Bairstow's Method
Description
Example
P(x) = 2x5 − 9x4 + 15x3 + 65x2 − 267x + 234 = 0
Insert coefficients
2
STO 9
-9
STO .0
15
STO .1
65
STO .2
-267
STO .3
234
STO .4
Initialization
9.014
STO 8
1
STO 0
STO 1
Alternatively use:
MATRIX 1
Run program
GSB A
-52.0000
RCL 0
1.5000
RCL 1
-4.5000
RCL 9
2.0000
RCL .0
-12.0000
RCL .1
42.0000
RCL .2
-52.0000
GSB B
1.5000
x↔y
-3.0000
Conclusion
2x5 − 9x4 + 15x3 + 65x2 − 267x + 234 =
(x2 + 1.5x − 4.5)(2x3 − 12x2 + 42x − 52)
Solutions
For x2 + 1.5x − 4.5 = 0:
x_1 = 1.5
x_2 = −3
Initialize guess
1
STO 0
STO 1
Alternatively use:
MATRIX 1
Run program again
GSB A
-4.0000
RCL 0
-4.0000
RCL 1
13.0000
RCL 9
2.0000
RCL .0
-4.0000
GSB B
Error 0
⇐
-9.0000
CHS
9.0000
√x
3.0000
x↔y
2.0000
RCL .0
-4.0000
RCL 9
2.0000
÷
-2.0000
CHS
2.0000
Conclusion
2x3 − 12x2 + 42x − 52 =
(x2 − 4x + 13)(2x − 4)
Solutions
For x2 − 4x + 13 = 0:
x3 = 2 + 3i
x4 = 2 − 3i
For 2x − 4 = 0:
x5 = 2
Summary
Factors
2x5 − 9x4 + 15x3 + 65x2 − 267x + 234 =
(x2 + 1.5x − 4.5)(x2 − 4x + 13)(2x − 4) =
(x − 1.5)(x + 3)(x2 − 4x + 13)2(x − 2) =
(2x − 3)(x − 2)(x + 3)(x2 − 4x + 13)
Solutions
x1 = 1.5
x2 = 2
x3 = −3
x4 = 2 + 3i
x5 = 2 − 3i
Program Resources
Labels
Name |
Description |
|
A |
Bairstow's method |
|
B |
Solve quadratic equation |
|
0 |
Partial Derivatives |
|
1 |
Polynomial Division |
|
Storage Registers
Name |
Description |
|
Name |
Description |
|
0 |
p |
|
5 |
b = b_{i} |
|
1 |
q |
|
7 |
b' = b_{i+1} |
|
2 |
c = c_{j} |
|
8 |
index = 9.fff |
|
3 |
c' = c_{j+1} |
|
(i) |
|
|
4 |
c'' = c_{j+2} |
|
I |
|
|
Program
Line |
Display |
Key Sequence |
|
Line |
Display |
Key Sequence |
|
Line |
Display |
Key Sequence |
|
000 |
|
|
|
025 |
42, 6,25 |
f ISG I |
|
050 |
30 |
− |
|
001 |
42,21,11 |
f LBL A |
|
026 |
22 0 |
GTO 0 |
|
051 |
45 0 |
RCL 0 |
|
002 |
42 32 |
f ∑ |
|
027 |
42 49 |
f L.R. |
|
052 |
43 33 |
g R⬆ |
|
003 |
45 8 |
RCL 8 |
|
028 |
44,40, 0 |
STO + 0 |
|
053 |
20 |
× |
|
004 |
44 25 |
STO I |
|
029 |
34 |
x↔y |
|
054 |
30 |
− |
|
005 |
42,21, 0 |
f LBL 0 |
|
030 |
44,40, 1 |
STO + 1 |
|
055 |
44 24 |
STO (i) |
|
006 |
45 5 |
RCL 5 |
|
031 |
43 1 |
g →P |
|
056 |
42, 6,25 |
f ISG I |
|
007 |
45 3 |
RCL 3 |
|
032 |
43 34 |
g RND |
|
057 |
22 1 |
GTO 1 |
|
008 |
44 4 |
STO 4 |
|
033 |
43,30, 0 |
g TEST x≠0 |
|
058 |
43 32 |
g RTN |
|
009 |
45,20, 1 |
RCL × 1 |
|
034 |
22 11 |
GTO A |
|
059 |
42,21,12 |
f LBL B |
|
010 |
30 |
− |
|
035 |
45 8 |
RCL 8 |
|
060 |
45 0 |
RCL 0 |
|
011 |
45 2 |
RCL 2 |
|
036 |
2 |
2 |
|
061 |
2 |
2 |
|
012 |
44 3 |
STO 3 |
|
037 |
26 |
EEX |
|
062 |
16 |
CHS |
|
013 |
45,20, 0 |
RCL × 0 |
|
038 |
3 |
3 |
|
063 |
10 |
÷ |
|
014 |
30 |
− |
|
039 |
16 |
CHS |
|
064 |
36 |
ENTER |
|
015 |
44 2 |
STO 2 |
|
040 |
30 |
− |
|
065 |
36 |
ENTER |
|
016 |
45 24 |
RCL (i) |
|
041 |
44 8 |
STO 8 |
|
066 |
43 11 |
g x² |
|
017 |
45 7 |
RCL 7 |
|
042 |
44 25 |
STO I |
|
067 |
45,30, 1 |
RCL − 1 |
|
018 |
45,20, 1 |
RCL × 1 |
|
043 |
0 |
0 |
|
068 |
11 |
√x̅ |
|
019 |
30 |
− |
|
044 |
36 |
ENTER |
|
069 |
30 |
− |
|
020 |
45 5 |
RCL 5 |
|
045 |
42,21, 1 |
f LBL 1 |
|
070 |
34 |
x↔y |
|
021 |
44 7 |
STO 7 |
|
046 |
45 24 |
RCL (i) |
|
071 |
43 36 |
g LSTΧ |
|
022 |
45,20, 0 |
RCL × 0 |
|
047 |
45 1 |
RCL 1 |
|
072 |
40 |
+ |
|
023 |
30 |
− |
|
048 |
43 33 |
g R⬆ |
|
073 |
43 32 |
g RTN |
|
024 |
44 5 |
STO 5 |
|
049 |
20 |
× |
|
|
|
|
|