MHB Is Any Real Zero of This Polynomial Greater Than M+1?

[Euge]

Here is this week's POTW:

-----
Let $f(x)\in \Bbb R[x]$ be a monic polynomial. Prove that if $M$ is the greatest of the absolute values of its coefficients, then no real zero of $f$ can exceed $M + 1$.

-----

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Euge]

No one answered this week's problem. You can find my solution below.

Spoiler

Write $f(x) = x^n + a_{n-1}x^{n-1} + a_{n-2}x^{n-2} + \cdots + a_1 x + a_0$. The entries of the bottom row of the synthetic division of $f(x)$ by $x - (M + 1)$ are the numbers $1, b_1, b_2,\ldots, b_n$ where $b_1 = a_{n - 1} + M + 1$ and $b_k = (M + 1)b_{k-1} + a_{n-k}$ for $2 \le k \le n$. Then $$b_1 \ge a_{n-1} + |a_{n-1}| + 1 \ge a_{n-1} - a_{n-1} + 1 = 1,$$ and if $b_k \ge 1$ for some $k \in \{2,\ldots, n-1\}$, then $$b_{k+1} = (M+1)b_k + a_{n-k} \ge M + 1 + a_{n-k} \ge |a_{n-k}| + a_{n-k} + 1 \ge 1.$$ So inductively all the $b$'s are greater than or equal to $1$. Since the entries in the bottom row are nonnegative and $M + 1 > 0$, by the boundedness for polynomials, every real zero of $f(x)$ is less than or equal to $M + 1$.