#### Answer

The given statement is true.

#### Work Step by Step

The remainder theorem is explained as:
When the function $P\left( x \right)$ is divided by $x-r$ then the remainder is $P\left( r \right)$.
Consider an example of a polynomial function:
$P\left( x \right)=6{{x}^{3}}-13{{x}^{2}}-79x+140$
From the remainder theorem, $P\left( 2 \right)$ is the remainder when $6{{x}^{3}}-13{{x}^{2}}-79x+140$ is divided by $\left( x-2 \right)$.
The constant term of the divisor is $-2$ with the opposite sign written to left.
The coefficients of the dividend $6,-13,-79,140$ are written to the right.
Now, use synthetic division to find the remainder.
The synthetic division is as follows.
$\begin{array}{c|rrrr}{2}&6&-13&-79&{140}\\&&12&-2&{2} \cdot {-81}={-162}\\\hline&6&-1&{-81}&\left({140}\right)+{-162}={-22}\end{array}$
So, the remainder is \[-22\]
When, $\left( x-2 \right)$ is a factor then $P\left( 2 \right)=-22$.
Hence, for this function, when $x-2$ is a factor of some polynomial $P\left( x \right)$, then $P\left( 2 \right)=0$ is false.
Now, consider another example of polynomial function:
$P\left( x \right)={{x}^{3}}+6{{x}^{2}}-x-30$
From the remainder theorem, $P\left( 2 \right)$ is the remainder when${{x}^{3}}+6{{x}^{2}}-x-30$ is divided by $\left( x-2 \right)$.
The constant term of the divisor is $-2$ with the opposite sign written to left.
The coefficients of the dividend $1,6,-1,-30$ are written to the right.
Now, use synthetic division to find the remainder.
The synthetic division is as follows.
$\begin{array}{c|rrrr}{2}&1&6&-1&{-30}\\&&2&16&{2} \cdot {15}={30}\\\hline&1&8&{15}&\left({-30}\right)+{30}={0}\end{array}$
So, the remainder is $0$.
When, $\left( x-2 \right)$ is a factor then$P\left( 2 \right)=0$.
Therefore, the given statement is true.