Difference between revisions of "2017 AMC 10A Problems/Problem 24"
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==Solution 6== | ==Solution 6== | ||
Simple polynomial division is a feasible method. Even though we have variables, we can equate terms at the end of the division so that we can cancel terms. Doing the division of <math>\frac{f(x)}{g(x)}</math> eventually brings us the final step <math>(1-a)x^3 + (b-1)x^2 + 90x + c</math> minus <math>(1-a)x^3 - (a-a^2)x^2 + (1-a)x + 10(1-a)</math> after we multiply <math>f(x)</math> by <math>(1-a)</math>. Now we equate coefficients of same-degree <math>x</math> terms. This gives us <math> 10(1-a) = c, b-1 = a - a^2, 1-a = 90 \Rightarrow a = -89, c = 900, b = -8009</math>. We are interested in finding <math>f(1)</math>, which equals <math>1^4 + 1^3 -8009\cdot1^2 + 100\cdot1 + 900 = \boxed{\textbf{(C)}\,-7007}</math>. ~skyscraper | Simple polynomial division is a feasible method. Even though we have variables, we can equate terms at the end of the division so that we can cancel terms. Doing the division of <math>\frac{f(x)}{g(x)}</math> eventually brings us the final step <math>(1-a)x^3 + (b-1)x^2 + 90x + c</math> minus <math>(1-a)x^3 - (a-a^2)x^2 + (1-a)x + 10(1-a)</math> after we multiply <math>f(x)</math> by <math>(1-a)</math>. Now we equate coefficients of same-degree <math>x</math> terms. This gives us <math> 10(1-a) = c, b-1 = a - a^2, 1-a = 90 \Rightarrow a = -89, c = 900, b = -8009</math>. We are interested in finding <math>f(1)</math>, which equals <math>1^4 + 1^3 -8009\cdot1^2 + 100\cdot1 + 900 = \boxed{\textbf{(C)}\,-7007}</math>. ~skyscraper | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/wXvRNC-48Lk | ||
==See Also== | ==See Also== |
Revision as of 18:32, 6 January 2020
Contents
Problem
For certain real numbers ,
, and
, the polynomial
has three distinct roots, and each root of
is also a root of the polynomial
What is
?
Solution 1
must have four roots, three of which are roots of
. Using the fact that every polynomial has a unique factorization into its roots, and since the leading coefficient of
and
are the same, we know that
where is the fourth root of
. Substituting
and expanding, we find that
Comparing coefficients with , we see that
(Solution 1.1 picks up here.)
Let's solve for and
. Since
,
, so
. Since
,
, and
. Thus, we know that
Taking , we find that
Solution 1.1
A faster ending to Solution 1 is as follows.
We shall solve for only and
. Since
,
, and since
,
. Then,
Solution 1.2 (Fast)
Let the term be the linear term that we are solving for in the equation
. Now, we know that
must have
, because only
. In addition, we know that, by distributing,
. Therefore,
, and all the other variables are quickly solved for.
Solution 2
We notice that the constant term of and the constant term in
. Because
can be factored as
(where
is the unshared root of
, we see that using the constant term,
and therefore
.
Now we once again write
out in factored form:
.
We can expand the expression on the right-hand side to get:
Now we have .
Simply looking at the coefficients for each corresponding term (knowing that they must be equal), we have the equations:
and finally,
.
We know that is the sum of its coefficients, hence
. We substitute the values we obtained for
and
into this expression to get
.
Solution 3
Let and
be the roots of
. Let
be the additional root of
. Then from Vieta's formulas on the quadratic term of
and the cubic term of
, we obtain the following:
Thus .
Now applying Vieta's formulas on the constant term of , the linear term of
, and the linear term of
, we obtain:
Substituting for in the bottom equation and factoring the remainder of the expression, we obtain:
It follows that . But
so
Now we can factor in terms of
as
Then and
Hence .
Solution 4 (Risky)
Let the roots of be
,
, and
. Let the roots of
be
,
,
, and
. From Vieta's, we have:
The fourth root is
. Since
,
, and
are common roots, we have:
Let
:
Note that
This gives us a pretty good guess of
.
Solution 5
First off, let's get rid of the term by finding
. This polynomial consists of the difference of two polynomials with
common factors, so it must also have these factors. The polynomial is
, and must be equal to
. Equating the coefficients, we get
equations. We will tackle the situation one equation at a time, starting the
terms. Looking at the coefficients, we get
.
The solution to the previous is obviously
. We can now find
and
.
,
and
. Finally
,
Solving the original problem,
.
Solution 6
Simple polynomial division is a feasible method. Even though we have variables, we can equate terms at the end of the division so that we can cancel terms. Doing the division of eventually brings us the final step
minus
after we multiply
by
. Now we equate coefficients of same-degree
terms. This gives us
. We are interested in finding
, which equals
. ~skyscraper
Video Solution
See Also
2017 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 10 Problems and Solutions |
2017 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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.