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

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Answer 1 · 2015-11-01T20:21:44

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

$(x^2-3xy+2y^2+3x-6y-8)/(x-y+4)$

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

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

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

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

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

So:

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

Then:

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

or

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

So we find:

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

Or:

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

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

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

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