Let P(x) = 1 + x + x^2 + x^3 + x^4 + x^5. Wha | vedclass

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(B) We have $P(x) = 1 + x + x^2 + x^3 + x^4 + x^5 = \frac{1 - x^6}{1 - x}$.Note that $P(x) = 0$ for $x = \omega^k$ where $\omega = e^{i \frac{2\pi}{6}

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$6$

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(B) We have $P(x) = 1 + x + x^2 + x^3 + x^4 + x^5 = \frac{1 - x^6}{1 - x}$.Note that $P(x) = 0$ for $x = \omega^k$ where $\omega = e^{i \frac{2\pi}{6}}$ and $k \in \{1, 2, 3, 4, 5\}$.We want to find the remainder $R(x)$ when $P(x^{12})$ is divided by $P(x)$.$P(x^{12}) = 1 + (x^{12}) + (x^{12})^2 + (x^{12})^3 + (x^{12})^4 + (x^{12})^5$.For any root $\alpha$ of $P(x)$,we have $\alpha^6 = 1$.Thus,$\alpha^{12} = (\alpha^6)^2 = 1^2 = 1$.Substituting $x = \alpha$ into $P(x^{12})$,we get $P(\alpha^{12}) = 1 + 1 + 1^2 + 1^3 + 1^4 + 1^5 = 6$.Since $P(x^{12}) = P(x)Q(x) + R(x)$,and $P(\alpha) = 0$,we have $R(\alpha) = P(\alpha^{12}) = 6$ for all $5$ roots of $P(x)$.Since $R(x)$ is a polynomial of degree at most $4$ and it takes the value $6$ at $5$ distinct points,$R(x)$ must be the constant polynomial $6$.

Let $P(x) = 1 + x + x^2 + x^3 + x^4 + x^5$. What is the remainder when $P(x^{12})$ is divided by $P(x)$?

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