Question

If $p\left(x\right)$ and $g\left(x\right)$ are any two polynomials with $g\left(x\right)\ne 0$, then we can find polynomial $q\left(x\right)$ and $r\left(x\right)$ such that $p\left(x\right)=q\left(x\right)g\left(x\right)+r\left(x\right)$ where

• A. $r\left(x\right)=0$ always
• B. deg $r\left(x\right)<$ deg $q\left(x\right)$
• C. $r\left(x\right)=0$ or deg $r\left(x\right)<$ deg $g\left(x\right)$
• D. $r\left(x\right)\ne 0$ always

Answer 1

Answer: (C)

$r\left(x\right)=0$ or deg $r\left(x\right)<$ deg $g\left(x\right)$

Let $p\left(x\right)$ and $g\left(x\right)$ be two polynomials

If $g\left(x\right)$ is any polynomial then it can divide $p\left(x\right)$ by $q\left(x\right)$ where $0<q\left(x\right)$ and may get a remainder say $r\left(x\right)$.

If $g\left(x\right)$ perfectly divides $p\left(x\right)$ by $q\left(x\right)$, then $r\left(x\right)=0$.

It is obvious that $degr\left(x\right)<degg\left(x\right)$.

$\therefore$ we can find polynomial $q\left(x\right)$ and $r\left(x\right)$ such that

$p\left(x\right)=q\left(x\right)q\left(x\right)+r\left(x\right)$, where $r\left(x\right)=0$ or $degr\left(x\right)<degg\left(x\right)$