Problem:
The graph of y=f(x), where f(x) is a polynomial of degree 3 , contains points A(2,4),B(3,9), and C(4,16). Lines AB,AC, and BC intersect the graph again at points D,E, and F, respectively, and the sum of the x-coordinates of D,E, and F is 24 . What is f(0) ?
Answer Choices:
A. −2
B. 0
C. 2
D. 524​
E. 8
Solution:
Let g(x)=f(x)−x2. Then g(2)=g(3)=g(4)=0, so for some constant aî€ =0,g(x)=a(x−2)(x−3)(x−4). Thus the coefficients of x3 and x2 in f(x) are a and 1−9a, respectively, so the sum of the roots of f(x) is 9−a1​. If L(x) is any linear function, then the roots of f(x)−L(x) have the same sum. The given information implies that the sets of roots for three such functions are {2,3,x1​}, {2,4,x2​}, and {3,4,x3​}, where
24=x1​+x2​+x3​=3(9−a1​)−2(2+3+4)=9−a3​,
so a=−51​. Therefore f(x)=x2−51​(x−2)(x−3)(x−4), and f(0)=524​​. (In fact, D=(9,39),E=(8,40),F=(7,37), and the roots of f are 12,1+i, and 1−i.)
The problems on this page are the property of the MAA's American Mathematics Competitions