2010 AMC 10A Problems/Problem 21
The polynomial has three positive integer roots. What is the smallest possible value of ?
By Vieta's Formulas, we know that is the sum of the three roots of the polynomial . Again Vieta's Formulas tell us that is the product of the three integer roots. Also, factors into . But, since the polynomial has only three roots, two of the four prime factors must be multiplied so that we are left with three roots. To minimize , and should be multiplied, which means will be and the answer is . ~JimPickens
If you are feeling unconfident about , you can try to expand the expression which has roots , , and .
As we can see, , and since this is the least answer choice, we can be confident that the right option is .
We can expand as
We do not care about in this case, because we are only looking for . We know that the constant term is We are trying to minimize a, such that we have Since we have three positive solutions, we have as our factors. We have to combine two of the factors of , and then sum up the resulting factors. Since we are minimizing, we choose and to combine together. We get which gives us a coefficient of of Therefore or
We want the polynomial to have POSITIVE integer roots. That means we want to factor it in to the form We therefore want the prime factorization for . The prime factorization of is . We want the smallest difference of the roots since by Vieta's formulas, is the sum of the roots.
We proceed to factorize it in to . Therefore, our answer is = .
Suggestion for the author: The variables and are already defined as coefficients of the cubic polynomial. So, consider using different variables when describing how we want to factor this polynomial in the second sentence. Thanks!
We can check has the smallest possible sum because of the following reasons:
: We don't want to multiply by anything since that would make the sum of the roots too big.
: The smallest number should multiply since that would make the numbers optimally small. Therefore, we want times .
Video Solution 1
Video Solution 2
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