Question

How do you find all the real and complex roots of `x^4 – 13x^3 + x^2 – 5 = 0`?

Answer 1

Use a numerical method to find approximate roots:

`x_1 ~~ 12.9249`

`x_2 ~~ 0.383152 + 0.64252i`

`x_3 ~~ 0.383152 - 0.64252i`

`x_4 ~~ -0.691249`

Explanation:

`f(x) = x^4-13x^3+x^2-5`

By the rational root theorem, any rational zeros of `f(x)` are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-5` and `q` a divisor of the coefficient `1` of the leading term.

So the only possible zeros are:

`+-1`, `+-5`

None of these work, so `f(x)` has no rational zeros.

It is possible to solve algebraically, but the solution is horribly messy.

Instead, find approximations to the zeros using a numerical method known as the Durand-Kerner method.

Suppose the `4` zeros are `p, q, r, s`.

Choose initial approximations:

`p_0 = (0.4+0.9i)^0`
`q_0 = (0.4+0.9i)^1`
`r_0 = (0.4+0.9i)^2`
`s_0 = (0.4+0.9i)^3`

Then iterate using the formulas:

`p_(i+1) = p_i-f(p_i)/((p_i-q_i)(p_i-r_i)(p_i-s_i))`

`q_(i+1) = q_i-f(q_i)/((q_i-p_(i+1))(q_i-r_i)(q_i-s_i))`

`r_(i+1) = r_i-f(r_i)/((r_i-p_(i+1))(r_i-q_(i+1))(r_i-s_i))`

`s_(i+1) = s_i-f(s_i)/((s_i-p_(i+1))(s_i-q_(i+1))(s_i-r_(i+1))`

Using this method and iterating a few times, we find approximations:

`x_1 ~~ 12.9249`

`x_2 ~~ 0.383152 + 0.64252i`

`x_3 ~~ 0.383152 - 0.64252i`

`x_4 ~~ -0.691249`

Here's a C++ program that I used...