The Fundamental Theorem for Palindromic Polynomials

If all the roots of a polynomial (with real coefficients) lie on the 
unit circle in the complex plane it's possible to effectively cut the
degree of the polynomial in half.  First, we can easily divide out any
real roots on the unit circle, i.e., we can divide out all powers of 
(x+1) and (x-1).  We're left with a polynomial having only complex 
roots.

In general we know that the complex roots of a polynomial with real
coefficients occur in conjugate pairs.  Thus, the degree of our
reduced polynomial must be even.  Moreover, if two conjugate roots
r1,r2 lie on the unit circle, then r1*r2 = 1.  If ALL the roots of a
polynomial lie on the unit circle, then we can substitute "1" for each
pair of conjugate roots in the symmetric polynomial expressions for
the coefficients.  The result is that the coefficients must be
palindromic.

For example, consider an 8th degree polynomial with real coefficients

  x^8 + Ax^7 + Bx^6 + Cx^5 + Dx^4 + Cx^3 + Bx^2 + Ax + 1  =  0

The coefficients form a palindrome (1ABCDCBA1) so this is called a
_palindromic polynomial_.  We know the complex roots occur in conjugate 
pairs, and if the coefficients are palindromic we know the roots occur 
in inverse pairs.  However, the inverse pairing is not necessarily the 
same as the conjugate pairing.  

Of course, if a given root's inverse is also its conjugate, then both
lie on the unit circle.  On the other hand, if a root's inverse is NOT
it's conjugate, then it must be one of a set of four related roots
that satisfy a palindromic quartic 

                   x^4 + Ax^3 + Bx^2 + Ax + 1 = 0

with A^2-4B+8 < 0.  Conversely, if a given quartic has  A^2-4B+8 > 0,
then the roots must all be on the unit circle (in which case the
quartic splits into two palindromic quadratics).  


Fundamental Theorem of Algebra for Palindromic Polynomials With Real
Coefficients:  Any palindromic polynomial with real coefficients can
be factored into a product of a constant times linear, quadratic, and
quartic palindromic polynomials with real coefficients.  


Moreover, if the degree of the original polynomial is odd then it 
is divisible by the linear palindrome (x+1).  If the degree of the 
original polynomial is even then it factors into quadratic and 
quartic palindromes.  The roots of the quadratic factors of the
form x^2+Ax+1 either are or are not on the unit circle, depending 
on whether A^2-4 is positive or negative, and the roots of the 
quartic factors x^4+Ax^3+Bx^2+Ax+1 either are or are not on the 
unit circle depending on whether A^2-4B+8 is positive or negative.

For the 8th degree polynomial above, if all the roots lie on the unit 
circle, then we have the four inverse pairs

   r1*r2 = 1        r3*r4 = 1        r5*r6 = 1        r7*r8 = 1

Notice that if we define four variables

    y1 = r1+r2      y2 = r3+r4      y3 = r5+r6     y4 = r7+r8

then these values of yi are the roots of the quartic

        y^4 + Ay^3 + (B-4)y^2 + (C-3A)y + (D-2B+2)  =  0

Of course, given a root of this equation we can easily determine the
corresponding roots of the original 8th degree polynomial because,
for example, we have y1 = r1+(1/r1) = r2+(1/r2), which implies that
r1 and r2 are the roots of the quadratic

                    r^2 - (y1)r + 1 = 0

This can be applied to any polynomial whose roots all lie on the unit
circle, i.e., any palindromic polynomial.  In fact, it can be applied
to any polynomial that can be converted to palindromic form.  For
example, any quartic polynomial

           Ax^4 + Bx^3 + Cx^2 + Dx + A(D/B)^2  =  0

can be converted to palindromic form, so the roots can be determined
using only quadratic equations.  The four roots are

        r  =  (-B+-sqrt[K]+-sqrt{J+-2Bsqrt[K]})/4A

where 
             K =  B^2 - 4AC + 8(D/B)A^2
             J = 2B^2 - 4AC - 8(D/B)A^2

and the "+-" signs have any of the four signatures (++-), (-++),
(+--), or (--+).