5. 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)(iii) deg r(x) = 0

Solutions: According to the division algorithm, dividend p(x) and divisor g(x) are two polynomials, where g(x)≠0. Then we can find the value of quotient q(x) and remainder r(x), with the help of below given formula; Dividend = Divisor × Quotient + Remainder ∴ p(x) = g(x)×q(x)+r(x) Where r(x) = 0 or degree of r(x)< degree of g(x). Now let us proof the three given cases as per division algorithm by taking examples for each. (i) deg p(x) = deg q(x) Degree of dividend is equal to degree of quotient, only when the divisor is a constant term. Let us take an example, p(x) = 3x2+3x+3 is a polynomial to be divided by g(x) = 3. So, (3x2+3x+3)/3 = x2+x+1 = q(x) Thus, you can see, the degree of quotient q(x) = 2, which also equal to the degree of dividend p(x). Hence, division algorithm is satisfied here. (ii) deg q(x) = deg r(x) Let us take an example, p(x) = x2 + 3 is a polynomial to be divided by g(x) = x – 1. So, x2 + 3 = (x – 1)×(x) + (x + 3) Hence, quotient q(x) = x Also, remainder r(x) = x + 3 Thus, you can see, the degree of quotient q(x) = 1, which is also equal to the degree of remainder r(x). Hence, division algorithm is satisfied here. (iii) deg r(x) = 0 The degree of remainder is 0 only when the remainder left after division algorithm is constant. Let us take an example, p(x) = x2 + 1 is a polynomial to be divided by g(x) = x. So, x2 + 1 = (x)×(x) + 1 Hence, quotient q(x) = x And, remainder r(x) = 1 Clearly, the degree of remainder here is 0. Hence, division algorithm is satisfied here.