Bairstow Method
This is another iterative method to find the roots of any polynomial
equation P_{n}(x) = 0 given in the form.
x^{n} + a_{1 }x^{n1} + .
. . + a_{n1}x + a_{n} = 0
Unlike BergeVieta method here a second degree polynomial is
taken as a factor from P_{n}(x) are written as
(x^{2} + px + q) Q_{n2}(x) + Rx + S = 0
where Q_{n2}(x), deflated polynomial,
is a polynomial of degree (n2) of the form
x^{n2} + b_{1 }x^{n3} + .
. . + b_{n3}x + b_{n2} = 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 p_{0} and q_{0}
are initial approximations for p and q then the iterative
process can be written as
p_{i+1} = p_{i} + dp and q_{i+1}
= q_{i} + dq
i = 0, 1, 2, . . .
where
dp =  (RS_{q}  SR_{q}) / (R_{p} S_{q}
 R_{q}S_{p})
dq =  (R_{p}S  RS_{p}) / (R_{p}S_{q}
 R_{q}S_{p})
where R_{p}, R_{q}, S_{ p} and
S_{q} are partial derivatives with respect to p and
q evaluated at p_{i} and q_{i} .
Now by comparing the two nth degree polynomials defined
above, we have
a_{1} 
= 
b_{1} + p 
Þ 
b_{1} 
= 
a_{1}  p 
a_{2} 
= 
b_{2} + pb_{1} + q 
Þ 
b_{2} 
= 
a_{2}  pb_{1}  q 
.
.
. 

.
.
. 

.
.
. 

.
.
. 
a_{i} 
= 
b_{i }+ pb_{i1}+ qb_{i2} 
Þ 
b_{i} 
= 
a_{i}  pb_{i1} qb_{i2} 
.
.
. 

.
.
. 

.
.
. 

.
.
. 
a_{n1} 
= 
R + pb_{n2}+ qb_{n3} 
Þ 
R = bn 
= 
a_{n1}  pb_{n2 } qb_{n3} 
a_{n} 
= 
S + qb_{n2} 
Þ 
S 
= 
a_{n } qb_{n2} 
i.e., 






b_{i} 
= 
a_{i}  pb_{i1}  qb_{i2} 



i=1,2,...n 
with b_{0} = 1, b_{1} = 0 
The last two equations here are
R = b_{n1
}S = b_{n} + pb_{n1}
The partial derivatives R_{p} , R_{q}, S_{p}
and S_{q} can be determined by differentiating above equation
with respect to p and q we get
( ¶b_{i
}/ ¶p) = b_{ i1} + p(¶b_{i1
}/ ¶p) + q(¶b_{i2
}/ ¶p)
( ¶b_{i
}/ ¶q) = b_{ i2} + p(¶b_{i1
}/ ¶q) + q(¶b_{i2
}/ ¶q)
with
(¶b_{0 }/ ¶p)
= (¶b_{1 }/ ¶
p) = (¶b_{0 }/ ¶q)
= (¶b_{1 }/ ¶q)
= 0.
substituting ( ¶b_{i
}/ ¶p) =  c_{ i1}
i = 1, 2, . . ., n
gives c_{i1} = b_{i1}  pc_{i2}  qc_{i3}
similarly if c_{i2} =  (¶b_{i
}/ ¶ q)
then c_{i2} = b_{i2}  pc_{i3}  qc_{i4}
thus we have
c_{i} = b_{i}  pc_{i1}  qc_{i2}
i = 1, 2, . . ., n
where c_{0} = 1, c1 = 0
Then
R_{p}
= c_{n2}
S_{p}
= b_{n1}  c_{n1}  pc_{n2}
R_{q}=
c_{n3}
S_{q}=
(c_{n2} + pc_{n3})
that is
dp =  
bc_{n3}  b_{n1} c_{n2}


dq =  
b_{n1}(c_{n1}  b_{n1})  b_{n}c_{n2}




c^{2}_{n2}  c_{n3}(c_{n1}
 b_{n1})


c^{2}_{n2}  c_{n3}(c_{n1}
 b_{n1})

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 Q_{n2}(x).
The coefficients of Q_{n2} (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 
x^{3}  11x^{2} + 32x
 22 = 0 
Solution 
Exapmple 2 
x^{3}x  3 = 0 
Solution 
Exapmple 3 
x^{4}+x^{3 }+ 2x^{2}
+ x + 1 = 0 
Solution 
Exapmple 4 
x^{6
} x^{4
} x^{3}
 1 = 0 
Solution 
Exapmple 5 
x^{3 } 3.7x^{2
}+ 6.2x
 4.069 = 0 
Solution 
Exapmple 6 
x^{4 } x  10 = 0 
Solution 
Problems
to workout

Work out here :
