Bairstow's Method

Description

By Thomas Klemm, February 2022 in the MoHPC - General Software Library

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 → b1.5000
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  6
 1 q  7 b' = bᵢ₊₁
 2 c = cᵢ  8 index = 9.fff
 3 c' = cⱼ₊₁ (i)
 4 c'' = cⱼ₊₂ I
 5 b = bᵢ

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
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 ×