Past exam of the algebra 1 and differential equations 1 course of the University of Oxford / 2014 / 2 / c / iv / Solution

The fixed points of reciprocal root pairing are $1$ and $-1$, so part (iii) alone does not establish their parity. Use the polynomial identity itself.

Write $P(x)=(x-1{)}^{r}A(x)$, where $A(1)\ne 0$ and allow $r=0$ if one is not a root. After substituting in $P(x)=x^{2m}P(1/x)$ and canceling $(x-1{)}^r$ away from $x=1$, obtainLetting $x\to 1$ yields $A(1)=(-1{)}^{r}A(1)$, so $r$ is even.

Similarly write $P(x)=(x+1{)}^{s}B(x)$ with $B(-1)\ne 0$. This time cancellation gives $B(x)=x^{2m-s}B(1/x)$. Letting $x\to -1$ yields $B(-1)=(-1{)}^{2m-s}B(-1)$, so $s$ is even. ThusThe conclusion includes multiplicity zero. It is a consequence of the reciprocal identity in even degree, which is stronger at these fixed points than just pairing the roots.