Question

How do you find all the zeros of `f(x) = –3x^5 – 8x^4 +25x^3 – 8x^2 +x – 19`?

Answer 1

Use Durand-Kerner method to find numerical approximations:

`x_1 ~~ -4.61495`

`x_(2,3) ~~ 1.34883+-0.412784i`

`x_(4,5) ~~ -0.374682+-0.74117i`

Explanation:

`f(x) = -3x^5-8x^4+25x^3-8x^2+x-19`

By the rational root theorem, any rational zeros of `f(x)` must be expressible in the form `p/q` for integers `p`, `q` with `p` a divisor of the constant term `-19` and `q` a divisor of the coefficient `-3` of the leading term.

So the only possible rational zeros are:

`+-1/3, +-1, +-19/3, +-19`

Evaluating `f(x)` for each of these values we find none works. So `f(x)` has no rational zeros.

`color(white)()`
In common with most quintics, the zeros of this `f(x)` are not expressible in terms of `n`th roots. There are no algebraic solutions in terms of elementary functions, including trigonometric ones.

About the best we can do is use a numeric method such as Durand-Kerner to find approximations to the zeros:

Suppose the `5` zeros are: `p, q, r, s, t`.

Choose initial approximations for these zeros as follows:

`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`

`t_0 = (0.4+0.9i)^4`

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)(p_i-t_i))`

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

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

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

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

Keep iterating until the values are stable to the desired accuracy.

With this method, I found the following approximations for the zeros:

`x_1 ~~ -4.61495`

`x_(2,3) ~~ 1.34883+-0.412784i`

`x_(4,5) ~~ -0.374682+-0.74117i`

Here's a sample C++ program that implements the algorithm for the example quintic: