# What are the zeros of f(x) = 5x^7 − x + 216?

Mar 2, 2015

The first attempt to do is to try to factor that polinomy.

For the remainder theorem we have to calculate $f \left(h\right)$ for all the integer numbers that divide $216$. If $f \left(h\right) = 0$ for a number h, so this is a zero.

The divisors are:

$\pm 1 , \pm 2 , \ldots$

I tried some small of them, that did't work, and the other were too big.
So this polinomy can't be factorized.

We have to try another way!

Let's try to study the function.

The domain is $\left(- \infty , + \infty\right)$, the limits are:

${\lim}_{x \rightarrow \pm \infty} f \left(x\right) = \pm \infty$

and so, there are not asymptotes of any type (obliqual, horizontal or vertical).

The derivative is:

$y ' = 35 {x}^{6} - 1$

and let's study the sign:

$35 {x}^{6} - 1 \ge 0 \Rightarrow {x}^{6} \ge \frac{1}{35} \Rightarrow$

$x \le - {\left(\frac{1}{35}\right)}^{\frac{1}{6}} \vee x \ge {\left(\frac{1}{35}\right)}^{\frac{1}{6}}$,

(the numbers are $\cong \pm 0.55$)

so the function growths before $- {\left(\frac{1}{35}\right)}^{\frac{1}{6}}$ and after ${\left(\frac{1}{35}\right)}^{\frac{1}{6}}$, and decrease in the middle of the two.

So: the point $A \left(- {\left(\frac{1}{35}\right)}^{\frac{1}{6}} , \cong 216\right)$ is a local maximum and the point $B \left({\left(\frac{1}{35}\right)}^{\frac{1}{6}} , \cong 215\right)$ is a local minumum.

Since their ordinate are positive, these point are over the x-axis,
so the function cuts the x-axis in only one point, as you can see:

graph{5x^7-x+216 [-34.56, 38.5, 199.56, 236.1]}

graph{5x^7-x+216 [-11.53, 10.98, -2.98, 8.27]}

So there is only one zero!