Factoring $x^4-11x^2y^2+y^4$ I am brushing up on my precalculus and was wondering how to factor the expression
$$
x^4-11x^2y^2+y^4
$$
Thanks for any help!
 A: We notice this looks a bit like $(x^2-y^2)^2$, so we write
\begin{align*}
x^4 - 11x^2y^2 + y^4
&= (x^4 - 2x^2y^2+y^4) - 9x^2y^2 \\
&= (x^2-y^2)^2 - (3xy)^2 \\
&= (x^2 + 3xy - y^2)(x^2 - 3xy - y^2).
\end{align*}
It factors further, but no longer over the integers (there will be square roots involved).
We also could have used $(x^2+y^2)^2$ instead of $(x^2-y^2)^2$, to get
\begin{align*}
x^4 - 11x^2y^2 + y^4
&= (x^4 + 2x^2y^2+y^4) - 13x^2y^2 \\
&= (x^2+y^2)^2 - (\sqrt{13}\;xy)^2 \\
&= (x^2 + \sqrt{13}\;xy + y^2)(x^2 - \sqrt{13}\; xy + y^2).
\end{align*}
The two factorizations may look different, but they are the same four linear factors combined in two different ways.
A: $$x^{ 4 }-11x^{ 2 }y^{ 2 }+y^{ 4 }={ \left( { x }^{ 2 }-{ y }^{ 2 } \right)  }^{ 2 }-9{ x }^{ 2 }{ y }^{ 2 }=\left( { x }^{ 2 }-{ y }^{ 2 }-3xy \right) \left( { x }^{ 2 }-{ y }^{ 2 }+3xy \right) $$
A: For emphasis, if you have a quadratic or quadratic form, $a x^2 + bx + c$ or $a x^2 + bxy + c y^2,$it factors over the integers if and only if the discriminant from the Quadratic Formula, $b^2 - 4 a c,$ is an integer squared. 
Things are rather different for quartics with only EVEN exponents, either 
$f^2 x^4 + g x^2 + h^2$ or $f^2 x^4 + g x^2 y^2 + h^2 y^4$ to keep it easy. No worry about the discriminant, but the only thing that can work (once no rational roots) is, where $fh$ could be positive or negative,
$$ (fx^2 + \gamma xy + h y^2 )  (fx^2 - \gamma xy + h y^2 )  = f^2 x^4 + g x^2 y^2 + h^2 y^4.  $$
Here we just need to be able to solve for $\gamma$ in
$$ 2fh - \gamma^2 = g,  $$
or
$$ 2fh - g = \gamma^2.  $$
To repeat, we can negate $fh$ at need, useful if $g$ is negative. We see this in the original question and the answer by 6005, where we can choose either $ 2 + 11 = 13$ or $-2 + 11 = 9.$
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$$ (x^2 + 2 xy + 2 y^2)(x^2 - 2xy + 2 y^2) = x^4 + 4 y^4, $$
$$ (x^2 +  xy +  y^2)(x^2 - xy +  y^2) = x^4 + x^2 y^2 + y^4. $$
The latter gives a real analytic alternative to the superellipse.  

A: $$
\begin{align}
x^4-11x^2y^2+y^4
&=\left(x^2-y^2\right)^2-(3xy)^2\\
&=\left(x^2-3xy-y^2\right)\left(x^2+3xy-y^2\right)\\
&=\left(x-\tfrac{3-\sqrt{13}}2y\right)\left(x-\tfrac{3+\sqrt{13}}2y\right)\left(x+\tfrac{3-\sqrt{13}}2y\right)\left(x+\tfrac{3+\sqrt{13}}2y\right)
\end{align}
$$
A: $$(x^2 + a y^2) (x^2 + b y^2) = x^4-11x^2y^2+y^4$$
Hence,
$$a + b = -11 \qquad \qquad \qquad a b = 1$$
In SymPy,
>>> a, b = symbols('a b')
>>> solve_poly_system([a + b + 11, a*b - 1], a, b)
             ____             ____               ____             ____  
    11   3*\/ 13     11   3*\/ 13       11   3*\/ 13     11   3*\/ 13   
[(- -- - --------, - -- + --------), (- -- + --------, - -- - --------)]
    2       2        2       2          2       2        2       2      

Thus,
$$\left( x^2 + \left(\frac{-11 - 3 \sqrt{13}}{2}\right) y^2 \right) \left( x^2 + \left(\frac{-11 + 3 \sqrt{13}}{2}\right) y^2 \right) = x^4-11x^2y^2+y^4$$
