Question

Can ${\mathrm{x}}^2-1$ be the quotient on division of ${\mathrm{x}}^6+2{\mathrm{x}}^3+\mathrm{x}-1$ by a polynomial in $\mathrm{x}$ of degree $5$?

Answer 1

Assumption:

Let us assume that $({\mathrm{x}}^2–1)$ be the quotient when a degree $6$ polynomial is divided by a degree $5$ polynomial $\left(1\right)$

As per our assumption,

$(\mathrm{degree}6\mathrm{polynomial})=({\mathrm{x}}^2–1)(\mathrm{degree}5\mathrm{polynomial})+\mathrm{r}\left(\mathrm{x}\right)$[ Since, $(\mathrm{a}=\mathrm{bq}+\mathrm{r})$]

$=(\mathrm{degree}7\mathrm{polynomial})+\mathrm{r}\left(\mathrm{x}\right)$ [ Since, $({\mathrm{x}}^2\mathrm{term}\times {\mathrm{x}}^5\mathrm{term}={\mathrm{x}}^7\mathrm{term})$]

$=(\mathrm{degree}7\mathrm{polynomial})$

From the above equation, we get to know that, our assumption is contradicted.

When a degree $6$ polynomial is divided by degree $5$ polynomial the quotient will be of degree$1$.

Hence, no,${\mathrm{x}}^2–1$ cannot be the quotient on division of${\mathrm{x}}^6+2{\mathrm{x}}^3+\mathrm{x}–1$ by a polynomial in $\mathrm{x}$ of degree $5$.