Numerical Analysis Questions and Answers – Bairstow’s Method

This set of Numerical Analysis Multiple Choice Questions & Answers (MCQs) focuses on “Bairstow’s Method”.

1. What will be the roots of the polynomial using Bairstow’s method where f(x) = x4-3x3+3x2-3 and r = 0.1, s = 0.1?
a) 1.8525, -0.7253, 0.9364+1.1644i, 0.9364-1.1644i
b) 1.2225, -0.7253, 0.9364+1.1644i, 0.9364-1.1644i
c) 1.8525, -0.7253, 0.4164+1.1644i, 0.9364-1.1644i
d) 1.8525, -0.7253, 0.9364+1.1644i, 0.9364-2.1644i
View Answer

Answer: a
Explanation: Given,

Iteration ai, n bi, n ci, n Δr Δs r s
1 a4=1
a3=-3
a2=3
a1=0
a0=-3
b4=1
b3=-2.9
b2=2.81
b1=-0.009
b0=-2.7199
c4=1
c3=-2.8
c2=2.63
c1=-0.026
1.1162 1.0452 1.2162 1.1452
2 a4=1
a3=-3
a2=3
a1=0
a0=-3
b4=1
b3=-1.7838
b2=1.9758
b1=0.3601
b0=-0.2994
c4=1
c3=-0.5676
c2=2.4307
c1=2.6662
-0.095 0.2274

1.1212 1.3726
3 a4=1
a3=-3
a2=3
a1=0
a0=-3
b4=1
b3=-1.8788
b2=2.2662
b1=-0.0382
b0=0.0677
c4=1
c3=-0.7577
c2=2.7893
c1=2.0491
0.0059 -0.0286 1.1271 1.344
4 a4=1
a3=-3
a2=3
a1=0
a0=-3
b4=1
b3=-1.8729
b2=2.2331
b1=-0.0003
b0=0.0009
c4=1
c3=-0.7458
c2=2.7364
c1=2.0815
0 -0.0003 1.1271 1.3437
5 a4=1
a3=-3
a2=3
a1=0
a0=-3
b4=1
b3=-1.8729
b2=2.2327
b1=0
b0=0
c4=1
c3=-0.7458
c2=2.7358
c1=2.0815
0 0 1.1271 1.3437

Now determining the roots using r and s
x1,2 = \(\frac{r±\sqrt{r^2+4s}}{2}\)
x1,2 = \(\frac{1.1271±\sqrt{1.2704+4⋅1.3437}}{2}\)
x1,2 = \(\frac{1.1271±2.5778}{2}\)
x1 = 1.8525, x2 = -0.7253
Now, dividing the polynomial
(x4-3x3+3x2-3)/(x2-1.1271x-1.3437)
=x2-1.8729x+2.2327.
This quotient polynomial is a quadratic polynomial, so roots can be found using quadratic formula
x3,4 = \(\frac{-b±\sqrt{b^2-4ac}}{2a}\)

x 3,4 = \(\frac{1.8729±\sqrt{3.5077-4⋅1⋅2.2327}}{2⋅1}\)
x3,4 = \(\frac{1.8729±\sqrt{-5.4231}}{2}\)
x3,4 = \(\frac{1.8729±2.3288i}{2}\)
x3 = 0.9364+1.1644i, x4 = 0.9364-1.1644i
So the roots of the equation are 1.8525, -0.7253, 0.9364+1.1644i, 0.9364-1.1644i.

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2. What will be the roots of the polynomial using Bairstow’s method where f(x) = 12x3+13x2+12x+4 and r = -1, s = -1?
a) -0.3076+0.7857i, 0.3076-0.7857i, -0.4682
b) 0.3076+0.7857i, -0.3076-0.7857i, -0.4682
c) -0.3076+0.7857i, -0.3076-0.7857i, 0.4682
d) -0.3076+0.7857i, -0.3076-0.7857i, -0.4682
View Answer

Answer: d
Explanation: Given,

Iteration ai, n bi, n ci, n Δr Δs r s
1 a3=12
a2=13
a1=12
a0=4
b3=12
b2=1
b1=-1
b0=4
c3=12
c2=-11
c1=-2
0.2552 0.3172 -0.7448 -0.6828
2 a3=12
a2=13
a1=12
a0=4
b3=12
b2=4.0621
b1=0.7814
b0=0.6446
c3=12
c2=-4.8759
c1=-3.7801
0.167 0.0027 -0.5778 -0.68
3 a3=12
a2=13
a1=12
a0=4
b3=12
b2=6.066
b1=0.3346
b0=-0.3184
c3=12
c2=-0.868
c1=-7.324
-0.0398 -0.0308 -0.6177 -0.7108
4 a3=12
a2=13
a1=12
a0=4
b3=12
b2=5.5881
b1=0.019
b0=0.0163
c3=12
c2=-1.8237
c1=-7.384
0.0025 -0.0012 -0.6152 -0.712
5 a3=12
a2=13
a1=12
a0=4
b3=12
b2=5.6182
b1=0.0001
b0=-0.0001
c3=12
c2=-1.7637
c1=-7.4589
0 0 -0.6152 -0.712
6 a3=12
a2=13
a1=12
a0=4
b3=12
b2=5.618
b1=0
b0=0
c3=12
c2=-1.764
c1=-7.4588
0 0 -0.6152 -0.712

Now determining the roots using r and s
x1, 2 = \(\frac{r±\sqrt{r^2+4s}}{2}\)
x1, 2 = \(\frac{-0.6152±\sqrt{0.3784+4⋅(-0.712)}}{2}\)
x1, 2 = \(\frac{-0.6152±1.5715i}{2}\)
x1 = -0.3076+0.7857i, x2 = -0.3076-0.7857i

Now, dividing the polynomial
(12x3+13x2+12x+4) / (x2+0.6152x+0.712)
=12x+5.618 
This quotient polynomial is a linear function
x3 = -5.618/12 = -0.4682
So the roots of the equation are -0.3076+0.7857i, -0.3076-0.7857i, -0.4682.

3. What will be the roots of the polynomial using Bairstow’s method where f(x) = x2-x+1 and r = -1, s = -1?
a) 0.866i, 0.5-0.866i
b) 0.5+0.866i, 0.5-0.866i
c) 0.5+0.866, 0.5-0.866i
d) 0.5+0.866i, 0.5-0.866
View Answer

Answer: b
Explanation: Given,
x2-x+1 = 0
Let the initial approximation be r=-1 and s=-1
Here a2=1, a1=-1, a0=1
This quotient polynomial is a quadratic polynomial, so roots can be found using quadratic formula
x1, 2 = \(\frac{-b±\sqrt{b^2-4ac}}{2a}\)
x1, 2 = \(\frac{1±\sqrt{1-4⋅1⋅1}}{2⋅1}\)

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x1, 2 = (1±√-3)/2
x1, 2 = (1±1.7321i)/2
x1 = 0.5+0.866i, x2 = 0.5-0.866i
So the roots of the equation are 0.5+0.866i, 0.5-0.866i.

4. What will be the roots of the polynomial using Bairstow’s method where f(x) = x4-2x+5 and r = -1,
s = -1?
a) 0.8212i, 1.0688-0.8212i, -1.0688+1.2689i, -1.0688-1.2689i
b) 1.0688+0.8212i, 0.8212i, -1.0688+1.2689i, -1.0688-1.2689i
c) 1.0688+0.8212i, 1.0688-0.8212i, -1.0688+1.2689i, -1.0688-1.2689i
d) 1.0688, 1.0688-0.8212i, -1.0688+1.2689i, -1.0688-1.2689i
View Answer

Answer: c
Explanation: Given,

Iteration ai, n bi, n ci, n Δr Δs r s
1 a4=1
a3=0
a2=0
a1=-2
a0=5
b4=1
b3=-1
b2=0
b1=-1
b0=6
c4=1
c3=-2
c2=1
c1=0
-11 -6 -12 -7
2 a4=1
a3=0
a2=0
a1=-2
a0=5
b4=1
b3=-12
b2=137
b1=-1562
b0=17790
c4=1
c3=-24
c2=418
c1=-6410
10.8196 123.3576 -1.1804 116.3576
3 a4=1
a3=0
a2=0
a1=-2
a0=5
b4=1
b3=-1.1804
b2=117.751
b1=-278.3477
b0=14034.7915
c4=1
c3=-2.3609
c2=236.8954
c1=-832.6879
0.6058 -57.1154 -0.5746 59.2422
4 a4=1
a3=0
a2=0
a1=-2
a0=5
b4=1
b3=-0.5746
b2=59.5725
b1=-70.2761
b0=3574.5895
c4=1
c3=-1.1493
c2=119.4751
c1=-207.0181
0.3055 -29.3898 -0.2692 29.8525
5 a4=1
a3=0
a2=0
a1=-2
a0=5
b4=1
b3=-0.2692
b2=29.9249
b1=-18.0891
b0=903.2011
c4=1
c3=-0.5383
c2=59.9223
c1=-50.2868
0.1677 -14.9321 -0.1014 14.9204
6 a4=1
a3=0
a2=0
a1=-2
a0=5
b4=1
b3=-0.1014
b2=14.9306
b1=-5.0273
b0=228.2802
c4=1
c3=-0.2028
c2=29.8716
c1=-11.083
0.1167 -7.5988 0.0153 7.3216
7 a4=1
a3=0
a2=0
a1=-2
a0=5
b4=1
b3=0.0153
b2=7.3218
b1=-1.7761
b0=58.5803
c4=1
c3=0.0306
c2=14.6439
c1=-1.3284
0.1296 -3.9886 0.1449 3.333
8 a4=1
a3=0
a2=0
a1=-2
a0=5
b4=1
b3=0.1449
b2=3.354
b1=-1.031
b0=16.0297
c4=1
c3=0.2898
c2=6.729
c1=0.91
0.2573 -2.417 0.4022 0.9161
9 a4=1
a3=0
a2=0
a1=-2
a0=5
b4=1
b3=0.4022
b2=1.0779
b1=-1.198
b0=5.5055
c4=1
c3=0.8044
c2=2.3175
c1=0.471
1.4434 -2.669 1.8456 -1.753
10 a4=1
a3=0
a2=0
a1=-2
a0=5
b4=1
b3=1.8456
b2=1.6534
b1=-2.1838
b0=-1.9288
c4=1
c3=3.6913
c2=6.7131
c1=3.7355
0.2411 0.1532 2.0867 -1.5998
11 a4=1
a3=0
a2=0
a1=-2
a0=5
b4=1
b3=2.0867
b2=2.7546
b1=0.4097
b0=1.4481
c4=1
c3=4.1734
c2=9.8635
c1=14.3153
0.0533 -0.2242 2.1401 -1.824
12 a4=1
a3=0
a2=0
a1=-2
a0=5
b4=1
b3=2.1401
b2=2.7558
b1=-0.006
b0=-0.0394
c4=1
c3=4.2801
c2=10.0915
c1=13.7832
-0.0025 0.0074 2.1375 -1.8167
13 a4=1
a3=0
a2=0
a1=-2
a0=5
b4=1
b3=2.1375
b2=2.7523
b1=0
b0=0
c4=1
c3=4.275
c2=10.0736
c1=13.7663
0 0 2.1375 -1.8167
14 a4=1
a3=0
a2=0
a1=-2
a0=5
b4=1
b3=2.1375
b2=2.7523
b1=0
b0=0
c4=1
c3=4.275
c2=10.0736
c1=13.7662
0 0 2.1375 -1.8167
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Now determining the roots using r and s
x1, 2 = \(\frac{r±\sqrt{r^2+4s}}{2}\)
x1, 2 = \(\frac{2.1375±\sqrt{4.569+4⋅ (-1.8167)}}{2}\)
x1, 2 = (2.1375±1.6424i)/2
x1 = 1.0688+0.8212i, x2 = 1.0688-0.8212i
Now, dividing the polynomial
(x4-2x+5)/(x2-2.1375x+1.8167)
=x2+2.1375x+2.7523 
This quotient polynomial is a quadratic polynomial, so roots can be found using quadratic formula
x3,4 = \(\frac{-b±\sqrt{b^2-4ac}}{2a}\)
x3,4 = \(\frac{-2.1375±\sqrt{4.569-4⋅1⋅2.7523}}{2⋅1}\)
x3,4 = \(\frac{-2.1375±\sqrt{-6.4403}}{2}\)
x3,4 = (-2.1375±2.5378i)/2
x3 = -1.0688+1.2689i, x4 = -1.0688-1.2689i
So the roots of the equation are 1.0688+0.8212i, 1.0688-0.8212i, -1.0688+1.2689i, -1.0688-1.2689i.

5. What will be the roots of the polynomial using Bairstow’s method where f(x) = 0.7x3-4x2+2 and r = -1, s = -1?
a) 0.2394, -0.669, 5.624
b) 0.7594, -0.669, 7.624
c) 0.7594, -0.669, 5.624
d) 0.7594, 0.669, 5.624
View Answer

Answer: b
Explanation: Given,

Iteration ai, n bi, n ci, n Δr Δs r s
1 a3=0.7
a2=-4
a1=0
a0=2
b3=0.7
b2=-4.7
b1=4
b0=2.7
c3=0.7
c2=-5.4
c1=8.7
1.0182 2.1404 0.0182 1.1404
2 a3=0.7
a2=-4
a1=0
a0=2
b3=0.7
b2=-3.9873
b1=0.7257
b0=-2.534
c3=0.7
c2=-3.9745
c1=1.4517
0.0751 -0.6101 0.0933 0.5303
3 a3=0.7
a2=-4
a1=0
a0=2
b3=0.7
b2=-3.9347
b1=0.004
b0=-0.0862
c3=0.7
c2=-3.8693
c1=0.014
-0.003 -0.0223 0.0903 0.508
4 a3=0.7
a2=-4
a1=0
a0=2
b3=0.7
b2=-3.9368
b1=0
b0=0.0001
c3=0.7
c2=-3.8735
c1=0.0057
0 0 0.0903 0.508
5 a3=0.7
a2=-4
a1=0
a0=2
b3=0.7
b2=-3.9368
b1=0
b0=0
c3=0.7
c2=-3.8735
c1=0.0057
0 0 0.0903 0.508

Now determining the roots using r and s
x1, 2 = \(\frac{r±\sqrt{r^2+4s}}{2}\)
x1, 2 = \(\frac{0.0903±\sqrt{0.0082+4⋅0.508}}{2}\)
x1, 2 = (0.0903±1.4284)/2

x1 = 0.7594, x2 = -0.669
Now, dividing the polynomial
(0.7x3-4x2+2)/(x2-0.0903x-0.508)
=0.7x-3.9368 
This quotient polynomial is a linear function
x3 = 3.9368/0.7 = 5.624
So the roots of the equation are 0.7594, -0.669, 5.624.

6. Which one is the incorrect option, while finding the roots of fn(x) with respect to the possibilities of
quotient polynomial?
a) fn-2(x) is a third order or higher order polynomial
b) fn-2(x) is a quadratic function
c) fn-2(x) is a linear function
d) Directly getting the roots
View Answer

Answer: d
Explanation: If fn-2(x) is a third order or higher order polynomial then we apply the Bairstow's technique another time to the quotient polynomial, if it is quadratic function we can simply apply the quadratic formula, if it is a linear function then the remaining single root becomes x=-b/a.

7. Quadratic formula for quadratic function is \(\frac{-b±\sqrt{b^2-4ac}}{2a}\).
a) False
b) True
View Answer

Answer: a
Explanation: If quotient formula is a quadratic function we apply the quadratic formula which is \(\frac{-b±\sqrt{b^2-4ac}}{2a}\). Solving with the help of the quadratic formula the quadratic should be in the form of “ax2 + bx + c = 0”, for the formula to be applied successfully.

8. Bairstow’s method is an algorithm used to find the roots of a polynomial of arbitrary degree.
a) True
b) False
View Answer

Answer: a
Explanation: Bairstow’s method, as again an iterative approach to find the roots of a polynomial equation. This methods divides the polynomial of any arbitrary degree by a quadratic function, which further gives us a new polynomial and a remainder.

9. How is the iteration stopping error denoted in Bairstow’s method?
a) εa,r
b) εa,s
c) εs 
d) ε
View Answer

Answer: c
Explanation: If |εa, r| > εs or |εa, s| > εs, where εs is the iteration stopping error the values of the roots can be determined by x = \(\frac{r±\sqrt{r^2+4s}}{2}\). Where r and s are guessed values. Bairstow’s method is used to evaluate new values for r and s. The previous values of r and s
can serve as the starting guesses.

10. What will be the roots of the polynomial using Bairstow’s method where f(x) = 13x2+12x+4 and r = -1, s = -1?
a) -0.4615+0.3077i, -0.4615-0.3077i
b) -0.4615+0.3077i, 0.3077i
c) 0.3077i, -0.4615-0.3077i
d) -0.4615-0.3077i, -0.4615-0.3077i
View Answer

Answer: a
Explanation: Given,
13x2+12x+4 = 0
Let the initial approximation be r=-1 and s=-1
Here a2=13, a1=12, a0=4
This quotient polynomial is a quadratic polynomial, so roots can be found using quadratic formula
x1, 2 = \(\frac{-b±\sqrt{b^2-4ac}}{2a}\)
x1, 2 = \(\frac{-12±\sqrt{144-4⋅13⋅4}}{2⋅13}\)
x1, 2 = \(\frac{-12±\sqrt{-64}}{2⋅13}\)
x1, 2 = \(\frac{-12±8i}{2⋅13}\)
x1 = -0.4615+0.3077i, x2 = -0.4615-0.3077i
So the roots of the equation are -0.4615+0.3077i, -0.4615-0.3077i.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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