## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

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.