Difference between revisions of "2022 AMC 12A Problems/Problem 21"
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==Solution 3== | ==Solution 3== | ||
− | Let <math>x^{1011} = u</math>, now we can rewrite our polynomial as <math>u^2+u+1</math>. Using the quadratic formula to solve for the roots of this polynomial, we have <cmath>x^{1011} = \frac{-1\pm i\sqrt{3}}{2}</cmath> Looking at our answer choices, we want to find a polynomial whose roots satisfy this expression. Since the expression <math>x^6+x^3+1</math> is in a similar form to our original polynomial, except with <math>x^3</math> in place of <math>x^{1011}</math>, this would be a good place to start. Solving for the roots of <math>x^3</math> in a similar fashion, <cmath>x^3= \frac{-1\pm i\sqrt{3}}{2}</cmath> for the solution we are testing. Now notice that we can rewrite the roots of <math>x^3</math> as <cmath>x^3 = \operatorname{cis}{\frac{2\pi}{3}}, \operatorname{cis}{\frac{4\pi}{3}}</cmath> Both of which are third roots of unity. We want to now check if this value of <math>x^3</math> satisfies <math>x^{1011} = \frac{-1\pm i\sqrt{3}}{2}</math>. Notice that <math>x^{1011} = (x^{3})^{112\cdot3}\cdot x^3</math>, and since both values of <math>x^3</math> are roots of unity, we can simplify the expression we want | + | Let <math>x^{1011} = u</math>, now we can rewrite our polynomial as <math>u^2+u+1</math>. Using the quadratic formula to solve for the roots of this polynomial, we have <cmath>x^{1011} = \frac{-1\pm i\sqrt{3}}{2}</cmath> Looking at our answer choices, we want to find a polynomial whose roots satisfy this expression. Since the expression <math>x^6+x^3+1</math> is in a similar form to our original polynomial, except with <math>x^3</math> in place of <math>x^{1011}</math>, this would be a good place to start. Solving for the roots of <math>x^3</math> in a similar fashion, <cmath>x^3= \frac{-1\pm i\sqrt{3}}{2}</cmath> for the solution we are testing. Now notice that we can rewrite the roots of <math>x^3</math> as <cmath>x^3 = \operatorname{cis}{\frac{2\pi}{3}}, \operatorname{cis}{\frac{4\pi}{3}}</cmath> Both of which are third roots of unity. We want to now check if this value of <math>x^3</math> satisfies <math>x^{1011} = \frac{-1\pm i\sqrt{3}}{2}</math>. Notice that <math>x^{1011} = (x^{3})^{112\cdot3}\cdot x^3</math>, and since both values of <math>x^3</math> are roots of unity, we can simplify the expression we want satisfied to the expression to <math>x^{1011}=x^3</math>. Since both values of <math>x^3</math> are also values of <math>x^{1011}</math>, the roots for our <math>x^6+x^3+1</math> are also roots of <math>x^{2022}+x^{1011}+1</math>, meaning that <cmath>x^6+x^3+1 | x^{2022}+x^{1011}+1</cmath> so Therefore, the answer is <math>\boxed{E}</math>. |
- DavidHovey | - DavidHovey | ||
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Put the value <math>x = -1</math>. This gives <math>P(-1) = (-1)^{2022} + (-1)^{1011} + 1 = 1</math>. | Put the value <math>x = -1</math>. This gives <math>P(-1) = (-1)^{2022} + (-1)^{1011} + 1 = 1</math>. | ||
This automatically eliminates choices <math>A</math> , <math>C</math> and <math>D</math> since they do not form a factor of <math>1</math> at <math>x = -1</math>. | This automatically eliminates choices <math>A</math> , <math>C</math> and <math>D</math> since they do not form a factor of <math>1</math> at <math>x = -1</math>. | ||
+ | |||
+ | |||
+ | Solution <math>5.1</math>: | ||
Now, put <math>x = \omega</math> (omega) at choice <math>B</math>. We get that, | Now, put <math>x = \omega</math> (omega) at choice <math>B</math>. We get that, | ||
<cmath>\omega^2 + \omega + 1 = 0</cmath> | <cmath>\omega^2 + \omega + 1 = 0</cmath> | ||
Thus choice B has <math>(x - \omega)</math> as a factor. If choice <math>B</math> were to be a factor of <math>P(x)</math>, <math>(x - \omega)</math> would also have to be a factor of <math>P(x)</math>, which is clearly not the case, as <math>P(\omega) \neq 0</math>. | Thus choice B has <math>(x - \omega)</math> as a factor. If choice <math>B</math> were to be a factor of <math>P(x)</math>, <math>(x - \omega)</math> would also have to be a factor of <math>P(x)</math>, which is clearly not the case, as <math>P(\omega) \neq 0</math>. | ||
This eliminates choice <math>B</math>, leaving us with answer <math>\boxed{E}</math>. | This eliminates choice <math>B</math>, leaving us with answer <math>\boxed{E}</math>. | ||
+ | |||
+ | |||
+ | Solution <math>5.2</math>(Mods): | ||
+ | Plug in <math>2</math> for <math>x</math> at choice <math>B</math>. <math>2^2+2*2+1=7</math>. | ||
+ | The cycle of mods for <math>2</math> and <math>7</math> is <math>2</math>, <math>4</math>, <math>1</math>. Both <math>2022</math> and <math>1011</math> are divisible by <math>3</math>, so we have <math>1</math> + <math>1</math> + <math>1</math> = <math>3</math>. <math>3</math> is not divisible by <math>7</math>, so we eliminate choice <math>B</math>. This leaves us with <math>\boxed{(E) x^{6}+x^{3}+1}</math>. | ||
+ | |||
~SouradipClash_03 | ~SouradipClash_03 | ||
+ | |||
+ | Solution <math>5.2</math> ~ Bread 10 | ||
==Solution 6 (Elimination but slightly different)== | ==Solution 6 (Elimination but slightly different)== | ||
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Through quadratic formula, we find the roots of this quadratic as <math>\dfrac{-1+i\sqrt{3}}{2}</math> and <math>\dfrac{-1-i\sqrt{3}}{2}</math>. | Through quadratic formula, we find the roots of this quadratic as <math>\dfrac{-1+i\sqrt{3}}{2}</math> and <math>\dfrac{-1-i\sqrt{3}}{2}</math>. | ||
− | We notice that these roots can be written nicely in polar form <math>\ | + | We notice that these roots can be written nicely in polar form <math>\cos(\dfrac{2\pi}{3})+i\sin(\dfrac{2\pi}{3})</math> or <math>\cos(\dfrac{4\pi}{3})+i\sin(\dfrac{4\pi}{3})</math>. |
− | We plug either one of these and see that the polynomial doesn't equal <math>0</math> suggesting that <math>B</math> is not the correct answer choice. | + | We plug either one of these (preferably the root on the left as it is smaller) and see that the polynomial doesn't equal <math>0</math> suggesting that <math>B</math> is not the correct answer choice. |
As we only have one answer choice left, we choose <math>\boxed{E}</math> | As we only have one answer choice left, we choose <math>\boxed{E}</math> | ||
+ | |||
+ | ~Batmanstark | ||
+ | |||
+ | ==Video Solution by Math-X (Smart and Simple)== | ||
+ | https://youtu.be/7yAh4MtJ8a8?si=OkLQzwiqd9AnIS5K&t=6374 ~Math-X | ||
== Video Solution== | == Video Solution== |
Latest revision as of 02:14, 11 April 2024
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 satisfied to 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
.
Solution :
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 (Mods):
Plug in
for
at choice
.
.
The cycle of mods for
and
is
,
,
. Both
and
are divisible by
, so we have
+
+
=
.
is not divisible by
, so we eliminate choice
. This leaves us with
.
~SouradipClash_03
Solution ~ Bread 10
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
~Batmanstark
Video Solution by Math-X (Smart and Simple)
https://youtu.be/7yAh4MtJ8a8?si=OkLQzwiqd9AnIS5K&t=6374 ~Math-X
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 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.