Chapter 6.6, Problem 40HP

To determine

To proof: the Factor Theorem.

Explanation of Solution

Given information:

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 $f(a)=0$ .

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 $P(x)$ .

Proof:

Let the polynomial be $f(x)$ which is divided by $(x-c)$ then $f(c)=0$ .

Let a and b are two real numbers .

Using remainder theorem we get ,

  $f(x)=(x-c)q(x)+f(c)$ where q(x) is the quotient polynomial

Since, $f(c)=0$ we get,

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

This implies $(x-c)$ is a factor of the polynomial $f(x)$ .

Hence proved.