Chapter 8.8, Problem 27WE

To determine

To show: $P(x)=0$ has no roots greater than m if P ( m ) and the coefficients of Q ( x ) are all non-negative, and so m is an upper bound for the roots.

Explanation of Solution

Given information :

  $P(x)=(x-m)Q(x)+P(m)$ , where m is a positive number.

Assumption: the leading coefficient of P ( x ) is positive.

Proof :

As per the problem,

Assume the leading coefficient of P ( x ) is positive.

Now,

  $P(x)=(x-m)Q(x)+P(m)$

Where $m>0$ .

Consider $P(m)\ge 0$ and all the coefficients of Q ( x ) are also non negative.

Let x be any number such that

  $\begin{array}{l}x>m\\ x-m>0\end{array}$

Asall the coefficients of Q ( x ) are also non negative, Q ( x ) is a sum of non-negative terms ( $x-m>0$ ), so $Q(x)\ge 0$

The leading coefficient of P ( x ) is positive and is equal to the leading coefficient of Q ( x ), $Q(x)>0$

As $x-m>0$ , $Q(x)>0$ and $P(m)\ge 0$ , so

  $\begin{array}{l}P(x)=(x-m)Q(x)+P(m)\ge (x-m)Q(x)\\ (x-m)Q(x)>0\end{array}$

For $x=m$

  $(x-m)Q(x)=0$

So, $P(x)=0$ has no roots greater than m .

Therefore, m is an upper bound for the roots.