This document discusses methods for finding the roots of polynomial equations, including Muller's method, Bairstow's method, and the Bairstow algorithm. It provides details on how each method works, such as deriving the coefficients of a parabola through three points for Muller's method or using synthetic division and solving simultaneous equations to estimate changes in values for Bairstow's method. An example is also shown applying Bairstow's method to find the roots of a 5th order polynomial equation.

1. Maria Fernanda Vergara Mendoza
Petroleum Engineering
UIS-COLOMBIA

2.  In this chapter, you will learn some methods to find the
roots of polynomial equations of the general form:
 Where n= the order of the polynomial; a= constant
coefficients.
 RULES:
 For an nth-order equation, there are n real or complex
roots.
 If n is odd, there is at least one real root
 The complex roots exsist in conjugate pairs (a+bi and
a-bi), i=√(-1)

3. Muller’s Bairstow’s
Method Method

4.  The Muller’s method, is like the secant method, just that
this one projects a parabola through three points unlike
secant method, who projects a straight line.
 This method consists of deriving the coefficients of the
parabola that goes through the three points.

5.  Write the parabolic equation in this form:
P( x) a( x xi 1 ) 2 b( x xi 1 ) c
 The coefficients a, b, and c can be evaluated by
substituting each of the three points to give:
f ( xi 1 ) a( xi 1 xi 1 ) 2 b( xi 1 xi 1 ) c
f ( xi ) a( xi xi 1 ) 2 b( xi xi 1 ) c
f ( xi 1 ) a( xi 1 xi 1 ) 2 b( xi 1 xi 1 ) c

6.  Two of the terms of f ( xi 1 ) are zero, it can be solved
for c=f(xi+1).
f ( xi 1 ) f ( xi 1 ) a ( xi 1 xi 1 ) 2 b( xi 1 xi 1 )
f ( xi ) f ( xi 1 ) a ( xi xi 1 ) 2 b( xi xi 1 )
 Using algebraic manipulations, we solve the remaining
coefficients:
hi 1 xi xi 1
hi xi 1 xi
f ( xi ) f ( xi 1 )
i 1
xi xi 1
f ( xi 1 ) f ( xi )
i
xi 1 xi

7.  These can be substituted to give:
(hi 1 hi )b (hi 1 hi ) 2 a hi 1 i 1 hi i
2
hi b hi a hi i
 The results can be summarized as
i i 1
a b ahi i c f ( xi 1 )
hi hi 1
 Once you know the approximate coefficients you have
to find the approximated root using the quadratic
equation :
2c
xi 2 xi 1
b b 2 4ac

8.  The error can be calculated as:
 There is a problem with xi 2 equation, this equation
yields two roots, in this method the sign is chosen with
this strategies:
 1. If only real roots are being located, we choose the
two original points that are nearest the new root
estimate, xi+2 .
 If both real and complex roots are being evaluated, a
sequential approach is employed. That means: xi, xi+1,
xi+2 take the place of xi-1, xi, xi+1

9. If you have as initial values xi 1 4.5 xi 5 .5 xi 1 5
respectively, find the root of the equation:
f ( x) x 3 13x 12
FIRST: Evalue the equation in its initial values
f (4.5) 20.625
f (5.5) 82.875
f (5) 48

10. SECOND: This values are used to calculate:
hi 1 5 . 5 4 .5 1
hi 5 5 .5 0 .5
82.875 20.625
i 1 62.25
5 . 5 4 .5
48 82.875
i 69.75
5 5.5
THIRD: Find the a, b, c coefficients:
69.75 62.25
a 15
0 .5 1
b 15( 0.5) 69.75 62.75
c 28

11. 2(48)
xi 2 5 3.976487
62.25 31.54451
The error is:
1.023513
a 100% 25.74
3.976487
This is a huge error, so its necesary to do other iterations:
xi 1 5 .5 xi 5 xi 1 3.976487
Repeat the calculations and get a low percent of error:
Iteration Xr Ea%
0 5 --
1 3.976487 25.74
2 4.00105 0.6139
3 4 0.0262
4 4 0.0000119

12.  Is an iterative approach related loosely to both the
Muller and Newton Raphson methods.
 It is based on the idea of synthetic division of the
given polynomial by a quadratic function and can
be used to find all the roots of a polynomial.
 The idea is to do a synthetic division of the
polynomial Pn(x) by the quadratic factor (x2 - rx -
s).

13.  The synthetic division can be extended to quadratic
factors:
Pn ( x) ( x 2 rx s)Qn 2 ( x) R
x 2 rx s bn x n 2 bn 1 x n 3
... b3 x b2 residue
residue b1 ( x r ) ... b0
 When you multiply and match factors have:
an bn 1 bn an
an 1 bn 1 rbn bn 1 an 1 rbn
an 2 bn 2 rbn 1 sbn bn 2 an 2 rbn 1 sbn
an 3 bn 3 rbn 2 sbn 1 bn 3 an 3 rbn 2 sbn 1
: :
a1 b1 rb2 sb3 b1 a1 rb2 sb3
a0 b0 rb1 sb2 b0 a0 rb1 sb2

14.  The idea is to find values of r and s, making
b1 and b0 zero.
 The method works taking an initial
approach (r0, s0) and getting better
approaches (rk, sk), this is an iterative
procedure, the process ends when the
residue of dividing the polynomial by (x2 -
rkx - sk) its zero.
 B1=f(s, r)
 B0=g(s, r)

15.  Because both bo and b1 are functions of both r and s,
they can be expanded using a Taylor series:
b1 b1
b1 (r r, s s ) b1 r s
r s
b0 b0
b0 (r r, s s ) b0 r s
r s
 The changes, Δr and Δs, can be estimated by setting the
expansion equal to zero:
b1 b1
b1 r s
r s
b0 b0
b0 r s
r s

16.  “If the partial derivatives of the b’s can be determined,
these are a system of two equations that can be solved
simultaneously for the two unknowns, Δr and Δs.”
 According to Bairstow, the partial derivatives can be
obtained by a synthetic division of the b’s.
cn bn
cn 1 bn 1 rcn
cn 2 bn 2 rcn 1 scn
:
cn k bn k rcn ( k 1) scn ( k 2)

17. b1 b2 b b1 b2 b
r b2 s 3 c2 r b3 s 3 c3
r r r s s s
b0 b b b0 b b
r 1 b1 s 2 c1 r 0 b2 s 2 c2
r r r s s s
 Then the system of equations can be written
as:
c2 r c3 s b1
c1 r c2 s b0

18.  APPROXIMATED ERROR
r s
a ,r .100% a,s .100%
r s
 When both of these error estimates fall below a
stopping criterion, the values of the roots can be
determined by:
r r 2 4s
x
2

19.  Employ Bairstow’s method to determine the roots of the
polynomial
f 5 ( x) x5 3.5x 4 2.75x3 2.125x 2 3.875x 1.25
 Use initial guesses of r=s=-1 and iterate to a level of tolerance
of 1%
SOLUTION:
 b5=1 b4=-4.5 b3=6.25 b2=0.375 b1=-10.5
b0=11.375
 c5=1 c4=-5.5 c3=10.75 c2=-4.875 c1=-16.375
Thus, the simultaneous equations to solve Δr and Δs are:
4.875 r 10.75 s 10.5
16.375 r 4.875 s 11.375

20.  Which can be solved for Δr=0.3558 and Δs=1.1381.
 r=-0.6442
 S=0.1381
 And the approximate errors are:
0.3558 1.1381
a ,r .100% 55.23% a,s .100% 824.1%
0.6442 0.1381
 The computation can be continued with the result that
after four iterations the metod converges on velues of
r=-0.5 and s=0.5
0.5 ( 0.5) 2 4(0.5)
x 0.5, 1
2

21.  CHAPRA, Steven C. “Numerical methods for
engineers”, Fifth edition. Mc Graw Hill.
 CARRILLO, Eduardo. “Raices de polinomios”.
PPT.