Question

p(x)= g(x)q(x)+r(x) where, r(x)=0 or*

degree r(x) >degree g(x)

degree r(x)= degree g(x)

degree r(x) < d​

Answer 1

Answer:

(i) deg p(x) = deg q(x)

We know the formula,

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 4x2 by 2.

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)

4x 2 =2(2x 2 )

Hence, the division algorithm is satisfied.

(ii) deg q(x) = deg r(x)

Let us assume the division of x 3+x by x 2,

Here, p(x) = x3+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.

(iii) deg r(x) = 0

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 +1g(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+1x

4 +1=x 4 +1

Hence, the division algorithm is satisfied.