2022 AMC 12A Problems/Problem 21

Revision as of 18:00, 10 April 2023 by Batmanstark (talk | contribs) (Solution 6 (Elimination but slightly different))


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 $E$


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 $P.$


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 $\boxed{E}$.


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



Video Solution by ThePuzzlr


~ MathIsChess

See Also

2022 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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