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∑4P(zi)=i=1∑4zi2−i=1∑4zi+4=(i=1∑4zi)2−2i<j∑zizj−i=1∑4zi+4
Use the formulas for sum and product of the roots to obtain ∑i=14P(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∑4P(zi)=i=1∑4zi2−i=1∑4zi+4
and, as above, ∑i=14P(zi)=6.
The problems on this page are the property of the MAA's American Mathematics Competitions