Chapter 3.4, Problem 97E

a.

To determine

To show:that $P\left(z\right)>0$ .

Explanation of Solution

Given information:

The polynomial is $P\left(x\right)=\left(x-b\right)·Q\left(x\right)+r$ .

Proof:

Substitute the x by z,

  $\begin{array}{l}P\left(z\right)=\left(z-b\right)·Q\left(z\right)+r\text{ [substitute x by z]}\\ r\ge 0\\ z>b\\ Q\left(z\right)>0\\ P\left(z\right)>0\end{array}$

b.

To determine

To prove: the when divide $x-b$ by (with $b>0$ ) using synthetic division and if the row that contains the quotient and remainder has no negative entry, then b is an upper bound for the real zeros of P.

Explanation of Solution

Given information:

The polynomial is $P\left(x\right)=\left(x-b\right)·Q\left(x\right)+r$ .

Proof:

  $\begin{array}{l}P\left(x\right)=\left(x-b\right)·Q\left(x\right)+r\\ P\left(-x\right)=\left(-x-b\right)·Q\left(-x\right)+r\text{ [substitute x by -x]}\end{array}$

When the polynomial divided by the $x-b$ so the given polynomial row has no negative entry ,

So b is the upper bound for the real zeros of P.

c.

To determine

To prove: the when divide $x-a$ by (with $a<0$ ) using synthetic division and if the row that contains the quotient and remainder has entries that are alternately nonpositive and nonnegative , then a is an lower bound for the real zeros of P.

Explanation of Solution

Given information:

The polynomial is $P\left(x\right)=\left(x-b\right)·Q\left(x\right)+r$ .

Proof:

  $P\left(x\right)=\left(x-b\right)·Q\left(x\right)+r$

When the polynomial divided by the $x+a$ so the given polynomial row has alternative negative and positiveentries ,

Soa is the lower bound for the real zeros of P.