# Techniques for forming square factorizations

Say you have the polynomial $$x^4 + 2 + x^{-4}$$

Looking at it, you see you can do \begin{align*} x^4 + 1 + 1 + x^{-4} & =x^2( x^2 + x^{-2} ) + x^{-2}( x^2 + x^{-2} )\\ &= \left( x^2 + x^{-2} \right)^2. \end{align*} Another one is \begin{align*} x^2 + \frac{1}{2} + \frac{1}{16x^2} &= x^2 + \frac{1}{4} + \frac{1}{4} + \frac{1}{16x^2} \\ &=x^2\left( 1 + \frac{1}{4x^2} \right) + \frac{1}{4}\left( 1 + \frac{1}{4x^2} \right)\\ &=\left( x^2 + \frac{1}{4}\right) \left( 1 + \frac{1}{4x^2} \right)\\ (x^2) \left( 1 + \frac{1}{4x^2} \right)^2. \end{align*}

So the question is, I've been doing this by "inspection" - are there any techniques for recognizing when this type of factorization is possible or how to do more easily?

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You have $A(x) + B + C(x)$, set $a = \sqrt{A(x)}, b = \sqrt{C(x)}$ and check if $2ab = B$. (So that $(a+b)^2 = a^2 + 2ab + b^2$.) In your fist example $a=x^2, b=x^{-2}$, in the second $a=x, b=\frac{1}{4x}$. – Eelvex Mar 22 '11 at 14:37
math.stackexchange.com/questions/21232/… – anonymous Mar 22 '11 at 15:02

For trinomials:

• If you have $(p(x))^2 + K + (q(x))^2$, then check to see if $K = \pm 2p(x)q(x)$. if so, then $(p(x))^2 + K + (q(x))^2 = (p(x)\pm q(x))^2$.

• In your first example, you have $$x^4 + 2 + x^{-4} = (x^2)^2 + 2 + (x^{-2})^2.$$ Taking $p(x) = x^2$, $q(x) = x^{-2}$, we have $2p(x)q(x) = 2$, which is precisely the middle term, so $x^2+2+x^{-4} = (x^2 + x^{-2})^2$.
• In your second example, you have $$x^2 + \frac{1}{2} + \frac{1}{16x^2} = (x)^2 + \frac{1}{2} + \left(\frac{1}{4x}\right)^2.$$ Taking $p(x) = x$ and $q(x) = \frac{1}{4x}$, you have $2p(x)q(x) = \frac{2x}{4x} = \frac{1}{2}$, again exactly the middle term, so you get $$x^2 + \frac{1}{2} + \frac{1}{16x^2} = \left(x + \frac{1}{4x}\right)^2.$$
• More generally, if you have $(p(x))^2 + Kp(x) + L$, then see if you can find two expressions, $s(x)$ and $t(x)$, which when multiplied give $L$ and when added give $K$, $s(x)t(x) = L$, $s(x)+t(x) = K$. Then $(p(x))^2 + Kp(x) +L = (p(x)+s(t))(p(x)+t(x))$.

• For example, $$2x^4 + (\sqrt{2}-1)x^2 + x.$$ You may notice that the leading term is $(\sqrt{2}x^2)^2$, and that you have a "middle term" of $\sqrt{2}x^2$. So rewriting a bit you get $$(\sqrt{2}x^2)^2 + (\sqrt{2}x^2)1 + (x-x^2).$$ This suggests looking for $s(x)$ and $t(x)$ with $s(x)+t(x) = 1$ and $s(x)t(x)=x-x^2 = x(1-x)$. This in turn quickly suggests $s(x) = x$ and $t(x)=1-x$, which gives $$2x^4 + (\sqrt{2}-1)x^2 + x = \left(\sqrt{2}x^2 + x\right)\left(\sqrt{2}x^2 + (1-x)\right).$$

But generally, there is some substantial amount of "inspect and notice" going on. The more of these you do, the more you will notice patterns and be able to "notice" the things that one needs to notice.

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