2017 AMC 10A Problems/Problem 24
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 . (Using instead of makes the following computations less messy.) Substituting and expanding, we find that
Comparing coefficients with , we see that
Let's solve for and . Since , .
Since , .
(Solution 1.1 branches from here and takes a shortcut.)
.
Then, since , . Thus,
(Solution 1.2 branches from here and takes another shortcut)
Taking , we find that
Solution 1.1
A faster ending to Solution 1 is as follows.
Solution 1.2
Also a faster ending to Solution 1 is as follows.
To find we just need to find the sum of the coefficients which is
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
Solution 7
We first note that where is the quotient function and is the remainder function.
Clearly, because every single root in is also in , thus implying divides . So, we wish to find .
Such an expression for is pretty clean here as we can obtain , so we rewrite . Well, now we need to know how is expressed in order to obtain . This motivates us to long divide to obtain the quotient function. After simple long division . In addition, what is left over, namely , has a constant piece of (you'll see in a few sentences why we only care about particularly the constant piece).
Now we can write: .
Now, as we have already established for ALL that means or the constant piece is , so , in which we obtain . We now plug this back into our equation for to get . ~triggod
General Notes
for any polynomial is simply the sum of the coefficients of the polynomial.
must have real . Both and have all real coefficients, and so odd-degree must have an odd number of real roots, and even-degree must have an even number of real roots, so 's single additional root must be real.
, and is a good number sense fact to know. It's interesting because and the 3 nearest primes to are .
Video Solution by Pi Academy
https://youtu.be/xd_01X989Q0?si=Q7LINw1XRjoF1xSq
~ Pi Academy
Video Solution 2
https://www.youtube.com/watch?v=MBIiz0mroqk (By Richard Rusczyk)
https://youtu.be/3dfbWzOfJAI?t=4412
~ pi_is_3.14
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 |
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