Question 11 Mark
Answer the following and justify:
Can $x^2-1$ be the quotient on division of $x^6+2 x^3+x-1$ by a polynomial in $x$ of degree 5 ?
Can $x^2-1$ be the quotient on division of $x^6+2 x^3+x-1$ by a polynomial in $x$ of degree 5 ?
Answer
View full question & answer→Let the divisor of degree 5 is $g(x)=a x^5+b x^4+c x^3+d x^2+e x+1$
Dividend $=p(x)=x^6+2 x^3+x-1$,
$q(x)=x^2-1$ and let remainder be $r(x)$
So, by Euclid,s division algorithm
$p(x)=g(x) q(x)+r(x)$
[deg $p(x)$ is 6] = [g(x) of deg 5] [q(x) degree 2] + $r(x)$ of degree less than 5
degree $p(x)=$ degree $g(x)+$ degree $q(x)+$ degree $r(x)$
$6=5+2+\text { any }$
So, degree of $q(x)$ can never be 2 it may be only one.
So $\left(x^2-1\right)$ can never be the quotient.
Dividend $=p(x)=x^6+2 x^3+x-1$,
$q(x)=x^2-1$ and let remainder be $r(x)$
So, by Euclid,s division algorithm
$p(x)=g(x) q(x)+r(x)$
[deg $p(x)$ is 6] = [g(x) of deg 5] [q(x) degree 2] + $r(x)$ of degree less than 5
degree $p(x)=$ degree $g(x)+$ degree $q(x)+$ degree $r(x)$
$6=5+2+\text { any }$
So, degree of $q(x)$ can never be 2 it may be only one.
So $\left(x^2-1\right)$ can never be the quotient.