# Polynomial Equations and Maximum Coefficients

If $\alpha$ is a root of a polynomial equation $P(x) = 0$, then $P(\alpha) = 0$, and if $M$ is the maximum coefficient value of $|a_{n}|, |a_{n-1}|, |a_{n-2}|, \cdots ,|a_2|, |a_1|$ then $nM \ge |a_{n}| + |a_{n-1}| + |a_{n-2}| + \cdots + |a_2| + |a_1|$. One of the absolute value inequalities can be used effectively for proving polynomial equation properties, $|x + y| \le |x| + |y|$. You can find the proof of absolute value inequalities.

## Worked Example

Let $\alpha$ be a root of $P(x) = x^n +A_{n-1} x^{n-1} + A_{n-2} x^{n-2} + \cdots + A_2 x^2 + A_1 x + A_0$.
Let $M$ be the maximum value of $|A_{n}|, |A_{n-1}|, |A_{n-2}|, \cdots , |A_2|, |A_1|$.

(a)    Show that $|\alpha |^n \ge M(|\alpha|^{n-1} + |\alpha|^{n-2} + \cdots + |\alpha| + 1)$, given $|a + b| \ge |a| + |b|$.

\begin{aligned} \require{AMSsymbols} \require{color} \displaystyle P(\alpha) &= 0 &\color{red} \text{since } \alpha \text{ is a root} \\ 0 &= \alpha^n + A_{n-1} \alpha^{n-1} + A_{n-2} \alpha^{n-2} + \cdots + A_1 \alpha + A_0 \\ \alpha^n &= -A_{n-1} \alpha^{n-1}-A_{n-2} \alpha^{n-2}-\cdots-A_1 \alpha-A_0 \\ |\alpha^n| &= |-A_{n-1} \alpha^{n-1}-A_{n-2} \alpha^{n-2}-\cdots-A_1 \alpha-A_0| \\ |\alpha|^n &= |A_{n-1} \alpha^{n-1} + A_{n-2} \alpha^{n-2} + \cdots + A_1 \alpha + A_0| \\ |\alpha|^n &\le |A_{n-1} \alpha^{n-1}| + |A_{n-2} \alpha^{n-2}| + \cdots + |A_1 \alpha| + |A_0| &\color{red} |x + y| \le |x| + |y|\\ &= |A_{n-1}||\alpha^{n-1}| + |A_{n-2}||\alpha^{n-2}| + \cdots + |A_1||\alpha| + |A_0| \\ &\le M|\alpha^{n-1}| + M|\alpha^{n-2}| + \cdots + M|\alpha| + M \\ &= M(|\alpha^{n-1}| + |\alpha^{n-2}| + \cdots + |\alpha| + 1) \\ &\le M(|\alpha|^{n-1} + |\alpha|^{n-2} + \cdots + |\alpha| + 1) &\color{red} |\alpha^n| \le |\alpha|^n \\ \therefore |\alpha|^n &\le M(|\alpha|^{n-1} + |\alpha|^{n-2} + \cdots + |\alpha| + 1) \end{aligned}

(b)    Show that $1 < | \alpha| < 1 + M$ for any root $\alpha$ of $P(x)$.

\begin{aligned} \displaystyle \require{AMSsymbols} \require{color}\ \newcommand\ddfrac[2]{\frac{\displaystyle #1}{\displaystyle #2}} |\alpha|^n &\le (|\alpha|^{n-1} + |\alpha|^{n-2} + \cdots + |\alpha| + 1) \\ |\alpha|^n &\le M \times \frac{|\alpha|^n-1}{|\alpha|-1} &\color{red} a + ar + ar^2 + \cdots + ar^{n-1} = \frac{a(r^2-1)}{r-1} \\ 1 &\le M \times \frac{1-\ddfrac{1}{|\alpha|^n}}{|\alpha|-1} &\color{red}\text{divide both sides by } |\alpha|^n \\ 1 &\le M \times \frac{1-\ddfrac{1}{|\alpha|^n}}{|\alpha|-1} < \frac{1}{|\alpha|-1} \\ 1 &< \frac{1}{|\alpha|-1} \\ (|\alpha|-1)^2 &< M \times (|\alpha|-1) &\color{red}\text{multipy both sides by } (|\alpha|-1)^2 \\ (|\alpha|-1)^2-M \times (|\alpha|-1) &<0 \\ (|\alpha|-1)( |\alpha|-1-M ) &< 0 \\ \therefore 1 &< |\alpha| < 1 + M \end{aligned}

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