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 aî€ =b. In each case, k=a+b. The choices for a and b are then (1,36),(−1,−36),(2,18),(−2,−18),(3,12), (−3,−12),(4,9), and (−4,−9), yielding k=37,−37,20,−20,15,−15,13, and -13 , respectively. Thus there are (B)8​ values for k. (Note that the ordered pairs (a,b) and (b,a) produce the same value of k, so the pairs (36,1),(−36,−1),(18,2),(−18,−2) and so forth can be ignored.)
The problems on this page are the property of the MAA's American Mathematics Competitions