2022 AMC 12A Problems/Problem 21
Contents
Problem
Let Which of the following polynomials is a factor of
?
Solution 1
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
.
Solution 2
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
~bluesoul
Solution 3
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
.
- DavidHovey
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
~pianoboy
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
.
~SouradipClash_03
Video Solution
includes review of factoring polynomials
~MathProblemSolvingSkills.com
Video Solution by ThePuzzlr
~ MathIsChess
See Also
2022 AMC 12A (Problems • Answer Key • Resources) | |
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 | |
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