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

May 17, 2016

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

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

Explanation:

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

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

Now $- {x}^{2} \times \left({x}^{2} - 3 x + 9\right) = - {x}^{4} + 3 {x}^{3} - 9 {x}^{2}$

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

$\left(- {x}^{4} + 6 {x}^{3} + 8 x + 12\right) - \left(- {x}^{4} + 3 {x}^{3} - 9 {x}^{2}\right) = 3 {x}^{3} + 9 {x}^{2} + 8 x + 12$

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

$3 {x}^{3} + 9 {x}^{2} + 8 x + 12 - 3 x \left({x}^{2} - 3 x + 9\right)$

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

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

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

$18 {x}^{2} - 19 x + 12 - 18 \left({x}^{2} - 3 x + 9\right) = - 19 x + 54 x + 12 - 162 = 35 x - 150$

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