Problem:
For how many ordered pairs (a,b) of integers does the polynomial x3+ax2+bx+6 have 3 distinct integer roots?
Answer Choices:
A. 5
B. 6
C. 8
D. 7
E. 4
Solution:
Let r,s, and t be the roots of x3+ax2+bx+6. Then
x3+ax2+bx+6=(x−r)(x−s)(x−t)=x3−(r+s+t)x2+(rs+st+tr)x−rst
so rst=−6,r+s+t=−a, and rs+st+tr=b. Because r,s, and t are distinct integers and rst=−6, the possible roots {r,s,t} are the sets {−1,1,6},{−1,2,3},{1,−2,3},{1,2,−3}, and {−1,−2,−3}. The corresponding ordered pairs (a,b) are (−6,−1),(−4,1),(−2,−5),(0,−7), and (6,11), respectively. This gives (A)5 possible ordered pairs (a,b).
The problems on this page are the property of the MAA's American Mathematics Competitions