Bounds on roots of polynomials

Here's an elementary bound. Let $p(x) = x^n - a_{n-1} x^{n-1} - ... - a_0$. If $|x| \ge 1$ then $m \ge n$ implies $|x|^m \ge |x|^n$, hence if $p(x) = 0$ then

$$|x|^n = \left| a_{n-1} x^{n-1} + ... + a_0 \right| \le |x|^{n-1} \left( |a_{n-1}| + ... + |a_0| \right)$$

hence

$$|x| \le \text{max}(1, |a_{n-1}| + ... + |a_0|).$$

In your case we can do much better because of the zero coefficients: we get

$$|x|^7 \le \text{max}(1, \frac{47}{3} |x|^3)$$

hence $|x|^4 \le \frac{47}{3} < 16$, or $|x| < 2$.

A technique useful in special cases is the Gauss-Lucas theorem, and a technique useful in general cases is Rouche's theorem.

If $a_n x^n + \dots + a_1 x + a_0 = 0$ we get the following useful bound by Lagrange:

$$|x| \le 2\max\left\{ \left(\frac{|a_{i}|}{|a_n|}\right)^{\frac1{n-i}} : 0 \le i < n \right\}.$$

The followinig proof by Chee-Keng Yap is simple: Let $\beta=\max\left\{\left(|a_{i}|/|a_n|\right)^{1/(n-i)} : 0 \le i < n\right\}$. Then we want to prove $|x|\le 2\beta$. Obviously this is true if $|x|\le \beta$, so we focus on $|x|>\beta$.

Since $a_n x^n = -(a_{n-1}x^{n-1}+\dots+a_0)$ we have $|a_n| |x|^n \le |a_{n-1}||x|^{n-1}+\dots+|a_0|$. Hence $$ 1 \le \sum_{i=1}^{n}{\frac{|a_{n-i}||x|^{n-i}}{|a_n||x|^n}} = \sum_{i=1}^{n}{\frac{|a_{n-i}|}{|a_n||x|^i}} \le \sum_{i=1}^{n}{\frac{\beta^{i}}{|x|^{i}}} < \frac{\beta/|x|}{1-\beta/|x|} \le 2\beta/|x|, $$ where for the last step we used $|x|>\beta$ and that $\frac{z}{1+z}\le 2z$ for $z\in[0,1]$.