Problem:
For how many values of the constant k will the polynomial x2+kx+36 have two distinct integer roots?
Answer Choices:
A. 6
B. 8
C. 9
D. 14
E. 16
Solution:
In order for the given polynomial to have two distinct integer roots, the polynomial must factor as the product (x+a)(x+b), where a and b are integers, with ab=36 and aξ =b. In each case, k=a+b. The choices for a and b are then (1,36),(β1,β36),(2,18),(β2,β18),(3,12), (β3,β12),(4,9), and (β4,β9), yielding k=37,β37,20,β20,15,β15,13, and -13 , respectively. Thus there are (B)8β values for k. (Note that the ordered pairs (a,b) and (b,a) produce the same value of k, so the pairs (36,1),(β36,β1),(18,2),(β18,β2) and so forth can be ignored.)
The problems on this page are the property of the MAA's American Mathematics Competitions