# 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

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

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

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

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

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

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

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

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

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

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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