Question

How to show that a polynomial has a reciprocal factor?

Answer 1

In algebra, the reciprocal polynomial, or reflected polynomial[1][2] p∗ or pR,[2][1] of a polynomial p of degree n with coefficients from an arbitrary field, such as

{\displaystyle p(x)=a_{0}+a_{1}x+a_{2}x^{2}+\cdots +a_{n}x^{n},\,\!} p(x) = a_0 + a_1x + a_2x^2 + \cdots + a_nx^n, \,\!
is the polynomial[3]

{\displaystyle p^{*}(x)=a_{n}+a_{n-1}x+\cdots +a_{0}x^{n}=x^{n}p(x^{-1}).} p^*(x) = a_n + a_{n-1}x + \cdots + a_0x^n = x^n p(x^{-1}).
Essentially, the coefficients are written in reverse order. They arise naturally in linear algebra as the characteristic polynomial of the inverse of a matrix.

In the special case that the polynomial p has complex coefficients, that is,

{\displaystyle p(z)=a_{0}+a_{1}z+a_{2}z^{2}+\cdots +a_{n}z^{n},\,\!} p(z) = a_0 + a_1z + a_2z^2 + \cdots + a_nz^n, \,\!
the conjugate reciprocal polynomial, p† given by,

{\displaystyle p^{\dagger }(z)={\overline {a_{n}}}+{\overline {a_{n-1}}}z+\cdots +{\overline {a_{0}}}z^{n}=z^{n}{\overline {p({\bar {z}}^{-1})}},} p^{\dagger}(z) = \overline{a_n} + \overline{a_{n-1}}z + \cdots + \overline{a_0}z^n = z^n\overline{p(\bar{z}^{-1})},
where {\displaystyle {\overline {a_{i}}}} \overline{a_i} denotes the complex conjugate of {\displaystyle a_{i}\,\!} a_i \,\!, is also called the reciprocal polynomial when no confusion can arise.

A polynomial p is called self-reciprocal or palindromic if p(x) = p∗(x). The coefficients of a self-reciprocal polynomial satisfy ai = an−i. In the conjugate reciprocal case, the coefficients must be real to satisfy the condition.