Reciprocal Equations

A polynomial $f(x) = a_0x^n + a_1x^{n-1} + \cdots + a_n$ is called reciprocal, and the equation $f(x)=0$ a reciprocal equation, if its coefficient sequence read forwards is the same as read backwards, either exactly or up to an overall sign flip. Precisely: $f(x)=0$ is reciprocal of Class One if $a_{n-k}=a_k$ for every $k$ (palindromic coefficients), and of Class Two if $a_{n-k}=-a_k$ for every $k$ (anti-palindromic coefficients). This coefficient symmetry is exactly the statement — via the “invert every root” transformation of Section 4.7 — that the roots of the equation come in reciprocal pairs $\alpha, 1/\alpha$.

A few structural facts fall out immediately from the symmetry and are worth knowing before solving: an odd-degree Class One equation always has $-1$ as a root (dividing it out leaves an even-degree Class One equation); an odd-degree Class Two equation always has $1$ as a root; and an even-degree Class Two equation always has both $1$ and $-1$ as roots. So the very first move for an odd-degree or Class-Two reciprocal equation is to peel off that guaranteed root by synthetic division, which always leaves behind an even-degree Class One equation of order $2m$ to finish off.

For an even-degree Class One equation of order $2m$, divide every term by $x^m$ and group symmetric pairs of powers together; the substitution $y = x + \frac{1}{x}$ (using $x^2+\frac1{x^2} = y^2-2$, $x^3+\frac1{x^3}=y^3-3y$, and so on) turns the whole thing into an ordinary degree-$m$ polynomial equation in $y$ — half the original degree. (For an even-degree Class Two equation, after removing the guaranteed roots $\pm 1$ you are again left with a Class One equation, so the same $y=x+1/x$ substitution finishes the job.)

Worked example. Solve $2x^4 - 9x^3 + 14x^2 - 9x + 2 = 0$.

The coefficients $2,-9,14,-9,2$ read the same forwards and backwards, so this is an even-degree ($n=4$) Class One reciprocal equation — no guaranteed root to remove first. Divide through by $x^2$:

Put $y = x + \frac1x$, so $x^2+\frac1{x^2} = y^2-2$. The equation becomes

$2(y^2-2) - 9y + 14 = 0 \;\Longrightarrow\; 2y^2 - 9y + 10 = 0 \;\Longrightarrow\; (2y-5)(y-2)=0,$ …