Factoring $x^4-11x^2y^2+y^4$

Question

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!

Answer 1

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.

Answer 2

$$x^{ 4 }-11x^{ 2 }y^{ 2 }+y^{ 4 }={ \left( { x }^{ 2 }-{ y }^{ 2 } \right) }^{ 2 }-9{ x }^{ 2 }{ y }^{ 2 }=\left( { x }^{ 2 }-{ y }^{ 2 }-3xy \right) \left( { x }^{ 2 }-{ y }^{ 2 }+3xy \right) $$

Answer 3

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 $-2 + 11 = 9.$

Favorites $$ (x^2 + 2 xy + 2 y^2)(x^2 - 2xy + 2 y^2) = x^4 + 4 y^4, $$ $$ (x^2 + xy + y^2)(x^2 - xy + y^2) = x^4 + x^2 y^2 + y^4. $$

The latter gives a real analytic alternative to the superellipse.

Answer 4

$$ \begin{align} x^4-11x^2y^2+y^4 &=\left(x^2-y^2\right)^2-(3xy)^2\\ &=\left(x^2-3xy-y^2\right)\left(x^2+3xy-y^2\right)\\ &=\left(x-\tfrac{3-\sqrt{13}}2y\right)\left(x-\tfrac{3+\sqrt{13}}2y\right)\left(x+\tfrac{3-\sqrt{13}}2y\right)\left(x+\tfrac{3+\sqrt{13}}2y\right) \end{align} $$