Difference between revisions of "2017 AMC 10A Problems/Problem 24"
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==Solution 5== | ==Solution 5== | ||
− | First off, let's get rid of the <math>x^4</math> term by finding <math>h(x)=f(x)-xg(x)</math>. This polynomial consists of the difference of two polynomials with <math>3</math> common factors, so it must also have these factors. The polynomial is <math>h(x)=(1-a)x^3 + (b-1)x^2 + 90x + c</math>, and must be equal to <math>(1-a)g(x)</math>. Equating the coefficients, we get <math>4</math> equations. We will tackle | + | First off, let's get rid of the <math>x^4</math> term by finding <math>h(x)=f(x)-xg(x)</math>. This polynomial consists of the difference of two polynomials with <math>3</math> common factors, so it must also have these factors. The polynomial is <math>h(x)=(1-a)x^3 + (b-1)x^2 + 90x + c</math>, and must be equal to <math>(1-a)g(x)</math>. Equating the coefficients, we get <math>4</math> equations. We will tackle the situation one equation at a time, starting the <math>x</math> terms. <cmath>90=1-a</cmath> The solution is obviously <math>a=-89</math>. We can now find b and c with ease. <cmath>b-1=a(1-a)=-89*90=-8010</cmath> and <math>b=-8009</math>. <cmath>c=10(1-a)=10*90=900</cmath> Finally, <math>f(1)=102+b+c=102+900-8009=\boxed{\textbf{(C)}\, -7007}</math>. |
==See Also== | ==See Also== |
Revision as of 23:43, 12 February 2017
Contents
[hide]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
Let's solve for and
. Since
,
, so
. Since
,
, and
. Thus, we know that
Taking , we find that
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 (Slight guessing)
Let the roots of be
,
, and
. Let the roots of
be
,
,
, and
. From Vieta's, we have:
The fourth root is
. Since
,
, and
are coon 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.
The solution is obviously
. We can now find b and c with ease.
and
.
Finally,
.
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.