Solve for all the roots of x using Bairstow's Method starting at ri = 0.25 and s1 = 8.75 of the polynomial f(x)=x*+8x +72x – 81

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section: Chapter Questions
Problem 14T
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Bairstow's Method

Stopping Criteria: 0.00005

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Solve for all the roots of x using Bairstow's Method starting at rị = 0.25 and s1 = 8.75 of the polynomial
f(x)=x*+8x +72x – 81
Transcribed Image Text:Solve for all the roots of x using Bairstow's Method starting at rị = 0.25 and s1 = 8.75 of the polynomial f(x)=x*+8x +72x – 81
Example
Find the roots of
f(x) = x' – 3x? + 4x – 2 = 0
-
From the given equation,
az = 1.0
a, = -3.0
a, = 4.0
a, = -2.0
We initially guess values for r and s. For this
problem, we'll have r, = 1.5 and s, = -2.5
Tabular Solution
1
-3
4
-2
, = 1.5
1.5
-2.25
-1.125
S1 = -2.5
-2.5
3.75
1
-1.5
-0.75
0.625
= 1.5
1.5
S, = -2.5
-2.5
1
-3.25
+ (-)
(0'0)Ar4 (1.0)As = 0.75
(-)
-3.25Ar + (0.0)As =-0.625
Solution
Solving simultaneously
Ar = 0.192308
and
As = 0.75
And
r2 = r; + Ar = 1.5+0.192308 =1.692308
S2 = S, + As = -2.5+0.75 = -1.75
Then we use the new set of values of r and s
until such time that...
Tabular Solution
i
Ar
As
1
-3
4
-2
r. =
1.999999
1
1.5
-2.5
0.192308
0.75
1.999999
-2.000001
S6 = -2
-2
2
1.692308
-1.75
0.278352
-0.144041
1
-1.000001
3
1.97066
-1.894041
0.034644
-0.110091
1.999999
1.999999
2
4
2.005304
-2.004132
-0.005317
0.004173
$6 = -2
-2
1.999987
-1.999959
0.000012
-0.000041
1
0.999998
6
1.999999
-2
Transcribed Image Text:Example Find the roots of f(x) = x' – 3x? + 4x – 2 = 0 - From the given equation, az = 1.0 a, = -3.0 a, = 4.0 a, = -2.0 We initially guess values for r and s. For this problem, we'll have r, = 1.5 and s, = -2.5 Tabular Solution 1 -3 4 -2 , = 1.5 1.5 -2.25 -1.125 S1 = -2.5 -2.5 3.75 1 -1.5 -0.75 0.625 = 1.5 1.5 S, = -2.5 -2.5 1 -3.25 + (-) (0'0)Ar4 (1.0)As = 0.75 (-) -3.25Ar + (0.0)As =-0.625 Solution Solving simultaneously Ar = 0.192308 and As = 0.75 And r2 = r; + Ar = 1.5+0.192308 =1.692308 S2 = S, + As = -2.5+0.75 = -1.75 Then we use the new set of values of r and s until such time that... Tabular Solution i Ar As 1 -3 4 -2 r. = 1.999999 1 1.5 -2.5 0.192308 0.75 1.999999 -2.000001 S6 = -2 -2 2 1.692308 -1.75 0.278352 -0.144041 1 -1.000001 3 1.97066 -1.894041 0.034644 -0.110091 1.999999 1.999999 2 4 2.005304 -2.004132 -0.005317 0.004173 $6 = -2 -2 1.999987 -1.999959 0.000012 -0.000041 1 0.999998 6 1.999999 -2
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