If $ax^2 +bxy+cy^2+5x-2y+3$ divided by $x-y+1$ has remainder $0$, determine $a$, $b$, and $c$.
I do not know how to approach this problem and would appreciate advice how to proceed.
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Chill2Macht
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Ledi
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So what did you try to do? – Oscar Lanzi Jul 23 '16 at 02:14
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Like in one variable, substitute suitable values of $z$ and $y$. – André Nicolas Jul 23 '16 at 02:32
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2Given constants $f,g,h,$ what is $$ (fx + gy + h))(x-y+1)? $$ – Will Jagy Jul 23 '16 at 02:36
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Can x=y or is that obviously disallowed? – Inquisitive Jul 23 '16 at 02:41
2 Answers
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Since there is no remainder the original equation will be the product of $x-y+1$ with another unknown equation. In order to create the $x^2$, $y^2$, and $xy$ terms we expect the form to be $dx+ey+f$.
$$ax^2+bxy+cy^2+5x−2y+3 = (x−y+1)(dx+ey+f)$$ $$ax^2+bxy+cy^2+5x−2y+3 = dx^2+(e-d)xy-ey^2+(d+f)x+(e-f)y+f$$
working right to left
$$f=3$$
$$e-f=-2, e=1$$
$$d+f=5, d=2$$
$$c=-e, c=-1$$
$$b=e-d, b=-1$$
$$a=d, a=2$$
Dominic Wynter
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Andrew Green
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In $ax^2+bxy+cy^2+5x−2y+3 = (x−y+1)(dx+ey+f)$, put $x=0, y=1$ to get $c+1 = 0$ and hence $c=-1$. Put $y=0, x=-1$ to get $a-2=0$ and hence $a=2$. Putting $x=1, y = 2$, we get $a+2b+4c+5-4+3 = 0$ and hence $b=-1$.