It's often helpful to have an interval that is guaranteed to
hold all the real roots of a polynomial.
This section states and proves such a theorem—
it's a beautiful proof,
which uses some important properties of the real numbers that have not yet appeared in this course.
In particular, if you're using a calculator to graph a polynomial,
then you probably want to set your window so you're guaranteed to see all the $x$-intercepts.
This section is for you!
Here's the outline for this section:
Let $\,P\,$ be a polynomial of degree $\,n\ge 1\,$ with real coefficients.
That is,
$$\cssId{s17}{P(x) = a_nx^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0}$$
where $\,a_n\ne 0\,$ and all the $\,a_i\,$ are real numbers.
Let $\displaystyle\,M := \text{max}\bigl(\,|a_{n-1}|,\ldots,|a_0|\,\bigr)\,.$ Let $\,r\,$ be any real root (zero) of $\,P\,,$ so that $\,P(r) = 0\ $. Then: $$\cssId{s22}{|r| \le 1 + \frac{M}{|a_n|}}$$ If we define $\displaystyle\,K := 1 + \frac{M}{|a_n|}\,,$ then: |
Let $\,P(x) = (x-1)(x+2)(x-3) = x^3 - 2x^2 - 5x + 6\,.$
It's clear from the factored form that the roots of $\,P\,$ are $\,1\,,$ $\,-2\,,$ and $\,3\,,$
so we'll be able to confirm the bounding interval obtained from the theorem.
The previous example is tweaked by changing the leading coefficient to $\,7\,.$
Let $\,P(x) = 7x^3 - 2x^2 - 5x + 6\,.$
$n\,$ (the degree) is $3$
$a_3\,$ (the leading coefficient) is $\,7\,$
$M = \text{max}\bigl(\,|-2|\,,\,|-5|\,,\,|6|\,\bigr) = 6$
$\displaystyle K \ =\ 1 + \frac{M}{|a_n|} \ =\ 1 + \frac{6}{|7|} \ =\ \frac{13}{7}\ \approx\ 1.9$
All the zeroes are guaranteed to lie in the interval $\,[-\frac{13}{7},\frac{13}{7}]\,.$
A quick trip to WolframAlpha shows that there is only one real root, which is approximately $\,-1.1\,.$
Here's a sketch from WolframAlpha that shows all the complex roots:
Let $\,r\,$ be a real root of $\,P\,,$ so $\,P(r) = 0\ $: $$\cssId{sb2}{a_nr^n + a_{n-1}r^{n-1} + \cdots + a_1r + a_0 = 0}$$ Dividing both sides of the equation by $\,a_n\ne 0\,$ gives the equivalent equation: $$\cssId{sb4}{r^n + \frac{a_{n-1}}{a_n}r^{n-1} + \cdots + \frac{a_1}{a_n}r + \frac{a_0}{a_n} = 0} \tag{1}$$
Define $\displaystyle\,M := \text{max}\bigl(\,|a_{n-1}|,\ldots,|a_0|\,\bigr)\,,$ so that $$ \cssId{sb6}{\text{max}\left(\left|\frac{a_{n-1}}{a_n}\right|,\ldots,\left|\frac{a_0}{a_n}\right|\right)} \cssId{sb7}{= \text{max}\left(\frac{|a_{n-1}|}{|a_n|},\ldots,\frac{|a_0|}{|a_n|}\right)} \cssId{sb8}{= \frac{\text{max}(|a_{n-1}|,\ldots,|a_0|)}{|a_n|}} \cssId{sb9}{= \frac{M}{|a_n|}} $$
We consider two cases: $|r|\le 1\,$ and $|r| > 1\,.$ In both cases, it must be shown that $\displaystyle\,|r| \le 1 + \frac{M}{|a_n|}\,.$
Assume $\,|r| \le 1\,.$
Since absolute value is a nonnegative quantity, $\displaystyle\,\frac{M}{|a_n|} \ge 0\,.$
Then, as desired, we have:
$$\cssId{sb17}{|r| \quad\le\quad 1 \quad\le\quad 1 + \frac{M}{|a_n|}}$$
Assume $|r| > 1$. Rewrite (1) as
$$\cssId{sb20}{r^n = -\frac{a_0}{a_n} - \frac{a_1}{a_n}r - \cdots - \frac{a_{n-1}}{a_n}r^{n-1}}\tag{2}$$
Then:
$$
\begin{alignat}{2}
\cssId{sb22}{|r^n|} \quad
&\cssId{sb23}{= \quad \left|-\frac{a_0}{a_n} - \frac{a_1}{a_n}r - \cdots - \frac{a_{n-1}}{a_n}r^{n-1}\right|} &&\cr\cr
&\cssId{sb24}{\le \quad \left|\frac{a_0}{a_n}\right| + \left|\frac{a_1}{a_n}r\right| + \cdots + \left|\frac{a_{n-1}}{a_n}r^{n-1}\right|} &\qquad\qquad&\cssId{sb25}{\text{triangle inequality}}\cr\cr
&\cssId{sb26}{= \quad \frac{|a_0|}{|a_n|} + \frac{|a_1|}{|a_n|}|r| + \cdots + \frac{|a_{n-1}|}{|a_n|}|r^{n-1}|} &\qquad\qquad& \cssId{sb27}{\text{since} \left|\frac{a}{b}\right| = \frac{|a|}{|b|} \text{ and } |ab| = |a|\cdot|b|}\cr\cr
&\cssId{sb28}{= \quad \frac{1}{|a_n|}\left(|a_0| + |a_1|\cdot|r| + \cdots + |a_{n-1}|\cdot|r^{n-1}|\right)} &&\cssId{sb29}{\text{factor out the common factor}} \cr\cr
&\cssId{sb30}{\le \quad \frac{1}{|a_n|}\left(M + M|r| + \cdots + M|r^{n-1}|\right)} &&\cssId{sb31}{\text{since $M$ is the max}}\cr\cr
&\cssId{sb32}{\le \quad \frac{1}{|a_n|}\left(M + M|r| + \cdots + M|r|^{n-1}\right)} &&\cssId{sb33}{\text{since $|r^i| = |r|^i$}}\cr\cr
&\cssId{sb34}{= \quad \frac{M}{|a_n|}(1 + |r| + \cdots + |r|^{n-1})} &&\cssId{sb35}{\text{factor out $M$}}\cr\cr
&\cssId{sb36}{= \quad \frac{M}{|a_n|}\left(\frac{|r|^n-1}{|r| - 1}\right)}&&\cssId{sb37}{\text{sum the geometric series}}\cr\cr
\end{alignat}
$$
Thus:
$$
\cssId{sb39}{|r^n| \le \frac{M}{|a_n|}\left(\frac{|r|^n-1}{|r| - 1}\right)} \tag{3}
$$
Multiplying or dividing both sides of an inequality by a positive number does not change the direction of
the inequality symbol.
Since, by hypothesis, $\,|r| >1\,,$ we have $\,|r| - 1 > 0\,.$
Also,
$$\cssId{sb42}{|r^n|}
\cssId{sb43}{= |r|^n}
\cssId{sb44}{> 1^n}
\cssId{sb45}{= 1}
\cssId{sb46}{> 0}
$$
So, dividing both sides of (3) by $|r^n| = |r|^n$ and multiplying by $\,|r| - 1\,$ gives
$$
\cssId{sb48}{|r| - 1}
\cssId{sb49}{\le \frac{M}{|a_n|}\frac{|r|^n - 1}{|r|^n}}
\cssId{sb50}{= \frac{M}{|a_n|}\left(1 - \frac{1}{|r|^n}\right)}
$$
Note:
$$
\cssId{sb52}{|r| > 1} \quad
\cssId{sb53}{\Rightarrow}\quad
\cssId{sb54}{|r|^n > 1} \quad
\cssId{sb55}{\Rightarrow}\quad
\cssId{sb56}{0 < \frac{1}{|r|^n} < 1}\quad
\cssId{sb57}{\Rightarrow}\quad
\cssId{sb58}{0 > -\frac{1}{|r|^n} > -1}\quad
\cssId{sb59}{\overbrace{\ \ \Rightarrow\ \ }^{\text{add $1$}}}\quad
\cssId{sb60}{1 > 1-\frac{1}{|r|^n} > 0}\quad
\cssId{sb61}{\Rightarrow}\quad
\cssId{sb62}{0 < 1-\frac{1}{|r|^n} < 1}
$$
so
$$\cssId{sb64}{|r| - 1 \le \frac{M}{|a_n|}}$$
from which we get the desired result that
$$\cssId{sb66}{ |r| \le 1 + \frac{M}{|a_n|}}$$
QED