# Factor $x^4 - 11x^2y^2 + y^4$

This is an exercise from Schaum's Outline of Precalculus. It doesn't give a worked solution, just the answer.

The question is:

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

$(x^2 - 3xy -y^2)(x^2 + 3xy - y^2)$

My question is:

How did the textbook get this?

I tried the following methods (examples of my working below):

1. U-Substitution.
2. Guess and Check.
3. Reversing the question (multiplying the answer out).

Here is my working for each case so far.

(1) U-Substitution.

I tried a simpler case via u-substitution.

Let $u = x^2$ and $v = y^2$. Then $x^4 - 11x^2y^2 + y^4 = u^2 -11uv + v^2$. Given the middle term is not even, then it doesn't factor into something resembling $(u + v)^2$ or $(u - v)^2$. Also, there doesn't appear to be any factors such that $ab = 1$ (the coefficient of the third term) and $a + b = -11$ (the coefficient of the second term) where $u^2 -11uv + v^2$ resembles $(u \pm a)(v \pm b)$ and the final answer.

(2) Guess and Check.

To obtain the first term in $x^4 - 11x^2y^2 + y^4$ it must be $x^2x^2 = x^4$. To obtain the third term would be $y^2y^2 = y^4$. So I'm left with something resembling:

$(x^2 \pm y^2)(x^2 \pm y^2)$.

But I'm at a loss as to how to get the final solution's middle term from guessing and checking.

(3) Reversing the question

Here I took the answer, multiplied it out, to see if the reverse of the factoring process would illuminate how the answer was generated.

$(x^2 - 3xy -y^2)(x^2 + 3xy - y^2) = [(x^2 - 3xy) - y^2][(x^2 + 3xy) - y^2]$

$= (x^2 - 3xy)(x^2 + 3xy) + (x^2 - 3xy)(-y^2) + (x^2 + 3xy)(-y^2) + (-y^2)(-y^2)$

$= [(x^2)^2 - (3xy)^2] + [(-y^2)x^2 + 3y^3x] + [(-y^2)x^2 - 3y^3x] + [y^4]$

$= x^4 - 9x^2y^2 - y^2x^2 + 3y^3x - y^2x^2 - 3y^3x + y^4$

The $3y^3x$ terms cancel out, and we are left with:

$x^4 - 9x^2y^2 - 2x^2y^2 + y^4 = x^4 - 11x^2y^2 + y^4$, which is the original question.

The thing I don't understand about this reverse process is where the $3y^3x$ terms came from. Obviously $3y^3x - 3y^3x = 0$ by additive inverse, and $a + 0 = a$, but I'm wondering how you would know to add $3y^3x - 3y^3x$ to the original expression, and then make the further leap to factoring out like terms (by splitting the $11x^2y^2$ to $-9x^2y^2$, $-y^2x^2$, and $-y^2x^2$).

\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-y^2-3xy)(x^2-y^2+3xy). \end{align*}

• Gah. You make it look so easy. Thanks for that. – jsmith95 Jun 27 '13 at 7:37
• "My question is: how did the book get this?" – The Chaz 2.0 Jul 24 '13 at 14:45

A start: Divide the original by $y^4$, and set $z=x/y$. We are trying to factor $z^4-11z^2+1$ as a product of two quadratics. Without loss of generality we may assume they both begin with $z^2$. So we are looking for a factorization of type $(z^2+az+b)(z^2+cz+d)$. By looking at the coefficient of $z^3$ in the product, we can see that $c=-a$, and you are on your way.

The next step is to see what $c=-a$ and the coefficient of $x$ in the product being $0$ tells us about the relationship between $b$ and $d$.

The idea works smoothly for, say $(11.3)x^2y^2$ replacing $11x^2y^2$, except of course the factorization involves square roots.

Another way: Divide the original polynomial by $x^2y^2$. We get $$\frac{x^4-11x^2y^2+y^4}{x^2y^2}=\frac{x^2}{y^2}-11+\frac{y^2}{x^2}.\tag{1}$$ Make the substitution $w=\frac{x}{y}+\frac{y}{x}$. Then $\frac{x^2}{y^2}+\frac{y^2}{x^2}=w^2-2$. The right-hand side of Expression (1) becomes $w^2-9$, which factors as $(w-3)(w+3)$. Thus (1) can be expressed as the product $$\left(\frac{x}{y}+\frac{y}{x}-3\right)\left(\frac{x}{y}+\frac{y}{x}+3\right).\tag{2}$$ Finally, multiply Expression (2) by $x^2y^2$, by multiplying each term by $xy$, and we get our factorization.

The idea generalizes immediately to $x^4-ax^2y^2+y^4$ where $a\ge 2$, and can be pushed well beyond.

Note that the answer is not unique. There are four linear factors, and you can get different quadratic factors by grouping them the other way. For instance: \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+y^2-\sqrt{13}xy)(x^2+y^2+\sqrt{13}xy). \end{align*} I admit that I got the given answer first (by the same method as this answer) , because 9 looks more like a square than 13.

Let $x^2=u$ and $y^2=v$, then expression becomes $u^2-11uv+v^2$. Let the roots of the quadratic equation $u^2-11uv+v^2=0$ be $\alpha v,\beta v$, then, $\alpha+\beta=11$ and $\alpha\beta=1$ . You can check easily that $\alpha,\beta>0$, so $\sqrt{\alpha},\sqrt{\beta}\in \Bbb R$

therefore, $$x^4-11x^2y^2+y^4=u^2-11uv+v^2=(u-\alpha v)(u-\beta v)=(x^2-\alpha y^2)(x^2-\beta y^2)$$ $$=(x-\sqrt{\alpha}y)(x+\sqrt{\alpha}y)(x-\sqrt{\beta}y)(x+\sqrt{\beta}y)$$ $$=(x-\sqrt{\alpha}y)(x+\sqrt{\beta}y).(x+\sqrt{\alpha}y)(x-\sqrt{\beta}y)\tag{Rearraging}$$ $$=(x^2+(\sqrt{\beta}-\sqrt{\alpha})xy-\sqrt{\alpha \beta}y^2))(x^2-(\sqrt{\beta}-\sqrt{\alpha})xy-\sqrt{\alpha \beta}y^2))\tag{1}$$

Now, WLOG, let $\beta>\alpha$, then $\sqrt{\beta}-\sqrt{\alpha}=\sqrt{\alpha+\beta-2\sqrt{\alpha\beta}}=3$ as $\alpha\beta=1,\alpha+\beta=11$

Therefore, Putting it in $(1)$ gives, $$x^4-11x^2y^2+y^4=(x^2+3xy-y^2)(x^2-3xy-y^2)$$