2022 AMC 12A Problems/Problem 21
Let Which of the following polynomials is a factor of ?
is equal to by difference of powers.
Therefore, the answer is a polynomial that divides but not .
Note that any polynomial divides if and only if is a factor of .
The prime factorizations of and are and , respectively.
Hence, is a divisor of but not .
By difference of powers, . Therefore, the answer is .
We simply test roots for each, as are multiples of three, we need to make sure the roots are in the form of , so we only have to look at .
If we look at choice , which works perfectly, the answer is just
Let , now we can rewrite our polynomial as . Using the quadratic formula to solve for the roots of this polynomial, we have Looking at our answer choices, we want to find a polynomial whose roots satisfy this expression. Since the expression is in a similar form to our original polynomial, except with in place of , this would be a good place to start. Solving for the roots of in a similar fashion, for the solution we are testing. Now notice that we can rewrite the roots of as Both of which are third roots of unity. We want to now check if this value of satisfies . Notice that , and since both values of are roots of unity, we can simplify the expression we want satisfiedto the expression to . Since both values of are also values of , the roots for our are also roots of , meaning that so Therefore, the answer is .
Solution 4 (Describe the Roots)
We know that a monic polynomial divides a monic polynomial if and only if all the roots of are roots of Since , the roots of are the rd roots of unity that aren't th roots of unity.
Now, note that:
1: The roots of polynomial are the primitive th roots of unity.
2: The roots of polynomial are the primitive cube roots of unity.
3: The roots of polynomial are the primitive th roots of unity.
4: The roots of polynomial are the primitive th roots of unity.
5: The roots of polynomial are the primitive th roots of unity.
However, since , , and don't divide , the roots of polynomial are not all rd roots of unity, and the same is true for polynomials and , eliminating choices , and Also, since divides , the roots of polynomial are all th roots of unity, eliminating choice That leaves choice , and we can confirm that this is correct by noticing that divides but not From that, we can see that the roots of polynomial are rd roots of unity but not th roots of unity, so they are all roots of Therefore, divides
Solution 5 (Simple Elimination)
Put the value . This gives . This automatically eliminates choices , and since they do not form a factor of at . Now, put (omega) at choice . We get that, Thus choice B has as a factor. If choice were to be a factor of , would also have to be a factor of , which is clearly not the case, as . This eliminates choice , leaving us with answer .
Solution 6 (Elimination but slightly different)
Like Solution 5, let which eliminates the choices of , , and as they do not divide as they form respectively by letting .
This leaves us with only choices, and . Notice that letting or 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 is quadratic so if answer choice 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 and .
We notice that these roots can be written nicely in polar form or .
We plug either one of these (preferably the root on the left as it is smaller) and see that the polynomial doesn't equal suggesting that is not the correct answer choice.
As we only have one answer choice left, we choose
includes review of factoring polynomials
Video Solution by ThePuzzlr
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