Question

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

Answer 1

This does not divide exactly as given, but we can investigate a little...

Explanation:

I think you do not "long divide" it as it has multiple variables, but we can have a go at dividing it.

Consider:

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

We would like to express this as a quotient and remainder of lower degree than that of the dividend.

Since the dividend is of mixed degree with maximum degree `2` and the divisor is of mixed degree with maximum degree `1`, we may be able to find a quotient of mixed degree with maximum `1`.

That is, there may be a quotient of the form:

`ax+by+c`

We would hope that any remainder would have degree `0`, but this may not be possible.

We find:

`(x-y+4)(ax+by+c)`

`= ax^2+(b-a)xy-by^2+(4a+c)x+(4b-c)y+4c`

If it is possible to match all of the terms of degree `2` in the dividend, then we must have:

`{ (a=1), (b-a=-3), (-b=2) :}`

This is satisfied by `a=1` and `b=-2` and we find:

`(x-y+4)(x-2y+c)`

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

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Choice 1

If we try to match the coefficient of `x` then we have:

`4+c = 3` and hence `c=-1`

Then we have:

`(x-y+4)(x-2y-1)`

`= x^2-3xy+2y^2+3x-7y-4`

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

So:

`(x^2-3xy+2y^2+3x-6y-8)/(x-y+4) = (x-2y-1)+(y-4)/(x-y+4)`

This is identical to the result we get if we treat `y` as a constant and long divide the quadratic in `x`:

`x^2+(-3y+3)x+(2y^2-6y-8)`

by the linear polynomial in `x`:

`x+(-y+4)`

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Choice 2

If we try to match the coefficient of `y` then we have:

`-8-c = -6` and hence `c=-2`

Then we have:

`(x-y+4)(x-2y-2)`

`= x^2-3xy+2y^2+2x-6y-8`

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

So:

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

This is identical to the result we get if we treat `x` as a constant and long divide the quadratic in `y`:

`2y^2+(-3x-6)y+(x^2+3x-8)`

by the linear polynomial in `y`:

`-y+(x+4)`

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Footnote

I suspect a typo in the question. I think the dividend should have been specified as:

`x^2-3xy+2y^2+2x-6y-8`

which would have resulted in an exact division with no remainder.

Answer 2

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

Explanation:

Calling

`n(x,y) = x^2 - 3 x y + 2 y^2 + 3 x - 6 y - 8` and
`d(x,y)=x - y + 4`

we ask for

`q(x,y) = x + a y + b` and
`r(x,y) = x +c y + d`

such that

`n(x,y) = d(x,y)q(x,y) + r(x,y)`

equating coefficients we obtain the conditions

#{ (-8 - 4 b - d = 0), (-6 - 4 a + b - c = 0), (2 + a = 0), (2 + b = 0) :} #
Solving we have

`{a = -2, b = -2, c = 0, d = 0}`

so

`q(x,y) = x-2y-2` and
`r(x,y) = x`