Civil Engineering Reference
In-Depth Information
original polynomial. The following is the form of the second polynomial
with the terms in the brackets being the remainder:
(
)
=
[
]
n
−
4
n
−
5
n
−
6
1
x
+
c x
+ ++ ++++
c x
c xc c
c
c
0
1
2
n
−
5
n
−
4
n
−
3
n
−
2
n
−
1
The solution may be set up in synthetic division form shown in Table 1.21:
Table 1.21.
Bairstow's method
−u
a
0
a
1
a
2
…….
a
n−3
a
n−2
a
n−1
a
n
0
−ub
0
−ub
1
−ub
n−4
−ub
n−3
−ub
n−2
−ub
n−1
−v
0
0
−vb
0
…….
−vb
n−5
−vb
n
−
4
−vb
n−3
−vb
n−2
−u
b
0
b
1
b
2
…….
b
n−3
b
n−2
b
n−1
b
n
0
−uc
0
−uc
1
−uc
n−4
−uc
n−3
−uc
n−2
−v
0
0
−vc
0
…….
−vc
n−5
−vc
n−4
−vc
n−3
c
0
c
1
c
2
…….
c
n−3
c
n−2
c
n−1
Using the preceding values, the approximations for the change in
u
and
v
values denoted Δ
u
and Δ
v
are as follows:
bc
bc
c
c
b
n
n
−
2
n
−
1
n
c
b
n
−
1
n
−
3
n
−
2
n
−
1
∆
u
=
and
∆
v
=
c
c
c
n
−
1
n
−
2
n
−
1
n
−
2
c
c
c
c
n
−
2
n
−
3
n
−
2
n
−
3
uuu and vvv
=+
∆
=+
∆
2
2
Continue the process until Δ
u
and Δ
v
are equal to zero. The two roots are
as follows by the quadratic equation:
=
−± −
uuv
2
4
(
)
2
x xv with x
++
12
,
2
Example 1.17
Bairstow's method
Find all the roots of the following polynomial using Bairstow's method.
5
4
3
2
fx
()==−− +
0
x
3
x
10
x
10
x
+
44
x
+
48
Begin by assuming
u =
1
.
5 and
v =
1
.
5 to perform the synthetic division
shown in Table 1.22.
