Civil Engineering Reference
In-Depth Information
−
57 208 1 561
19 421 0 353
57 208 19 421
19 421
.
−
.
(
)(
)
−−
(
)(
)
.
.
−
57 20
.80353
.
1 561 19 421
.
.
∆
v
=
=
(
)(
)
−
(
)
(
)
−
.
.
−
57 208 9 546
.
.
19 421 19 42
.
.
1
.
9 546
.
10 11
923 27
.
=
−
=−
0 0110
.
.
uuu
vvv
=+ =
∆
∆
1 9758 0 0236 1 999
2 0102 0 0110 1 999
.
+
.
=
.
2
=+ =
.
−
.
=
.
2
It appears the values are
u =
2 and
v =
2. Repeat the process using the
revised values for
u
and
v
shown in Table 1.25.
Table 1.25.
Example 1.17 Bairstow's method
−
2
1
−
3
−
10
10
44
48
−
2
10
4
−
48
0
−
2
−
2
10
4
−
48
1
−
5
−
2
24
0
0
Since the remainders of the first division
b
n−
1
and
b
n
are both zero,
u =
2 and
v =
2 are the coefficients of the root quadratic. Substitute into the
quadratic equation to find the roots.
=
−± −
()
=− ±−=− ±
2
2242
2
(
)
2
x
++
22
x
with x
1
1
1
i
12
,
The first two roots are
x
1
= −
1
+ i
and
x
2
= −
1
− i.
The remaining values
are the coefficients of the factored polynomial.
(
)
−−+
(
)
2
3 2
22 5224
fx x
()=++
x
x
x
x
The remaining polynomial may be solved using the same method. This
time begin with
u
=
0 and
v
=
0 in Table 1.26.
Table 1.26.
Example 1.17 Bairstow's method
0
1
24
−5
−2
0
0
0
0
0
0
0
1
24
−5
−
2
0
0
0
0
1
−
5
−
2
