Now evaluate at \(x = \omega\):

February 22, 2026

$Now evaluate at \(x = \omega\):$

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Solution: We perform polynomial division or use the fact that the remainder when dividing by a quadratic is linear: let
x^4 + 3x^3 - 2x^2 + x + 5 = (x^2 - x + 1)Q(x) + ax + b
Let \(\omega\) be a root of \(x^2 - x + 1 = 0\). Then \(\omega^2 = \omega - 1\), and \(\omega^3 = \omega(\omega - 1) = \omega^2 - \omega = (\omega - 1) - \omega = -1\). So \(\omega^3 = -1\), and \(\omega^6 = 1\).
\omega^4 + 3\omega^3 - 2\omega^2 + \omega + 5 = a\omega + b
We compute each term:
\(\omega^4 = \omega \cdot \omega^3 = \omega(-1) = -\omega\)

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