2022 AMC 12A Problems/Problem 21

Problem

Let $P(x) = x^{2022} + x^{1011} + 1.$ Which of the following polynomials is a factor of $P(x)$ ?

$\textbf{(A)} \, x^2 -x + 1 \qquad\textbf{(B)} \, x^2 + x + 1 \qquad\textbf{(C)} \, x^4 + 1 \qquad\textbf{(D)} \, x^6 - x^3 + 1 \qquad\textbf{(E)} \, x^6 + x^3 + 1$

Solution 1

$P(x) = x^{2022} + x^{1011} + 1$ is equal to $\frac{x^{3033}-1}{x^{1011}-1}$ by difference of powers.

Therefore, the answer is a polynomial that divides $x^{3033}-1$ but not $x^{1011}-1$ .

Note that any polynomial $x^m-1$ divides $x^n-1$ if and only if $m$ is a factor of $n$ .

The prime factorizations of $1011$ and $3033$ are $3*337$ and $3^2*337$ , respectively.

Hence, $x^9-1$ is a divisor of $x^{3033}-1$ but not $x^{1011}-1$ .

By difference of powers, $x^9-1=(x^3-1)(x^6+x^3+1)$ . Therefore, the answer is $\boxed{E}$ .

Solution 2

We simply test roots for each, as $2022,1011$ are multiples of three, we need to make sure the roots are in the form of $e^{i\frac{k\pi}{9}}$ , so we only have to look at $D,E$ .

If we look at choice $E$ , $x=e^{i\frac{\pm2\pi}{9}}$ which works perfectly, the answer is just $E$

~bluesoul

Solution 3

Let $x^{1011} = u$ , now we can rewrite our polynomial as $u^2+u+1$ . Using the quadratic formula to solve for the roots of this polynomial, we have $x^{1011} = \frac{-1\pm i\sqrt{3}}{2}$ Looking at our answer choices, we want to find a polynomial whose roots satisfy this expression. Since the expression $x^6+x^3+1$ is in a similar form to our original polynomial, except with $x^3$ in place of $x^{1011}$ , this would be a good place to start. Solving for the roots of $x^3$ in a similar fashion, $x^3= \frac{-1\pm i\sqrt{3}}{2}$ for the solution we are testing. Now notice that we can rewrite the roots of $x^3$ as $x^3 = \operatorname{cis}{\frac{2\pi}{3}}, \operatorname{cis}{\frac{4\pi}{3}}$ Both of which are third roots of unity. We want to now check if this value of $x^3$ satisfies $x^{1011} = \frac{-1\pm i\sqrt{3}}{2}$ . Notice that $x^{1011} = (x^{3})^{112\cdot3}\cdot x^3$ , and since both values of $x^3$ are roots of unity, we can simplify the expression we want satisfiedto the expression to $x^{1011}=x^3$ . Since both values of $x^3$ are also values of $x^{1011}$ , the roots for our $x^6+x^3+1$ are also roots of $x^{2022}+x^{1011}+1$ , meaning that $x^6+x^3+1 | x^{2022}+x^{1011}+1$ so Therefore, the answer is $\boxed{E}$ .

- DavidHovey

Solution 4 (Describe the Roots)

We know that a monic polynomial $q$ divides a monic polynomial $p$ if and only if all the roots of $q$ are roots of $p.$ Since $P(x)=x^{2022}+x^{1011}+1=\frac{x^{3033}-1}{x^{1011}-1}$ , the roots of $P$ are the $3033$ rd roots of unity that aren't $1011$ th roots of unity.

Now, note that:

1: The roots of polynomial $A$ are the primitive $6$ th roots of unity.

2: The roots of polynomial $B$ are the primitive cube roots of unity.

3: The roots of polynomial $C$ are the primitive $8$ th roots of unity.

4: The roots of polynomial $D$ are the primitive $18$ th roots of unity.

5: The roots of polynomial $E$ are the primitive $9$ th roots of unity.

However, since $6$ , $8$ , and $18$ don't divide $3033$ , the roots of polynomial $A$ are not all $3033$ rd roots of unity, and the same is true for polynomials $C$ and $D$ , eliminating choices $A$ , $C$ and $D.$ Also, since $3$ divides $1011$ , the roots of polynomial $B$ are all $1011$ th roots of unity, eliminating choice $B.$ That leaves choice $\boxed{E}$ , and we can confirm that this is correct by noticing that $9$ divides $3033$ but not $1011.$ From that, we can see that the roots of polynomial $E$ are $3033$ rd roots of unity but not $1011$ th roots of unity, so they are all roots of $P.$ Therefore, $E$ divides $P.$

~pianoboy

Solution 5 (Simple Elimination)

Put the value $x = -1$ . This gives $P(-1) = (-1)^{2022} + (-1)^{1011} + 1 = 1$ . This automatically eliminates choices $A$ , $C$ and $D$ since they do not form a factor of $1$ at $x = -1$ . Now, put $x = \omega$ (omega) at choice $B$ . We get that, $\omega^2 + \omega + 1 = 0$ Thus choice B has $(x - \omega)$ as a factor. If choice $B$ were to be a factor of $P(x)$ , $(x - \omega)$ would also have to be a factor of $P(x)$ , which is clearly not the case, as $P(\omega) \neq 0$ . This eliminates choice $B$ , leaving us with answer $\boxed{E}$ .

~SouradipClash_03

Solution 6 (Elimination but slightly different)

Like Solution 5, let $x=-1$ which eliminates the choices of $A$ , $C$ , and $D$ as they do not divide $P(1)=1$ as they form $3, 0, 3$ respectively by letting $x=-1$ .

This leaves us with only $2$ choices, $B$ and $E$ . Notice that letting $x=0$ or $x=1$ still make these answer choices work and the other values will leave large numbers for us to check which is not feasible in a 75 minutes math competition.

However, we notice answer choice $B$ is quadratic so if answer choice $B$ divides the given polynomial, then the roots of the quadratic must also be roots of the polynomial.

Through quadratic formula, we find the roots of this quadratic as $\dfrac{-1+i\sqrt{3}}{2}$ and $\dfrac{-1-i\sqrt{3}}{2}$ .

We notice that these roots can be written nicely in polar form $\cos(\dfrac{2\pi}{3})+i\sin(\dfrac{2\pi}{3})$ or $\cos(\dfrac{4\pi}{3})+i\sin(\dfrac{4\pi}{3})$ .

We plug either one of these (preferably the root on the left as it is smaller) and see that the polynomial doesn't equal $0$ suggesting that $B$ is not the correct answer choice.

As we only have one answer choice left, we choose $\boxed{E}$

Video Solution

includes review of factoring polynomials

https://youtu.be/UIxePPu0Zus

~MathProblemSolvingSkills.com

Video Solution by ThePuzzlr

https://youtu.be/YRcaIrwA2AU

~ MathIsChess

2022 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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