Chapter 6.6, Problem 12PPS

a.

To determine

To calculate: the value of $f(2)$ .

Answer to Problem 12PPS

The value is $0$ .

Explanation of Solution

Given information:

  $f(x)=x^3-5x+2$ .

Formula used:

Synthetic Substitution :- Synthetic division is used to find the value of a function.

Remainder Theorem:- If a polynomial $P(x)$ is divided by $x-r$ , the remainder s a constant $P(r)$ , and $P(x)=Q(x)(x-r)+P(r)$ ,

Where $Q(x)$ is a polynomial with degree one less than $P(x)$ .

Calculation:

Consider ,

  $f(x)=x^3-5x+2$

By using Remainder Theorem, $f(2)$ should be the remainder when the polynomial is divided by $x-2$ .

  $\begin{array}{l}\overline{)2}\begin{array}{cccc}1& 0& -5& \text{ 2}\end{array}\\ \begin{array}{ccc}2& \text{ 4}& -2\end{array}\\ \overline{\begin{array}{ccc}1& 2& -1\end{array}\hspace{.2em}|}\text{ 0}\end{array}$

The remainder is $0$ .

Therefore, by synthetic substitution, $f(2)=0.$

b.

To determine

To calculate: the value of $f(-5)$ .

Answer to Problem 12PPS

The value is $-98$ .

Explanation of Solution

Given information:

  $f(x)=x^3-5x+2$ .

Formula used:

Synthetic Substitution :- Synthetic division is used to find the value of a function.

Remainder Theorem:- If a polynomial $P(x)$ is divided by $x-r$ , the remainder s a constant $P(r)$ , and $P(x)=Q(x)(x-r)+P(r)$ ,

Where $Q(x)$ is a polynomial with degree one less than $P(x)$ .

Calculation:

Consider ,

  $f(x)=x^3-5x+2$

By using Remainder Theorem, $f(-5)$ should be the remainder when the polynomial is divided by $x+5$ .

  $\begin{array}{l}-\overline{)5}\begin{array}{cccc}1& 0& -5& \text{ 2}\end{array}\\ \begin{array}{ccc}-5& \text{ 25}& -100\end{array}\\ \overline{\begin{array}{ccc}1& -5& \text{ 20}\end{array}\hspace{.2em}|}-\text{98}\end{array}$

The remainder is $-98$ .

Therefore, by synthetic substitution, $f(-5)=-98.$