Problem:
Consider the polynomials P(x)=x6−x5−x3−x2−x and Q(x)=x4−x3−x2−1. Given that z1​,z2​,z3​, and z4​ are the roots of Q(x)=0, find P(z1​)+P(z2​)+ P(z3​)+P(z4​).
Solution:
Apply the division algorithm for polynomials to obtain
P(x)=Q(x)(x2+1)+x2−x+1
Therefore
i=1∑4​P(zi​)=i=1∑4​zi2​−i=1∑4​zi​+4=(i=1∑4​zi​)2−2i<j∑​zi​zj​−i=1∑4​zi​+4
Use the formulas for sum and product of the roots to obtain ∑i=14​P(zi​)=1+2− 1+4=6.
OR
Since, for each root w of Q(x)=0, we have w4−w3−w2−1=0, conclude that w4−w3=w2+1, and then w6−w5=w4+w2=w3+2w2+1. Thus P(w)=w3+2w2+1−w3−w2−w=w2−w+1. Therefore
i=1∑4​P(zi​)=i=1∑4​zi2​−i=1∑4​zi​+4
and, as above, ∑i=14​P(zi​)=6​.
The problems on this page are the property of the MAA's American Mathematics Competitions