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 r1r2 = 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

 

 

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

 

 

with A2 - 4B + 8 < 0. Conversely, if a given quartic has A2 - 4B + 8 > 0, then the roots must all be on the unit circle (in which case the quartic splits into two palindromic quadratics).

 

Now we state what might be called the 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 x2 + Ax + 1 either are or are not on the unit circle, depending on whether A2 - 4 is positive or negative, and the roots of the quartic factors x4 + Ax3 + Bx2 + Ax + 1 either are or are not on the unit circle depending on whether A2 - 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

 

 

Notice that if we define four variables

 

 

then these values of yi are the roots of the quartic

 

 

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

 

 

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

 

 

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

 

 

where

 

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

 

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