# 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 $-y^2x^2$).

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\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*}

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Gah. You make it look so easy. Thanks for that. – jsmith95 Jun 27 '13 at 7:37
+1 and more Adi. ;-) – Babak S. Jun 27 '13 at 17:38
"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.

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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.

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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, $$x^4-11x^2y^2+y^4=(x^2+3xy-y^2)(x^2-3xy-y^2)$$

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