9.3. A trinomial of degree $p$ is a function of the form $f(x)=x^{p}+a x^{q}+1$, where $p, q-$ are natural numbers, $q<p$, and $a$ is an arbitrary real number (possibly equal to 0). Find all factorizations of the polynomial $x^{12}+1$ into a product of two trinomials.

ANSWER: $\left(x^{4}+1\right)\left(x^{8}-x^{4}+1\right),\left(x^{6}-\sqrt{2} x^{3}+1\right)\left(x^{6}+\sqrt{2} x^{3}+1\right)$.

CRITERIA

A +/- grade was given for work in which a clear set of alternative options for the trinomial factors was described (with a complete proof of the non-realizability of all discarded cases), and most of these alternative options were fully analyzed, but one or two options were considered with errors in calculations and/or not completed or not considered at all, leading the author of the work to lose one of the two solutions.

A +/2 grade was given for work in which both correct answers were presented (usually obtained using the formulas for the sum of cubes and the difference of squares), but the proof that there are no other solutions is missing or hopelessly incorrect.

A -/+ grade was given for work in which one of the answers was presented (usually obtained using the formula for the sum of cubes or the difference of squares), but the proof that there are no other solutions is missing or hopelessly incorrect.
Solution. Let $\left(x^{p}+a x^{q}+1\right)\left(x^{r}+b x^{s}+1\right)=x^{12}+1(*)$.

Since the degrees of the monomials on both sides are equal, then $p+r=12$. The numbers $a$ and $b$ are not both zero (otherwise, after expanding the brackets and combining like terms, there would be at least three monomials in the left part).

First, consider the case when one of the numbers $a$ and $b$ is zero (let's assume $a=0$). Then the left part of the equality $(*)$ is $x^{p}+1+b x^{s+p}+b x^{s}+x^{12}+x^{r}$. Since each of the monomials $x^{p}$ and $x^{r}$ must cancel with one of the others, each of the numbers $p$ and $r$ must be equal to one of the numbers $s$ and $s+p$. Since $p<s+p$, then $p=s$ and $r=p+s=2 s$. Since $p+r=3 s=12$, we get $s=4, p=8, r=12, b=-1$, i.e., the first of the two answers.

Now consider the case when the numbers $a$ and $b$ are non-zero. In this case, expanding the brackets in the left part of the equality (*) gives nine monomials, the degrees of which are listed in the following table:

| 12 | $p+s$ | $p$ |
| :---: | :---: | :---: |
| $q+r$ | $q+s$ | $q$ |
| $r$ | $s$ | 0 |

Since all monomials, except $x^{12}$ and 1, must cancel each other out, all the degrees listed in the table, except 12 and 0, must be grouped into sets of equal numbers (and each set must consist of more than one number). In one set, there cannot be numbers, one of which is to the left and above the other (either strictly to the left or strictly above), because then the second number is strictly greater than the first. Therefore, a set cannot contain more than three numbers, and a set of three numbers can only consist of the numbers $r, q+s, p$. Therefore, the numbers in the table are divided into three groups: $r=q+s=p$, $s=q, q+r=p+s$. From this, we get $q=s=3$ and $p=r=6$. Substituting $x=1$ and $x=-1$ into the equality $(*)$, we get $2=(2+a)(2+b)=(2-a)(2-b)$, from which we find $a=-b= \pm \sqrt{2}$, i.e., the second of the two answers.