Reciprocal polynomial equations: symmetry, substitution, and applications

What is a reciprocal polynomial?

A polynomial is reciprocal, or palindromic, when its coefficients read the same in ascending and descending order. For example,

$$x^4+2x^3+3x^2+2x+1$$

has coefficient list $1,2,3,2,1$. If $P$ has degree $d$, the symmetry is expressed by $P(x)=x^dP(1/x)$ for nonzero $x$.

This structure can reduce the degree of an equation substantially. It does not guarantee a simple radical formula for equations of arbitrarily high degree, but it provides a systematic first reduction.

Even degree: divide by the middle power

Example. Solve

$$x^4-x^3-4x^2-x+1=0.$$

The constant term is one, so $x=0$ is not a solution. Divide by $x^2$:

$$x^2+\frac1{x^2}-\left(x+\frac1x\right)-4=0.$$

Set $X=x+1/x$. Since $x^2+x^{-2}=X^2-2$, the equation becomes

$$X^2-X-6=0,$$

giving $X=-2$ or $X=3$.

• $x+1/x=-2$ gives $(x+1)^2=0$, so $x=-1$.
• $x+1/x=3$ gives $x^2-3x+1=0$, so $x=(3\pm\sqrt5)/2$.

Thus the distinct solutions are

$$\boxed{x=-1,\quad\frac{3+\sqrt5}{2},\quad\frac{3-\sqrt5}{2}.}$$

The root $-1$ has multiplicity two; indeed the quartic factors as $(x+1)^2(x^2-3x+1)$.

Why the substitution works in general

After a reciprocal polynomial of degree $2m$ is divided by $x^m$, terms occur in pairs $x^j+x^{-j}$, together with a possible constant term.

Write $T_j=x^j+x^{-j}$. Then

$$T_0=2,\qquad T_1=X,\qquad T_j=XT_{j-1}-T_{j-2}.$$

By induction every $T_j$ is a polynomial in $X$. The original degree-$2m$ equation therefore reduces to one of degree at most $m$ in $X$. This is also an instance of expressing a symmetric polynomial in $x,1/x$ through their sum $X$ and product one.

Odd degree: first remove x + 1

Every odd-degree reciprocal polynomial has $P(-1)=0$: symmetric terms cancel in pairs. Hence it has a factor $x+1$.

For example,

$$x^5-5x^3-5x^2+1=(x+1)(x^4-x^3-4x^2-x+1).$$

The remaining quartic is the one already solved. Consequently the distinct roots are again $-1,(3+\sqrt5)/2,(3-\sqrt5)/2$, with $-1$ now having multiplicity three.

The zero coefficients of $x^4$ and $x$ are part of the palindromic coefficient list. Including them makes the symmetry visible.

Why the quotient remains reciprocal

If $P(x)=(x+1)Q(x)$ has degree $2m+1$, reciprocity gives

$$(x+1)Q(x)=x^{2m+1}\left(1+\frac1x\right)Q(1/x)=(x+1)x^{2m}Q(1/x).$$

Canceling for nonzero $x\ne-1$ proves $Q(x)=x^{2m}Q(1/x)$. Since this is a polynomial identity after clearing negative powers, it establishes the coefficient symmetry of the degree-$2m$ quotient.

Application 1: the fifth roots of unity

Problem. Find all complex numbers satisfying $z^5=1$.

Factor

$$z^5-1=(z-1)(z^4+z^3+z^2+z+1).$$

For the quartic factor, $z\ne0$. Divide by $z^2$ and put $X=z+1/z$:

$$X^2+X-1=0,\qquad X=\frac{-1\pm\sqrt5}{2}.$$

For either value of $X$, solve $z^2-Xz+1=0$. The five roots are

$$\boxed{1,\quad\frac{-1+\sqrt5\pm i\sqrt{10+2\sqrt5}}4,\quad\frac{-1-\sqrt5\pm i\sqrt{10-2\sqrt5}}4.}$$

These agree with the more compact trigonometric description

$$z=\cos\frac{2k\pi}{5}+i\sin\frac{2k\pi}{5},\qquad k=0,1,2,3,4.$$

The reciprocal substitution also derives the exact sine and cosine values rather than requiring them beforehand.

Application 2: prove a quartic is always positive

Problem. Show that for every real $x$,

$$2x^4-x^3+x^2-x+2>0.$$

At $x=0$, the expression is two. For nonzero $x$, division by the positive number $x^2$ preserves the inequality. With $X=x+1/x$, it becomes

$$2X^2-X-3>0.$$

First find the allowed range of X

A real nonzero $x$ satisfying $X=x+1/x$ exists exactly when $x^2-Xx+1=0$ has a real root. Its discriminant gives

$$X^2-4\ge0,\qquad\boxed{X\le-2\text{ or }X\ge2.}$$

Conversely, for every such $X$, the quadratic formula supplies real roots whose product is one, so they are nonzero. This proves the entire range, including attainability.

Now factor

$$2X^2-X-3=(X+1)(2X-3).$$

For $X\le-2$, both factors are negative. For $X\ge2$, both are positive. Hence their product is strictly positive throughout the allowed range, proving the original inequality.

The important step was to keep the substitution's range. Showing positivity for every real $X$ would be false and is unnecessary. Bounds from an inequality can help locate a range, but a complete range statement should also explain why every claimed value is attained.

Related: root placement in intervals and conjugate radical powers.