Polynomial Roots Evaluation
Consider the polynomials $ P(x) = x^6-x^5-x^3-x^2-x $ and $ Q(x)=x^4-x^3-x^2-1 $. Given that $ z_1, z_2, z_3 $, and $ z_4 $ are the roots of $ Q(x)=0 $, find $ P(z_1)+P(z_2)+P(z_3)+P(z_4) $.
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Solution
We perform polynomial division with $ P(x) $ as the dividend and $ Q(x) $ as the divisor, giving \[\begin{aligned} P(x) = x^6-x^5-x^3-x^2-x &= (x^2+1) (x^4-x^3-x^2+1) + (x^2-x+1)\\ & = (x^2+1)Q(x) + (x^2-x+1).\end{aligned}\]Thus, if $ z $ is a root of $ Q(x) = 0, $ then the expression for $ P(z) $ is especially simple, because \[\begin{aligned} P(z) &= \cancel{(z^2+1)Q(z)} + (z^2-z+1)\\& = z^2-z+1.\end{aligned}\]It follows that \[\sum_{i=1}^4 P(z_i) = \sum_{i=1}^4 (z_i^2 - z_i + 1).\]By Vieta's formulas, $ \sum_{i=1}^4 z_i = 1, $ and \[\sum_{i=1}^4 z_i^2 = \left(\sum_{i=1}^4 z_i\right)^2 - 2 \sum_{1 \le i < j \le 4} z_i z_j = 1^2 - 2 (-1) = 3.\]Therefore, \[\sum_{i=1}^4 P(z_i) = 3 - 1 + 4 = \boxed{6}.\]
Freeflows