Bairstow's Method uses the fact that all polynomials with real coefficients can be resolved into real quadratic factors and no more than one real linear factor. If the polynomial has complex zeroes, then the complex zeroes will form conjugate pairs which create real quadratic factors. In an n-degree polynomial, if n is odd there will always be a real linear factor, but otherwise all the zeroes could be complex.
The method begins by guessing the coefficients of a quadratic, then dividing the polynomial by this quadratic to obtain a quotient and remainder. Using a form of Newton's iterative method, the coefficients of the quadratic are refined such that the remainder approaches zero, with a given error margin, and the zeroes of the quadratic become zeroes of the polynomial, these being found by applying the usual quadratic formula.
The process reduces the degree of the polynomial by two, and is repeated until all relevant quadratics (and their zeroes) have been found. In this way all the zeroes of the polynomial can be found. We are left with a real zero if the degree was odd.
In problems of this type it is normal for first approximations of the quadratic coefficient to be given as well as some error margin. However, although an error margin has been provided, such approximations have not been provided in this case. But because the polynomial is a cubic (odd degree), there must be at least one real zero, and this can be found using Newton's iterative method.
f(3)=-1 and f(4)=11 so there is a real zero between x=3 and x=4, making x0=3 an approximate starting point for Newton's method. We need f'(x) (derivative of f(x)) to use this method=3x2-8x+3. f'(3)=27-24+3=6.
xn+1=xn-f(xn)/f'(xn) is the formula, so x1=3-(-1)/6=19/6=3.166...
x2=85/27=3.148148..., x3=3.147899... x3 and x4 are within the given error margin, so we can use synthetic division to derive the quadratic:
x4 | 1 -4 3 -1
1 x4 x42-4x4 | x43-4x42+3x4
1 x4-4 x42-4x4+3 | 0 (approximately) (because x43-4x42+3x4=1 approximately)
So the quadratic is x2+(x4-4)x+(x42-4x4+3).
Let's substitute actual values for the coefficients:
x4-4=-0.852101, x42-4x4+3=0.317672: quadratic is x2-0.852101x+0.317672=0.
(Approximations to these coefficients could have been 0.9 and 0.3, to start the iteration process for Bairstow's Method.)
Quadratic formula: x=½(0.852101±√(0.8521012-4×0.317672).
x=-0.42605±i√0.544613/2=-0.42605±0.36899i.
Although we've found the zeroes to be 3.147899, -0.42605+0.36899i, and -0.42605-0.36899i, we haven't yet actually used Bairstow's Method because we were lacking starting values for the quadratic. But we now seem to have two values r=-0.9 and s=0.3 as typical starting values for x2+rx+s. We can set up a table to get more accurate values for r and s, and we can expect to confirm the solution already determined above. This table performs synthetic quadratic division.
The second row contains the coefficients of the polynomial (cubic); the starting coefficients of the quadratic are r and s.
| i: |
0 |
1 |
2 |
3 |
| ai: |
1 |
-4 |
3 |
-1 |
| -r0=0.9 |
0 |
0.9 |
-2.79 |
-0.081 |
| -s0=-0.3 |
0 |
0 |
-0.3 |
0.93 |
| bi: |
1 |
-3.1 |
-0.09 |
-0.151 |
| -r0=0.9 |
0 |
0.9 |
-1.98 |
-2.133 |
| -s0=-0.3 |
0 |
0 |
-0.3 |
0.66 |
| ci: |
1 |
-2.2 |
-2.37 |
-1.624 |
Now we need to compute the differences Δa and Δb which take us to the next iteration of r and s.
cn-2Δa+cn-3Δb=bn-1 and (cn-1-bn-1)Δa+cn-2Δb=bn.
n=3 in this case (the degree of the polynomial).
Therefore: c1Δa+c0Δb=b2, (c2-b2)Δa+c1Δb=b3 are simultaneous equations for finding Δa and Δb.
Substituting computed values from the table:
-2.2Δa+Δb=-0.09, -2.28Δa-2.2Δb=-0.151.
(-4.84Δa+2.2Δb=-0.198)+(-2.28Δa-2.2Δb=-0.151)⇒
-7.12Δa=-0.349, Δa=0.049017 approx.
Δb=-0.09+2.2×0.049017=0.017837 approx.
r1=r0+Δa, s1=s0+Δb⇒r1=-0.850983 approx, s1=0.317837 approx.
The process is repeated using r1 and s1 in place of r0 and s0. We can see from the earlier solution that these values are better approximations.
This answer is sufficient to demonstrate the method, and I think it's safe to assume that the earlier solution is confirmed.
[See also:
https://www.mathhomeworkanswers.org/281500/bairstows-method
and
https://www.mathhomeworkanswers.org/280293/using-bairstows-method-determine-roots-polynomial-function for answers to the same problem.]