The polynomial $f(x)=x^{2007}+17x^{2006}+1$ has distinct zeroes $r_1,\ldots,r_{2007}$. A polynomial $P$ of degree $2007$ has the property that $P\left(r_j+\dfrac{1}{r_j}\right)=0$ for $j=1,\ldots,2007$. Determine the value of $P(1)/P(-1)$.
Let $P(x) = a_{2007}x^{2007} + … + a_0$.
$r_j + \frac{1}{r_j} = \frac{r_j^2 + 1}{r_j}$
So we need the map: $x + \frac{1}{x} \to x$.
We need a $u \to x$. Where $u(x + 1/x) = x$.
The inverse function is: $u = \frac{1}{2} \cdot( x \pm \sqrt{x^2 – 4})$ [USED WOLFRAM ALPHA]
So,
$P(x) = \bigg(\frac{1}{2} \cdot( x \pm \sqrt{x^2 – 4}) \bigg)^{2007} + 17 \bigg(\frac{1}{2} \cdot( x \pm \sqrt{x^2 – 4}) \bigg)^{2006} + 1$
But letting $x= 1$ or $-1$ makes it so that the answer is imaginary!
Best Answer
For first, $\frac{1}{r_i}+r_i = \frac{1}{r_j}+r_j$ is equivalent to $(r_i-r_j)(r_i r_j - 1)=0$. Since $f(x)$ is not a palyndromic polynomial, the set of numbers $r_i+\frac{1}{r_i}$ is exactly the set of roots of $P$.
Assuming that $P$ is a monic polynomial, Vieta's formulas give: $$ P(1) = \prod_{i=1}^{2007}\left(1-r_i-\frac{1}{r_i}\right),\qquad P(-1)=-\prod_{i=1}^{2007}\left(1+r_i+\frac{1}{r_i}\right)\tag{1} $$ hence it follows that: $$ \frac{P(1)}{P(-1)} = \prod_{i=1}^{2007}\frac{r_i^2-r_i+1}{r_i^2+r_i+1}=\prod_{i=1}^{2007}\frac{r_i-1}{r_i+1}\cdot\prod_{i=1}^{2007}\frac{r_i^3-1}{r_i^3+1}\tag{2} $$ where the first term of the RHS can be easily computed from $f(1)$ and $f(-1)$, and the whole expression just depends on the values of $f$ over the sixth roots of unity, due to $r_i^2+r_i+1 = (r_i-\omega^2)(r_i-\omega^4)$, where $\omega = \exp\left(\frac{2\pi i}{6}\right)$. Given $(2)$, it follows that: $$ \frac{P(1)}{P(-1)} = \frac{f(\omega)\cdot f(\omega^5)}{f(\omega^2)\cdot f(\omega^4)}=\frac{17\omega^2\cdot (-17\omega)}{(2-17\omega)\cdot(2+17\omega^2)}=\color{red}{\frac{1}{\frac{297}{578}-\frac{i\sqrt{3}}{2}}}.\tag{3}$$