# An upper bound for roots of polynomials

A natural question to ask when given a polynomial is: how do the roots and the coefficients interplay with one another? In fact a huge portion of polynomial theory and solving algebraic equations is finding out the details of this relationship. The quadratic formula is a prime example – It gives a way of finding roots by only using the values of the coefficients. An explicit relationship, an algebraic formula, is only possible for polynomials of degree less than $5$ which is a famous result pioneered by the work of Galois.

The interplay we are going to determine for polynomials of any degree is that you only need to look for roots in a neighborhood around $0$ with the neighborhood’s size depending only on the size of the polynomial’s coefficients.

Let $p \in \mathbb{R}[x]$ be a polynomial over $\mathbb{R}$ of degree $n$ with the representation $p(x) = a_nx^n +... a_1x + a_0$ where $a_n \neq 0$ and that $|a_i| \leq A$ for all $i = 1,..., a$ and some $A \in \mathbb{R}$. We will be considering our polynomial as a function $p : \mathbb{C} \longrightarrow \mathbb{C}$ and be setting $M \geq 0$ and $R = 1 + \frac{A + M}{|a_n|}$ where our choice of $R$ will be more apparent later.

What we are going to show is that for $x$ of a large enough magnitude our polynomial will have a magnitude as large as we like. We will also establish a simple bound on the magnitude of the roots of our polynomial $p$.

Suppose $|x| \geq R$, we have $|a_n x^n| = \frac{A + M}{R- 1}|x|^n$ by the definition of our constants and estimating $|a_{n-1}x^{n-1} + .... + a_0|$ gives $| a_{n-1}x^{n-1} + .... + a_0 | \leq A (|x^{n-1}| + .... + |x| + 1 ) = A\frac{|x|^n - 1}{|x| - 1 } < A \frac{|x|^n}{R-1}$ by using the sum of a geometric series and the triangle inequality.

Using the triangle inequality allows us to conclude $|a_nx^n| - |a_{n-1}x^{n-1} + ... + a_0 | \leq |p(x)|$ which gives us by using the previous estimation $|p(x)| \geq \frac{A+ M}{R-1} |x|^n - \frac{A}{R-1} |x|^n = \frac{M|x|^n}{R-1} \geq \frac{MR^n}{R-1} > M$ as both $|x| \leq R$ and $R^n > R > R - 1$ as $R > 1$.

Bringing this together we have shown for all $M \geq 0$ if we look at $|x| \geq R$ where $R = 1 + \frac{A + M}{|a_n|}$ we have $|p(x)| > M$.

Looking at the contrapositive of this statement lets us conclude if $|p(x)| \leq M$ for some $x \in \mathbb{C}$ and $M \geq 0$ we must have $|x| < R$ or equivalently $x$ living in the ball $B(0;R)$. If we take $x$ to be a root of $p$ then by setting $M = 0$ we have that $|x| < R = 1 +\frac{A}{|a_n|}$ giving us a bound on the size of our root in terms of the polynomials coefficients. This means all roots of $p$ will be found in the ball $B(0; a + \frac{A}{|a_n|} )$.