Question

How do you find all the zeros of `-x^5+3x^4+16x^3-2x^2-95x-44`?

Answer 1

Use Newton's method to find numeric approximations for the three Real zeros, then divide by the corresponding factors to get a quadratic for the Complex zeros.

Explanation:

`f(x) = -x^5+3x^4+16x^3-2x^2-95x-44`

You could try the rational root theorem first, which would allow you to infer that the only possible rational zeros of `f(x)` are the factors of `44`, viz:

`+-1`, `+-2`, `+-4`, `+-11`, `+-22`, `+-44`.

None of these work, so `f(x)` has no rational zeros, but in the process of trying you might find:

`f(-1) = 37`

`f(1) = -123`

`f(4) = 312`

`f(11) = -97163`

So `f(x)` changes sign at least `3` times and has at least `3` Real zeros.

We can use Newton's method to find good approximations for the Real roots by choosing suitable starting approximations `a_0` and iterating using the formula:

`a_(i+1) = a_i - f(a_i)/(f'(a_i))`

In our example, `f'(x) = -5x^4+12x^3+48x^2-4x-95`

Putting the iteration formula into a spreadsheet and using initial values `a_0 = -1`, `a_0 = 2` and `a_0 = 11`, I found the following approximations after a few iterations:

`-0.485335316717177`

`2.624730249302921`

`5.259365512110042`

To find the Complex zeros, you can either put together a more complicated spreadsheet, with separate columns for Real and Imaginary parts, or you can divide `f(x)` by the factors corresponding to the zeros we have found to get a quadratic and use the quadratic formula.