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

2 Answers
Jul 15, 2016

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#

#color(white)()#
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)#

#color(white)()#
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)#

#color(white)()#
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.

Jul 15, 2016

#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#