Difference between revisions of "2017 AMC 10A Problems/Problem 24"
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Hence <math>f(1)=91\cdot(-77)=\boxed{\textbf{(C)}\,-7007}</math>. | Hence <math>f(1)=91\cdot(-77)=\boxed{\textbf{(C)}\,-7007}</math>. | ||
+ | |||
+ | ==Solution 4 (Slight guessing)== | ||
+ | Let the roots of <math>g(x)</math> be <math>r_1</math>, <math>r_2</math>, and <math>r_3</math>. Let the roots of <math>f(x)</math> be <math>r_1</math>, <math>r_2</math>, <math>r_3</math>, and <math>r_4</math>. From Vieta's, we have: | ||
+ | <cmath>\begin{align*} | ||
+ | r_1+r_2+r_3=-a \ | ||
+ | r_1+r_2+r_3+r_4=-1 \ | ||
+ | r_4=a-1 | ||
+ | \end{align*}</cmath> | ||
+ | The fourth root is <math>a-1</math>. Since <math>r_1</math>, <math>r_2</math>, and <math>r_3</math> are coon roots, we have: | ||
+ | <cmath>\begin{align*} | ||
+ | f(x)=g(x)(x-(a-1)) \ | ||
+ | f(1)=g(1)(1-(a-1)) \ | ||
+ | f(1)=(a+12)(2-a) \ | ||
+ | f(1)=-(a+12)(a-2) \ | ||
+ | \end{align*}</cmath> | ||
+ | Let <math>a-2=k</math>: | ||
+ | <cmath>\begin{align*} | ||
+ | f(1)=-k(k+14) | ||
+ | \end{align*}</cmath> | ||
+ | Note that <math>-7007=-1001\cdot(7)=-(7\cdot(11)\cdot(13))\cdot(7)=-91\cdot(77)</math> | ||
+ | This gives us a pretty good guess of <math>\boxed{\textbf{(C)}\,-7007}</math> | ||
==See Also== | ==See Also== |
Revision as of 18:54, 8 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
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
or
.
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
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.