Factor the Polynomial

By putting $x=\omega$ , where $\omega$ is the cube root of unity, we note that $f(\omega) = \omega^{10}+\omega^5 + 1 = \omega + \omega^2 + 1 = 0$ . This means that $x^2+x+1$ is a factor of $f(x)$ . By long division we have $f(x) = (x^2+x+1)(x^8-x^7+x^5-x^4+x^3-x+1)$ (see note below), Let $g(x) = x^2 + x+1$ and $h(x) = x^8-x^7+x^5-x^4+x^3-x+1$ . Then $g(2) = 7$ and $h(2) = 151$ and $g(2) + h(2) = \boxed{158}$ .

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Note:

$\begin{aligned} f(x) & =x^{10} + x^5 + 1 \\ & = x^8(x^2 +x+1) - x^7(x^2+x+1) + x^5(x^2+x+1) - x^4(x^2+x+1) +x^3(x^2+x+1) - x(x^2 + x+1) + x^2+x+1 \\ & = (x^2+x+1)(x^8-x^7+x^5-x^4+x^3-x+1) \end{aligned}$

$f(x) = x^{10} + x^5 + 1 = (x^5)^2 + x^5 + 1 = \dfrac{(x^5)^3 - 1}{x^5 - 1} = \dfrac{(x^3)^5 -1}{x^5 -1} = \dfrac{(x^3-1)(x^{12} + x^9 + x^6 + x^3 + 1)}{x^5 - 1}$ $= \dfrac{(x-1)(x^2+x+1)(x^{12} + x^9 + x^6 + x^3 + 1)}{(x-1)(x^4 + x^3 + x^2 + x + 1)} = \dfrac{(x^2+x+1)(x^{12} + x^9 + x^6 + x^3 + 1)}{(x^4 + x^3 + x^2 + x + 1)}$ By long division, $x^{12} + x^9 + x^6 + x^3 + 1 = (x^4 + x^3 + x^2 + x + 1)(x^8 - x^7 + x^5 - x^4 + x^3 - x + 1)$ so that $f(x) = \dfrac{(x^2+x+1)(x^4 + x^3 + x^2 + x + 1)(x^8 - x^7 + x^5 - x^4 + x^3 - x + 1)}{(x^4 + x^3 + x^2 + x + 1)} \\ = (x^2 + x + 1)(x^8 - x^7 + x^5 - x^4 + x^3 - x + 1) = g(x)h(x)$ $g(2) + h(2) = 7 + 151 = 158$