# How do you divide (x^2 - 3xy + 2y^2 + 3x - 6y - 8) / (x-y+4)?

Nov 1, 2015

You can choose to end up with no $x$ in the remainder or no $y$...

$\frac{{x}^{2} - 3 x y + 2 {y}^{2} + 3 x - 6 y - 8}{x - y + 4}$

$= \left(x - 2 y - 1\right)$ with remainder $- 7 y - 4$

$= \left(x - 2 y - 2\right)$ with remainder $x$

#### Explanation:

Try to match the higher degree terms in the dividend with multiples of the divisor as follows:

$x \left(x - y + 4\right) = {x}^{2} - x y + 4 x$

$- 2 y \left(x - y + 4\right) = - 2 x y + 2 {y}^{2} - 8 y$

So:

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

Then:

$- 1 \left(x - y + 4\right) = - x + y - 4$

or

$- 2 \left(x - y + 4\right) = - 2 x + 2 y - 8$

So we find:

${x}^{2} - 3 x y + 2 {y}^{2} + 3 x - 6 y - 8 = \left(x - 2 y - 1\right) \left(x - y + 4\right) - 7 y - 4$

Or:

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

So no choice of quotient allows us to eliminate both $x$ and $y$ from the remainder.

I suspect the $+ 3 x$ in the question should have been $+ 4 x$