Algebra Problem From Yesterday’s Simulation – Mattia Giuri's bizarre blog

Given two polynomials $q(x) = x^4-x^3-x^2-1$ and $p(x) = x^6-x^5-x^3-x^2-x$ , let a, b, c, d be the roots of $q(x)$ , find

$p(a)+p(b)+p(c)+p(d)$

First, notice that $p(x) = (x^2+1)q(x)+x^2-x+1$

We know that plugging a, b, c, or d into $q(x)$ gives 0, so we have to find the sum of squares of the roots minus the sum of the roots plus 4.

By Vieta’s Theorem, the sum of the roots equals the negative ratio of the coefficients of degree 3 and 4, that is to say 1.

Also by Vieta’s Theorem, the sum of the products of any two roots equals the ratio of the coefficients coefficients of degree 2 and 4, that is to say -1.

So the sum of the squares equals $1^2-2(-1) = 3$

So the final sum equals

$3-1+4 = 6$