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}](b_76.gif) |
(1) |
Now write the original Polynomial as
![\begin{displaymath}
P(x)=(x^2+Bx+C)Q(x)+Rx+S
\end{displaymath}](b_77.gif) |
(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}](b_78.gif) |
(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}](b_79.gif) |
(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}](b_80.gif) |
(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}](b_81.gif) |
(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}](b_82.gif) |
(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}](b_83.gif) |
(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.
© 1996-9 Eric W. Weisstein
1999-05-26