Problem:
The graph of y=x6−10x5+29x4−4x3+ax2 lies above the line y=bx+c except at three values of x, where the graph and the line intersect. What is the largest of those values?
Answer Choices:
A. 4
B. 5
C. 6
D. 7
E. 8
Solution:
Let the three points of intersections have x-coordinates p,q, and r, and let f(x)=x6−10x5+29x4−4x3+ax2−bx−c. Then f(p)= f(q)=f(r)=0, and f(x)≥0 for all x, so f(x)=((x−p)(x−q)(x−r))2= (x3−Ax2+Bx−C)2, where A=p+q+r,B=pq+qr+rp, and C=pqr. The coefficient of x5 is −10=−2A, so A=5. The coefficient of x4 is 29=A2+2B= 25+2B, so B=2. The coefficient of x3 is −4=−2C−2AB=−2C−20, so C=−8. Thus f(x)=(x3−5x2+2x+8)2. Because the sums of the coefficients of the even and odd powers of x are equal, −1 is a zero of f(x). Factoring gives f(x)=((x+1)(x2−6x+8))2=((x+1)(x−2)(x−4))2, and the largest of the three zeros is 4.
The problems on this page are the property of the MAA's American Mathematics Competitions