xx(x^2-3x+9)=-x^4+3x^3-9x^2$

Subtracting $-x^4+3x^3-9x^2$ from $-x^4+6x^3+8x+12$ , we get

$(-x^4+6x^3+8x+12)-(-x^4+3x^3-9x^2)=3x^3+9x^2+8x+12$

Now in similar way $3x^3+9x^2+8x+12$ , $x^2-3x+9$ can go $3x$ times and remainder will be

$3x^3+9x^2+8x+12-3x(x^2-3x+9)$

= $3x^3+9x^2+8x+12-3x^3+9x^2-27x$

= $18x^2-19x+12$

Now in this, $x^2-3x+9$ goes $18$ times and remainder is

$18x^2-19x+12-18(x^2-3x+9)=-19x+54x+12-162=35x-150$

Hence when we divide $-x^4+6x^3+8x+12$ by $x^2-3x+9$ , the quotient is $-x^2+3x+18$ and remainder is $35x-150$

How do you divide (-x^4+6x^3+8x+12)/(x^2-3x+9)(-x^4+6x^3+8x+12)/(x^2-3x+9)?