How do you long divide # (x² - 3xy + 2y² + 3x - 6y - 8) / (x-y+4)#?
2 Answers
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
That is, there may be a quotient of the form:
#ax+by+c#
We would hope that any remainder would have degree
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
#{ (a=1), (b-a=-3), (-b=2) :}#
This is satisfied by
#(x-y+4)(x-2y+c)#
#= x^2-3xy+2y^2+(4+c)x+(-8-c)y+4c#
Choice 1
If we try to match the coefficient of
#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
#x^2+(-3y+3)x+(2y^2-6y-8)#
by the linear polynomial in
#x+(-y+4)#
Choice 2
If we try to match the coefficient of
#-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
#2y^2+(-3x-6)y+(x^2+3x-8)#
by the linear polynomial in
#-y+(x+4)#
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.
Explanation:
Calling
we ask for
such that
equating coefficients we obtain the conditions
Solving we have
so