# What are all the possible rational zeros for f(x)=x^5-6x^4+12x^2-8x+36 and how do you find all zeros?

Aug 27, 2016

Find there are no rational zeros. Use a numerical method to find approximations.

#### Explanation:

$f \left(x\right) = {x}^{5} - 6 {x}^{4} + 12 {x}^{2} - 8 x + 36$

By the rational roots 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 $36$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1 , \pm 2 , \pm 3 , \pm 4 , \pm 6 , \pm 9 , \pm 12 , \pm 18 , \pm 36$

None of these works, so $f \left(x\right)$ has no rational zeros.

About the best we can do is use the Durand-Kerner or similar numerical method to find approximations:

${x}_{1} \approx 5.63048$

${x}_{2} \approx 2.05347$

${x}_{3} \approx - 1.84949$

${x}_{4 , 5} \approx 0.0827684 \pm 1.29486 i$

See https://socratic.org/s/axpvv6Hk for some more description of the method.

For the current example, the above approximations were found using this C++ program... 