‘TRAPPING’ THE ROOTS OF A POLYNOMIAL;
AN INTERVAL GUARANTEED TO CONTAIN ALL REAL ROOTS

LESSON READ-THROUGH (Part 1 of 2)
by Dr. Carol JVF Burns (website creator)
Follow along with the highlighted text while you listen!
Thanks for your support!
 

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:

 
Theorem:   Bounding the Real Roots of a Polynomial
an interval guaranteed to contain all real roots
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:

all the real zeros of $\,P\,$ are contained in the interval $\,[-K,K]\,$

Notes on the theorem

 

EXAMPLES—HOW TO USE THE THEOREM

Example 1

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.

$n\,$ (the degree) is $3$

$a_3\,$ (the leading coefficient) is $\,1\,$

$M = \text{max}\bigl(\,|-2|\,,\,|-5|\,,\,|6|\,\bigr) = 6$

$\displaystyle K \ =\ 1 + \frac{M}{|a_n|} \ =\ 1 + \frac{6}{|1|} \ =\ 1 + 6 = 7$

All the zeroes are guaranteed to lie in the interval $\,[-K,K] = [-7,7]\,$ (which they do!)

Example 2

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:

 

TOOLS NEEDED TO PROVE THE THEOREM

  1. THE TRIANGLE INEQUALITY:
    For all real numbers $\,a\,$ and $\,b\,$:   $|a+b| \le |a| + |b|$
  2. PROPERTIES OF ABSOLUTE VALUE:
    For all real numbers $\,a\,$ and $\,b\,$:
    • $|ab| = |a|\cdot|b|$
    • $\displaystyle \left|\frac{a}{b}\right| = \frac{|a|}{|b|}$     ($b\ne 0$)
  3. Properties (1) and (2) extend to finite sums/products.
    For example:
    $|a + b + c| \le |a| + |b + c| \le |a| + |b| + |c|$
    and
    $|abc| = |a|\cdot|bc| = |a|\cdot|b|\cdot|c|$

    In particular: $|x^n| = |x|^n$
  4. Properties (1) and (2) also hold for complex numbers, where $\,|\cdot|\,$ is the modulus of a complex number.
    Complex numbers are explored in the next few sections.
  5. SUM OF A FINITE GEOMETRIC SERIES:
    For all real numbers $\,a\,$ and $\,x\,$:
    $\displaystyle a + ax + ax^2 + \cdots + ax^n = \frac{a(x^{n+1}-1)}{x-1}$
LESSON READ-THROUGH (Part 2 of 2)
by Dr. Carol JVF Burns (website creator)
Follow along with the highlighted text while you listen!
 

PROOF OF THE BOUNDING THE REAL ROOTS OF A POLYNOMIAL THEOREM

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 $\displaystyle\,|r| \le 1 + \frac{M}{|a_n|}\,$.

$|r| \le 1$

Assume $\,|r| \le 1\,$. Since absolute value is a nonnegative quantity, $\displaystyle\,\frac{M}{|a_n|} \ge 0\,$.
Then, as desired, we have: $$\cssId{sb17}{|r| \quad\le\quad 1 \quad\le\quad 1 + \frac{M}{|a_n|}}$$

$|r| > 1$

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

Master the ideas from this section
by practicing the exercise at the bottom of this page.


When you're done practicing, move on to:
introduction to complex numbers

On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.