Chapter 6.6, Problem 39HP

a.

To determine

To conclude: if the remainder is 0.

Answer to Problem 39HP

  $(x-c)$ is a factor of $f(x)$ .

Explanation of Solution

Given information:

Polynomial $f(x)$ is divided by $(x-c)$ .

Formula used:

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 $P(x)$ .

Factor Theorem: If $f(x)$ is a polynomial of degree $n\ge 1$ and a is any real number , then $(x-a)$ is a factor of $f(x)$ , if $f(a)=0$ .

Calculation:

Consider,

Polynomial $f(x)$ is divided by $(x-c)$

Since remainder is zero.

By factor theorem we concluded that

  $(x-c)$ is a factor of $f(x)$

Hence, $(x-c)$ is a factor of $f(x)$ .

b.

To determine

To conclude: if the remainder is 1.

Answer to Problem 39HP

  $(x-c)$ is not a factor of $f(x)$ .

Explanation of Solution

Given information:

Polynomial $f(x)$ is divided by $(x-c)$ .

Formula used:

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 $P(x)$ .

Factor Theorem: If $f(x)$ is a polynomial of degree $n\ge 1$ and a is any real number , then $(x-a)$ is a factor of $f(x)$ , if $f(a)=0$ .

Calculation:

Consider,

Polynomial $f(x)$ is divided by $(x-c)$

Since remainder is one.

By factor theorem we concluded that

  $(x-c)$ is not a factor of $f(x)$

Hence, $(x-c)$ is not a factor of $f(x)$ .

c.

To determine

To conclude: if the quotient is 1 and remainder is 0.

Answer to Problem 39HP

  $f(x)=-(x-c)$ .

Explanation of Solution

Given information:

Polynomial $f(x)$ is divided by $(x-c)$ .

Formula used:

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 $P(x)$ .

Factor Theorem: If $f(x)$ is a polynomial of degree $n\ge 1$ and a is any real number , then $(x-a)$ is a factor of $f(x)$ , if $f(a)=0$ .

Calculation:

Consider,

Polynomial $f(x)$ is divided by $(x-c)$

Since remainder is zero.

By remainder theorem we concluded that

  $\begin{array}{l}=f(x)=(-1)(x-c)+0\\ =f(x)=-(x-c)\end{array}$

Hence, $f(x)=-(x-c)$ .