Bairstow's Method

A procedure for finding the quadratic factors for the Complex Conjugate Roots of a Polynomial with Real Coefficients.

$\begin{displaymath}[x-(a+ib)][x-(a-ib)] = x^2+2ax+(a^2+b^2) \equiv x^2+Bx+C. \end{displaymath}$

(1)

Now write the original Polynomial as

$\begin{displaymath} P(x)=(x^2+Bx+C)Q(x)+Rx+S \end{displaymath}$

(2)

$\begin{displaymath} R(B+\delta B,C+\delta C)\approx R(B,C)+{\partial R\over\partial B}\,dB+{\partial R\over\partial C}\,dC \end{displaymath}$

(3)

$\begin{displaymath} S(B+\delta B,C+\delta C)\approx S(B,C)+{\partial S\over\partial B}\,dB+{\partial S\over\partial C}\,dC \end{displaymath}$

(4)

$\begin{displaymath} {\partial P\over\partial C} = 0 = (x^2+Bx+C){\partial Q\over... ...Q(x)+ {\partial R\over\partial C}+{\partial S\over\partial C} \end{displaymath}$

(5)

$\begin{displaymath} -Q(x) = (x^2+Bx+C){\partial Q\over\partial C}+{\partial R\over\partial C}+{\partial S\over\partial C} \end{displaymath}$

(6)

$\begin{displaymath} {\partial P\over\partial B} = 0 = (x^2+Bx+C){\partial Q\over... ...Q(x)+ {\partial R\over\partial B}+{\partial S\over\partial B} \end{displaymath}$

(7)

$\begin{displaymath} -xQ(x) = (x^2+Bx+C){\partial Q\over\partial B}+{\partial R\over\partial B}+{\partial S\over\partial B}. \end{displaymath}$

(8)

Now use the 2-D Newton's Method to find the simultaneous solutions.

References

Press, W. H.; Flannery, B. P.; Teukolsky, S. A.; and Vetterling, W. T. Numerical Recipes in C: The Art of Scientific Computing. Cambridge, England: Cambridge University Press, pp. 277 and 283-284, 1989.