Question

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

Answer 1

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`

Explanation:

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`