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

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

## closed as off-topic by heropup, user223391, Pragabhava, Watson, Stefan4024Sep 22 '16 at 22:29

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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – heropup, Community, Pragabhava, Watson, Stefan4024
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• should there be exponents? – Jorge Fernández Hidalgo Sep 22 '16 at 16:36
• When you work with higher degree polynomial and carry out multiplication and division you are not doing Linear Algebra. – P Vanchinathan Sep 22 '16 at 16:39
• – Jyrki Lahtonen Sep 22 '16 at 17:09
• Please improve your question by mentioning your attempts. There is an interesting interpolation approach related with the sixth roots of unity. – Jack D'Aurizio Sep 23 '16 at 0:45
• In particular, the remainder is $\color{red}{6}$ because the value of $Q(x)=P(x^{12})$ at any sixth root of unity is six. – Jack D'Aurizio Sep 23 '16 at 0:53

HINT: $$P(x) = \frac{x^6 - 1}{x-1}, \quad \text{when} \quad x \not = 1$$
• Secondary hint: $a^2-b^2$ – Hagen von Eitzen Sep 24 '16 at 9:24