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

Feb 27, 2016

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:

$f \left(x\right) = - {x}^{5} + 3 {x}^{4} + 16 {x}^{3} - 2 {x}^{2} - 95 x - 44$

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

$\pm 1$, $\pm 2$, $\pm 4$, $\pm 11$, $\pm 22$, $\pm 44$.

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

$f \left(- 1\right) = 37$

$f \left(1\right) = - 123$

$f \left(4\right) = 312$

$f \left(11\right) = - 97163$

So $f \left(x\right)$ 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:

${a}_{i + 1} = {a}_{i} - f \frac{{a}_{i}}{f ' \left({a}_{i}\right)}$

In our example, $f ' \left(x\right) = - 5 {x}^{4} + 12 {x}^{3} + 48 {x}^{2} - 4 x - 95$

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 $f \left(x\right)$ by the factors corresponding to the zeros we have found to get a quadratic and use the quadratic formula.