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

Solution:

According to the division algorithm, if p(x) and g(x) are two polynomials with g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that p(x) = g(x) × q(x) + r(x),
where r(x) = 0 or degree of r(x) < degree of g(x)
Degree of a polynomial is the highest power of the variable in the polynomial.
(i) deg p(x) = deg q(x)
Degree of quotient will be equal to degree of dividend when divisor is constant ( i.e., when any polynomial is divided by a constant).
Let us assume the division of 6x² + 2x + 2 by 2.
Here, p(x) = 6x² + 2x + 2
g(x) = 2
q(x) = 3x² + x + 1and 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)
6x² + 2x + 2 = (2) (3x² + x + 1) + 0
Thus, the division algorithm is satisfied.

(ii) deg q(x) = deg r(x)
Let us assume the division of x³ + x by x²,
Here, p(x) = x³ + x g(x) = x² q(x) = x and r(x) = x
Clearly, the 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)

x³ + x = (x² ) × x + x x³ + x = x³ + x
Thus, 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 x³ + 1 by x².
Here, p(x) = x³ + 1 g(x) = x² q(x) = x and r(x) = 1
Clearly, the degree of r(x) is 0. Checking for division algorithm,
p(x) = g(x) × q(x) + r(x) x³ + 1 = (x² ) × x + 1 x³ + 1 = x³ + 1
Thus, the division algorithm is satisfied.