How do you divide $\frac{x^{2} - 3 x y + 2 y^{2} + 3 x - 6 y - 8}{x - y + 4}$ $(x^2 - 3xy + 2y^2 + 3x - 6y - 8) / (x-y+4)$ ?

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Answer 1 · 2015-11-01T20:21:44

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

$\frac{x^{2} - 3 x y + 2 y^{2} + 3 x - 6 y - 8}{x - y + 4}$ $(x^2-3xy+2y^2+3x-6y-8)/(x-y+4)$

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

$= (x - 2 y - 2)$ $= (x-2y-2)$ with remainder $x$ $x$

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

$x (x - y + 4) = x^{2} - x y + 4 x$ $x(x-y+4) = x^2-xy+4x$

$- 2 y (x - y + 4) = - 2 x y + 2 y^{2} - 8 y$ $-2y(x-y+4) = -2xy+2y^2-8y$

So:

$(x - 2 y) (x - y + 4) = x^{2} - 3 x y + 2 y^{2} + 4 x - 8 y$ $(x-2y)(x-y+4) = x^2-3xy+2y^2+4x-8y$

Then:

$- 1 (x - y + 4) = - x + y - 4$ $-1(x-y+4) = -x+y-4$

or

$- 2 (x - y + 4) = - 2 x + 2 y - 8$ $-2(x-y+4) = -2x+2y-8$

So we find:

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

Or:

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

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

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

How do you divide x2−3xy+2y2+3x−6y−8x−y+4(x^2 - 3xy + 2y^2 + 3x - 6y - 8) / (x-y+4)?