Question

# Give examples of polynomials p(x),g(x), q(x) and r(x), which satisfy the division algorithm and(i)deg p(x)=deg q(x)(ii)deq q(x)=deq r(x)(iii)deg r(x)=0

Hint: Assume the values of p(x) and g(x) such that condition (i), (ii) and (iii) are satisfied and use the formula of division algorithm-
$\Rightarrow$ Dividend=Divisor × Quotient +Remainder

Here let us represent, p(x) =dividend=the number to be divided
g(x)=Divisor=the number by which dividend is divided
q(x)=quotient
And r(x)=remainder
Now we have to give examples of polynomials such that the division algorithm
$\Rightarrow$ Dividend=Divisor × Quotient +Remainder
Is satisfies and the given conditions are also satisfied
(i)Let us assume the division of $2x + 4$ by $2$
Then p(x) =$2x + 4$
g(x)= $2$
q(x)=$x + 2$
and r(x)=$0$
On using the division algorithm
$\Rightarrow$ Dividend=Divisor × Quotient +Remainder
$\Rightarrow$ p(x)=g(x) × q(x) +r(x)
On putting the given values we get,
$\Rightarrow 2x + 4 = 2 \times \left( {x + 2} \right) + 0$
On solving we get,
$\Rightarrow 2x + 4 = 2x + 4$
Hence the division algorithm is satisfied.
And here the degree of p(x) =$1$ =degree of q(x)
Hence (i) condition is also satisfied.
(ii) Let us assume the division of ${x^3} + x$ by${x^2}$
Then here, p(x) =${x^3} + x$
g(x)= ${x^2}$
q(x)=$x$
r(x)=$x$
It is clear that the degree of q(x)=1 and
Degree of r(x)=1
On using the division algorithm
$\Rightarrow$ Dividend=Divisor × Quotient +Remainder
$\Rightarrow$ p(x)=g(x) × q(x) +r(x)
On putting the given values we get,
$\Rightarrow {x^3} + x = \left( {{x^2} \times x} \right) + x$
On solving we get,
$\Rightarrow {x^3} + x = {x^3} + x$
Hence the division algorithm is satisfied.
And here the degree of r(x) =$1$ =degree of q(x)
Hence (ii) condition is also satisfied.
(iii)Let us assume the division of ${x^2} + 1$ by$x$
Then here, p(x) =${x^2} + 1$
g(x)= $x$
q(x)=$x$
r(x)=$1$
It is clear that the degree of r(x) =$0$ and
On using the division algorithm
$\Rightarrow$ Dividend=Divisor × Quotient +Remainder
$\Rightarrow$ p(x)=g(x) × q(x) +r(x)
On putting the given values we get,
$\Rightarrow {x^2} + 1 = \left( {x \times x} \right) + 1$
On solving we get,
$\Rightarrow {x^2} + 1 = {x^2} + 1$
Hence the division algorithm is satisfied.
And here the degree of r(x) =$0$

Hence (iii) condition is also satisfied.

Note: Here you can also assume any other polynomial for division but it is necessary that the chosen dividend and divisor be such that the conditions of the questions are satisfied.