Difference between revisions of "2017 AMC 10A Problems/Problem 24"
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r_4=a-1 | r_4=a-1 | ||
\end{align*}</cmath> | \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 | + | The fourth root is <math>a-1</math>. Since <math>r_1</math>, <math>r_2</math>, and <math>r_3</math> are common roots, we have: |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
f(x)=g(x)(x-(a-1)) \\ | f(x)=g(x)(x-(a-1)) \\ |
Revision as of 22:09, 3 March 2017
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
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 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. 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.