# How do you find all the zeros of x^5 - 3x^4 + 5x^3 + 9x^2 + x - 2?

Jun 10, 2016

Use a numerical method to find approximations:

${x}_{1} \approx 0.390503$

${x}_{2 , 3} \approx 2.01006 \pm 2.24677 i$

${x}_{4 , 5} \approx - 0.705308 \pm 0.257059 i$

#### Explanation:

$f \left(x\right) = {x}^{5} - 3 {x}^{4} + 5 {x}^{3} + 9 {x}^{2} + x - 2$

By the rational root theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 2$ and $q$ a divisor of the coefficient $1$ of the leading term.

So the only possible rational zeros are:

$\pm 1$, $\pm 2$

Neither of these is a zero, so $f \left(x\right)$ has no rational zeros.

In common with quintics in general, this $f \left(x\right)$ has no algebraic solution in terms of $n$th roots.

We can find rational approximations using a numeric method such as Durand-Kerner. For another example of such a quintic solution, see: https://socratic.org/s/avdSNDdg

In the current example we find approximations:

${x}_{1} \approx 0.390503$

${x}_{2 , 3} \approx 2.01006 \pm 2.24677 i$

${x}_{4 , 5} \approx - 0.705308 \pm 0.257059 i$

I used the following C++ program: