As a guide the solutions are: x=3.1479, 0.4261+0.3690i, 0.42610.3690i. This is what Bairstow’s Method should give us to 4 decimal places.
Bairstow’s Method is based on the fact that the zeroes of a polynomial with real coefficients can be complex and that these zeroes always come in pairs—a complex number and its conjugate. These pairs form quadratics with real coefficients. Dividing by a quadratic reduces the degree of the original polynomial by two. If there is no remainder after the division, the zeroes of the quadratic will be zeroes of the original polynomial. The process can then be repeated on the quotient polynomial until it is stripped to become a linear or quadratic expression which reveals the remaining zero/zeroes. In the given problem which is a cubic polynomial, there will be one real and at most two complex zeroes, hence only one quadratic needs to be found.
The initial quadratic equation is x²+x0.4 using the given values r=1 and s=0.4 for the quadratic x²+rx+s. Synthetic quadratic division is usually used for the next step, but algebraic division gives the same result: a quotient and a remainder. In this case the quotient will be a linear expression, the result of dividing a cubic by a quadratic. We get x5 as the quotient and 8.4x3 as the remainder. Below is the first iteration of the method, showing the details of quadratic synthetic division. In the last row of this table we have r₁ and s₁ which are the starting points for the second iteration. These values are negated and replace r₀ and s₀ in the third row. The division is repeated. The correction values are derived from the system:
c₁∆r₀+c₀∆s₀=b₂, (c₂b₂)∆r₀+c₁∆s₀=b₃ using simultaneous equation solutions.
r₁=r₀+∆r₀ and s₁=s₀+∆s₀ join the calculations in successive iterations.


a₀

a₁

a₂

a₃

r

s

1

4

3

1

1

0.4

0

1

5

8.4



0

0

0.4

2



b₀

b₁

b₂

b₃



1

5

8.4

11.4



0

1

6




0

0

0.4




c₀

c₁

c₂




1

6

14.8


0.31756757

0.09459459





The computations for further iterations are long and tedious, so are not included here. The final solutions for r and s are 0.852101 and 0.317672 to 6 decimal places making the final quadratic x²0.852101x+0.317672. The final value of b₁=3.147899, making x=3.147899 the real zero of the original cubic, x=3.1479 to 4 decimal places. The complex zeroes of the quadratic are given at the beginning of this solution.