Class IX Math NCERT Solutions For Polynomials Exercise– 2.3

Exercise– 2.3

1.  Find the remainder when x3 + 3x2 + 3x + 1 is divided by

         (i)   x + 1

         (ii)    

         (iii)   x

         (iv)   x + π

         (v)   5 + 2x

Ans. (i)   ∴ The zero of x + 1 is –1

                And by remainder theorem, when

                p(x) = x3 + 3x2 + 3x + 1 is divided by x + 1, then remainder is p(–1).

                ∴ p(–1) = (–1)3 + 3 (–1)2 + 3(–1) + 1

                = –1 + (3 × 1) + (–3) + 1

                = –1 + 3 – 3 + 1

                = 0

                Thus, the required = 0

                

                and p(x) = x3 + 3x2 + 3x + 1

                ∴ For divisor  remainder is given as

                

         (iii)   We have p(x) = x3 + 3x2 + 3x + 1 and the zero of x is 0

                p(0) = (0)3 + 3(0)2 + 3(0) + 1

                = 0 + 0 + 0 +1 = 1

                Thus, the required remainder = 1.

         (iv)   We have p(x) = x3 + 3x2 + 3x + 1 and zero of x + p = (–p)

                [∵ x + π = 0 ⇒ x = –π]

                ∴ p(–π) = (–5)3 + 3(–π)2 + 3(–π) + 1

                = –π3 + 3(π2) + (–3π) + 1

                = –π3 + 3π2 – 3π + 1

                Thus, the required remainder is –π3 + 3π2 – 3π + 1.

         (v)   We have (p(x) = x3 + 3x2 + 3x + 1 and zero of 5 + 2x is 

                 

2.  Find the remainder when x3 – ax2 + 6x – a is divided by x – a.

Ans. We have p(x) = x3 – ax2 + 6x – a

                ∵ Zero of x – a is a.

                [∵ x – a = 0 ⇒ x = a]

                ∴ p(a) = (a)3 a(a)2 + 6(a) – a = a3 – a3 + 6a – a

                = 0 + 5a = 5a

                ∴ Thus, the required remainder = 5a

3.  Check whether 7 + 3x is a factor of 3x3 + 7x.

Ans. We have p(x) = 3x3 + 7x and zero of 7 + 3x is 

                 

                i.e. the remainder is not 0.

                ∴ 3x3 – 7x is not divisible by 7 + 3x.

                Thus, (7 + 3x) is not a factor of 3x3 – 7x.