Question

(i) deg $p(x)=$deg $q(x)$ (ii) deg $q(x)=$deg $r(x)$ (iii) deg $r(x)=0$

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Solution

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Dividend = Divisor x quotient + Remainder

$p(x)=g(x)×q(x)+r(x)$

So here the degree of quotient will be equal to degree of dividend when the divisor is constant.

Let us assume the division of $4x_{2}$ by $2$.

Here, $p(x)=$$4x_{2}$

$g(x)=2$

$q(x)=$ $2x_{2}$ and $r(x)=0$

Here, $p(x)=$$4x_{2}$

$g(x)=2$

$q(x)=$ $2x_{2}$ and $r(x)=0$

Degree of $p(x)$ and $q(x)$ is the same i.e., $2$.

Checking for division algorithm,

$p(x)=g(x)×q(x)+r(x)$

$p(x)=g(x)×q(x)+r(x)$

$4x_{2}=2(2x_{2})$

Hence, the division algorithm is satisfied.

Let us assume the division of $x_{3}+x$ by $x_{2}$,

Here, p(x) = $x_{3}+x$, g(x) = $x_{2}$, q(x) = x and r(x) = x

Degree of q(x) and r(x) is the same i.e., 1.

Checking for division algorithm,

$p(x)=g(x)×q(x)+r(x)$

$x_{3}+x=x_{2}×x+x$

$x_{3}+x=x_{3}+x$

Hence, the division algorithm is satisfied.

$x_{3}+x=x_{3}+x$

Hence, the division algorithm is satisfied.

Degree of remainder will be 0 when remainder comes to a constant.

Let us assume the division of $x_{4}+1$ by $x_{3}$

Here, p(x) = $x_{4}+1$

g(x) = $x_{3}$

$q(x)=x$ and $r(x)=1$

Let us assume the division of $x_{4}+1$ by $x_{3}$

Here, p(x) = $x_{4}+1$

g(x) = $x_{3}$

$q(x)=x$ and $r(x)=1$

Degree of $r(x)$ is $0.$

Checking for division algorithm,

$p(x)=g(x)×q(x)+r(x)$

$x_{4}+1=x_{3}×x+1$

$x_{4}+1=x_{4}+1$

Hence, the division algorithm is satisfied.

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