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
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.
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
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
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 = \(\frac{1±\sqrt{1-4⋅1⋅1}}{2⋅1}\)
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
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 |
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
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
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
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
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
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
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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