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Find factors of $$x^2-6xy+y^2+3x-3y+4.$$

I could not find the factors of this polynomial. Can you please explain me how to proceed?

nonuser
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shadow Fire
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    If you rewrite it as $p(x,y)=x^2+(3-6y)x+(y^2-3y+4)$ you can "solve" it in the usual way as a quadratic in $x$. If there are "nice" factors, this will give them. – Mark Bennet Sep 03 '18 at 09:23
  • Can u solve further plz – shadow Fire Sep 03 '18 at 09:36
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    Please study our guide for new askers. Copy/pasting a homework assignment will not be received well. Make the question focused on a key concept, share your thoughts, or give other context (what related pieces of theory have been covered recently). That way the answerers can give an answer that helps you learn (rather than one that zips over your head). If you just want an answer, and don't want to learn, then this site is not for you. – Jyrki Lahtonen Sep 03 '18 at 09:40

4 Answers4

4

Hint: try with

$$ (x+ay+b)(x+cy+d)= x^2-6xy+y^2+3x-3y+4$$

Since this should be true for all $x,y$ you can put different values of $x,y$ into to calculate $a,b,c,d$. You should get a system of 4 equation.

Say for $y=0$ we get $$(x+b)(x+d)= x^2+3x+4$$ which should be true for all (real) $x$, but this is imposible since the discrimanant is $-7$. So you can't factor this expression (in real).

nonuser
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You cannot factor $f(x,y):=x^2-6xy+y^2+3x-3y+4$ over $\mathbb{Q}$, $\mathbb{R}$, or $\mathbb{C}$. If the base field is $\mathbb{Q}$ or $\mathbb{R}$, then the answer by greedoid covers it. We now assume that the base field is $\mathbb{C}$ (or any algebraically closed field of characteristic $0$).

To show that $f(x,y)$ is an irreducible element of $\mathbb{C}[x,y]$, you can first homogenize the polynomial with a dummy variable $z$ to get $$F(x,y,z)=x^2-6xy+y^2+3xz-3yz+4z^2\in\mathbb{C}[x,y,z]\,.$$ Then, $$\frac{\partial F}{\partial x}(x,y,z)=2x-6y+3z\,,$$ $$\frac{\partial F}{\partial y}(x,y,z)=-6x+2y-3z\,,$$ and $$\frac{\partial F}{\partial z}(x,y,z)=3x-3y+8z\,.$$ Thus, $\dfrac{\partial F}{\partial x}(x,y,z)=0$, $\dfrac{\partial F}{\partial y}(x,y,z)=0$, and $\dfrac{\partial F}{\partial z}(x,y,z)=0$ simultaneously if and only if $(x,y,z)=(0,0,0)$. From this link, we conclude that $F(x,y,z)$ is irreducible over $\mathbb{C}$, and so is $f(x,y)$.

It can be shown that $f(x,y)$ is reducible over a given base field $K$ if and only if $\text{char}(K)=2$ or $\text{char}(K)=23$. When $\text{char}(K)=2$, we have $$f(x,y)=(x+y)\,(x+y+1)\,.$$ When $\text{char}(K)=23$, we have $$f(x,y)=(7x+y+8)(10x+y+12)\,.$$

Batominovski
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Compute the determinant $$ \det\begin{bmatrix} 1 & -3 & 3/2 \\ -3 & 1 & -3/2 \\ 3/2 & -3/2 & 4 \end{bmatrix}=-23 $$ Since the determinant is nonzero, the polynomial represents a nondegenerate conic, so it is irreducible.

egreg
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  • is not it the quadratic form in $(x,y)$, so is not the determinant $\begin{vmatrix}1&-3\-3&1\end{vmatrix}=-8\ne 0$? Though the result is the same. – farruhota Sep 03 '18 at 10:28
  • @farruhota No, that determinant is $0$ when the conic is a parabola (degenerate or not). – egreg Sep 03 '18 at 10:38
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    @farruhota You can't use just only the coefficients of the quadratic terms in $x$ and $y$. For example, $x^2-y^2$ is reducible, but the determinant of $\begin{bmatrix}1&0\0&-1\end{bmatrix}$ is $-1\neq 0$. – Batominovski Sep 03 '18 at 10:44
  • @Batominovski, thank you, indeed, here it states "discriminant of non-homogenous form determines the parabola, ellipse or hyperbola", while "discriminant of homogenized form determines the non-degenerate or degenerate", so the given is non-degenerate hyperbola. – farruhota Sep 03 '18 at 11:25
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Hint:

Use the quadratic formula and treat $x$ as a constant. What you will get is a function with asymptotic behaviour, i.e., something like $xy=1$.

cansomeonehelpmeout
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