Problem:
All the roots of polynomial z6β10z5+Az4+Bz3+Cz2+Dz+16 are positive integers, possibly repeated. What is the value of B?
Answer Choices:
A. β88
B. β80
C. β64
D. β41
E. β40
Solution:
Because this polynomial has degree 6, there are 6 roots, counting multiplicities. By Vieta's formulas, the sum of the roots is 10 and their product is 16. The only way this can happen is for the roots, listed with repetitions, to be 1,1,2,2,2,2. Thus the polynomial is (zβ1)2(zβ2)4. By the Binomial Theorem, this polynomial equals
(z2β2z+1)β
(z4β8z3+24z2β32z+16)
When this product is expanded, the coefficient of z3 is B=β32β48β8=(A)β88β.
OR
Proceed as in the first solution. Then, using Vieta's formulas, observe that βB is the sum of the products of the roots taken 3 at a time, namely
(34β)(2β
2β
2)+(24β)(12β)(2β
2β
1)+(14β)(22β)(2β
1β
1)
This gives a total of 32+48+8=88. Therefore B=(A)β88β.
The problems on this page are the property of the MAA's American Mathematics Competitions