Answer the following and justify.

Solution:

(i) No. because whenever we divide a polynomial x6 + 2x3 + x -1 by a polynomial in x of degree 5, then we get quotient always as in linear form

i.e., polynomial in x of degree 1. Let divisor = a polynomial in x of degree 5

= ax5 + bx4 + cx3 + dx2 + ex + f

quotient = x2 -1

and            dividend = x6 + 2x3 + x -1

By division algorithm for polynomials,

Dividend = Divisor x Quotient + Remainder

= (ax5 + bx4 + cx3 + dx2 + ex + f)x(x2 -1) + Remainder

= (a polynomial of degree 7) + Remainder

[in division algorithm, degree of divisor > degree of remainder]

= (a polynomial of degree 7)

But              dividend = a polynomial of degree 6

So, division algorithm is not satisfied.

Hence, x2 -1 is not a required quotient.

(ii) Given that, Divisor px3 + gx2 + rx + s, p≠0

and dividend = ax2 + bx + c

We see that,

Degree of divisor > Degree of dividend

So, by division algorithm,

quotient = 0 and remainder = ax2 + bx + c

If degree of dividend < degree of divisor, then quotient will be zero and remainder as same as dividend.

(iii) If division of a polynomial p(x) by a polynomial g(x), the quotient is zero, then relation between the degrees ofp(x) and g(x) is degree of p(x) <

degree of g(x).

(iv) If division of a non-zero polynomial p(x) by a polynomial g(x), the remainder is zero, then g(x) is a factor of p(x) and has degree less than or

equal to the degree of p(x). e., degree of g(x) < degree of p(x).

(v) No, let p(x) = x2 + kx + k

If p(x) has equal zeroes, then its discriminant should be zero.

D = B2 -4AC = 0                                                                                 ,..(j)

On comparing p(x) with Ax2 + Bx + C, we get

A =1 B = k and C = k

∴                       (k)2-4(1)(k) = 0                                                                      [from Eq. (i)]

⇒                             k(k- 4)=0

⇒                                        k =0, 4

So, the quadratic polynomial p(x) have equal zeroes only at k =0, 4.