Problem:
Let p,q, and r be the distinct roots of the polynomial x3−22x2+ 80x−67. There exist real numbers A,B, and C such that
s3−22s2+80s−671=s−pA+s−qB+s−rC
for all real numbers s with s∈/{p,q,r}. What is A1+B1+C1?
Answer Choices:
A. 243
B. 244
C. 245
D. 246
E. 247
Solution:
Because
x3−22x2+80x−67=(x−p)(x−q)(x−r),
multiplying the given equation by the common denominator yields
1=A(s−q)(s−r)+B(s−p)(s−r)+C(s−p)(s−q)
This is now a polynomial identity that holds for infinitely many values of s, so it must hold for all s. This means the condition that s∈/ {p,q,r} can be removed.
Setting s=p yields 1=A(p−q)(p−r), so A1=(p−q)(p−r). Similarly, B1=(q−p)(q−r) and C1=(r−p)(r−q). Hence
A1+B1+C1=(p−q)(p−r)+(q−p)(q−r)+(r−p)(r−q)=(p+q+r)2−3(pq+qr+rp)
By Viète's Formulas p+q+r is the negative of the coefficient of x2 in the polynomial and pq+qr+rp is the coefficient of the x term, so the requested value is 222−3⋅80=(B)244. (The numerical values (p,q,r,A,B,C) are approximately (1.23,3.08,17.7,0.0329,−0.0371, 0.00416).
The problems on this page are the property of the MAA's American Mathematics Competitions