Chapter 8.8, Problem 28WE

To determine

To explain : the use of Exercise $27$ to find a lower bound for the negative roots of $P(x)=0$ .

Answer to Problem 28WE

  $-m$ is the lower bound for the negative roots of $P(x)=0$

Explanation of Solution

Given information :

Consider Exercise $27$ ,

  $P(x)=(x-m)Q(x)+P(m)$

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

The leading coefficient of P ( x ) is positive.

All roots of P ( x ) are less than or equal to m .

As per the problem,

  $P(x)=(x-m)Q(x)+P(m)$

Asx is a root of P ( x ), it can be assumed $-$ x is a root of P ( $-$ x ).

If x is a negative root of P ( x ), then $-$ x is a positive root of P ( $-$ x ).

So, $P(-x)=0$ and $-P(-x)=0$ .

Now, $P(x)\ne 0$

Divide P ( $-$ x ) by its leading coefficient (which is positive).

The resultant polynomial $P\text{'}(x)$ has positive leading coefficient, $1$ and positive roots which are exactly equal to the negative roots of P ( x ).

So, if $P\text{'}(x)=(x-m)Q(x)+R(m)$ , with $m>0$ , the leading coefficient of Q ( x ) being non- negative and $R(m)\ge 0$ , then all roots of $P\text{'}(x)\le m$ .

Thus $-m$ is the lower bound for the negative roots of $P(x)=0$