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Bairstow Method
This is another iterative method to find the roots of any polynomial equation Pn(x) = 0 given in the form.


xn + a1 xn-1 +  .  .  . + an-1x + an = 0

Unlike Berge-Vieta method here a second degree polynomial is taken as a factor from Pn(x) are written as 


(x2 + px + q) Qn-2(x) + Rx + S = 0

where Qn-2(x), deflated polynomial,   is a polynomial of degree (n-2) of the form


xn-2 + b1 xn-3 +  .  .  . + bn-3x + bn-2 = 0,

R and S are residues depend on p and q. Now the problem is to find p and q such that R(p,q) = 0, S(p,q) = 0. This is a system of two equations with two unknowns which we already discussed in the sections complex roots and also in system of equations.
According to their scheme if p0 and q0 are initial approximations for p and q then the  iterative process can be written as
pi+1 = pi + dp and  qi+1 = qi + dq                  i = 0, 1, 2, . . .
where                  dp = - (RSq - SRq) / (Rp Sq - RqSp)


                                     dq = - (RpS - RSp) / (RpSq - RqSp)
 

where Rp, Rq, S p and Sq are partial derivatives with respect to p and q evaluated at pi and qi .
Now by comparing the two nth degree polynomials defined above, we have

 

a1 = b1 + p Þ b1 = a1 - p
a2 = b2 + pb1 + q Þ b2 = a2 - pb1 - q
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ai = bi + pbi-1+ qbi-2 Þ bi = ai - pbi-1- qbi-2
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an-1 = R + pbn-2+ qbn-3 Þ R = bn = an-1 - pbn-2 - qbn-3
an = S + qbn-2 Þ S  = an - qbn-2
i.e.,
bi = ai - pbi-1 - qbi-2 i=1,2,...n

with b0 = 1, b-1 = 0
The last two equations here are
R = bn-1                        S = bn + pbn-1
The partial derivatives Rp , Rq, Sp and Sq can be determined by differentiating above equation with respect to p and q we get 
-(bi / p) = b i-1 + p(bi-1 / p) + q(bi-2 / p)

-(bi / q) = b i-2 + p(bi-1 / q) + q(bi-2 / q)

with 
(b0 / p) = (b-1 / p) = (b0 / q) = (b-1 / q) = 0. 

substituting   (bi / p) = - c i-1             i = 1, 2, . . ., n

gives ci-1 = bi-1 - pci-2 - qci-3

similarly if  ci-2 = - (bi / q)

then ci-2 = bi-2 - pci-3 - qci-4

thus we have 

ci = bi - pci-1 - qci-2                  i = 1, 2, .  .  ., n
where c0 = 1, c-1 = 0

Then 
           Rp = -cn-2

           Sp = bn-1 - cn-1 - pcn-2

           Rq= -cn-3

           Sq=  -(cn-2 + pcn-3)

that is
   dp = - 
bcn-3 - bn-1 cn-2
   dq = - 
bn-1(cn-1 - bn-1) - bncn-2
c2n-2 - cn-3(cn-1 - bn-1)
c2n-2 - cn-3(cn-1 - bn-1)

This process after convergence for p and q will separate a second degree polynomial from the given nth degree polynomial. The roots of this second degree equation can be obtained by the analytical formula for all kinds of roots. The same process can be repeated to get another second degree equation from the deflated polynomial Qn-2(x). The coefficients of Qn-2 (x) can be taken from the array b. This process is repeated until we get all n roots of the given equation. 

Worked out problems
  Find the roots of the following polynomial equations
 Exapmple 1   x3 - 11x2 + 32x - 22 = 0  Solution
 Exapmple 2  x3-x - 3 = 0  Solution
 Exapmple 3  x4+x3 + 2x2  + x + 1 = 0    Solution
 Exapmple 4  x6 - x4 - x3 - 1 = 0  Solution
 Exapmple 5  x3 - 3.7x2 + 6.2x - 4.069 = 0  Solution
 Exapmple 6  x4 - x - 10 = 0  Solution
Problems to workout

 
Work out  here :
Solution of Transcendental Equations | Solution of Linear System of Algebraic Equations | Interpolation & Curve Fitting
Numerical Differentiation & Integration | Numerical Solution of Ordinary Differential Equations
Numerical Solution of Partial Differential Equations