# How do you find all the zeros of f(x) = 9x^4 – 17x^3 + 2x^2 – 3x + 33?

Jun 3, 2016

Check that it has no rational zeros, then resort to a numerical method to find approximations.

#### Explanation:

$f \left(x\right) = 9 {x}^{4} - 17 {x}^{3} + 2 {x}^{2} - 3 x + 33$

By the rational root theorem, any rational zeros of $f \left(x\right)$ can be expressed in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $33$ and $q$ a divisor of the coefficient $9$ of the leading term.

So the only possible rational zeros are:

$\pm \frac{1}{9}$, $\pm \frac{1}{3}$, $\pm 1$, $\pm \frac{11}{9}$, $\pm 3$, $\pm \frac{11}{3}$, $\pm 11$, $\pm 33$

In addition, notice that if you invert the signs of the coefficients on the terms of odd degree, then all of the resulting coefficients are positive. Hence $f \left(x\right)$ has no negative zeros.

So the only possible rational zeros are:

$\frac{1}{9} , \frac{1}{3} , 1 , \frac{11}{9} , 3 , \frac{11}{3} , 11 , 33$

We find that $f \left(x\right) > 0$ for all of these values of $x$. So $f \left(x\right)$ has no rational zeros.

It is possible to solve such a quartic algebraically, but it gets very messy and is not necessarily very useful.

You can use a numerical method such as Durand-Kerner to find good approximations to the two pairs of Complex zeros that it does have.

See https://socratic.org/s/av2PtWLY for an example finding approximations for the zeros of a quintic.

In our example, we find approximations:

$x = 1.57163 \pm 0.656656 i$

$x = - 0.627188 \pm 0.932989 i$

An example C++ program that performs the calculations would be: