Show that $x-1$ is a factor of $P(x)$ where $P(x^5)+xQ(x^5)+x^2R(x^5)=\big(x^4+x^3+x^2+x+1\big)S(x)$

Question

For the problem below, my first step was to rewrite the right side as

$$\big(x^4+x^3+x^2+x+1\big)S(x)=\frac{x^5-1}{x-1}S(x)$$

From there I can isolate $P(x^5)$ which gives me

$$P(x^5)=\dfrac{S(x)(x^5-1)-x(x-1)\big(Q(x^5)-xR(x^5)\big)}{x-1}$$

I'm not sure how to move forward. To get $P(x)$ should I rewrite all the $x^5$ as $x$ and take the $5^{th}$ root of each $x$? The questions prior used the idea of roots of unity so perhaps I can use that somehow.

Thanks

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Answer 1

Hint: You know that if $P(1)=0$, then $x-1$ is a factor of $P$. Try plugging in special values of $x$ into the given functional equation. Try certain algebraic numbers with nice properties, such as, for example, the roots of unity.