Question

Which one of the following statements are correct?

• A. If $g\left(x\right)(\ne 0)$ and $f\left(x\right)$ are two polynomials $\in F\left(x\right)$, then there exists unique polynomials $q\left(x\right)$ and $r\left(x\right)\in F\left(x\right)$ such that $f\left(x\right)=g\left(x\right)q\left(x\right)+r\left(x\right)$ where $deg.r\left(x\right)<deg.g\left(x\right)$
• B. If $g\left(x\right)(\ne 0)$ and $f\left(x\right)$ are two polynomials $\in F\left(x\right)$, then there exists unique polynomials $q\left(x\right)$ and $r\left(x\right)\in F\left(x\right)$ such that $f\left(x\right)=g\left(x\right)q\left(x\right)+r\left(x\right)$ where $deg.r\left(x\right)\le deg.g\left(x\right)$
• C. If $g\left(x\right)(\ne 0)$ and $f\left(x\right)$ are two polynomials $\in F\left(x\right)$, then there exists unique polynomials $q\left(x\right)$ and $r\left(x\right)\in F\left(x\right)$ such that $f\left(x\right)=g\left(x\right)q\left(x\right)+r\left(x\right)$ where either $r\left(x\right)=0$ or $deg.r\left(x\right)<deg.g\left(x\right)$
• D. If $g\left(x\right)(\ne 0)$ and $f\left(x\right)$ are two polynomials $\in F\left(x\right)$, then there exists unique polynomials $q\left(x\right)$ and $r\left(x\right)\in F\left(x\right)$ such that $f\left(x\right)=g\left(x\right)q\left(x\right)+r\left(x\right)$ where $f\left(x\right)=0$ or $deg.r\left(x\right)=deg.g\left(x\right)$

Answer 1

The correct option is C If $g\left(x\right)(\ne 0)$ and $f\left(x\right)$ are two polynomials $\in F\left(x\right)$, then there exists unique polynomials $q\left(x\right)$ and $r\left(x\right)\in F\left(x\right)$ such that $f\left(x\right)=g\left(x\right)q\left(x\right)+r\left(x\right)$ where either $r\left(x\right)=0$ or $deg.r\left(x\right)<deg.g\left(x\right)$

By remainder theorem, we know, if $g\left(x\right)\ne 0$ and $f\left(x\right)$ are two polynomial $\in F\left(x\right)$ and $f\left(x\right)=g\left(x\right)q\left(x\right)+r\left(x\right)$, where $r\left(x\right)=0$ or deg. $r\left(x\right)<$deg. $g\left(x\right)$.

Thus option $C$ is correct.