Polynomial_Soln_2.2

 

NCERT Solutions for Class 9 Maths Exercise 2.2

Question 1. Find the value of the polynomial 5x – 4x2 + 3 at

(i) x = 0

(ii) x = – 1

(iii) x = 2

Solution:

Let p(x) = 5x – 4x2 + 3

(i) p(0) = 5(0) – 4(0)2 + 3 = 0 – 0 + 3 = 3

Thus, the value of 5x – 4x2 + 3 at x = 0 is 3.

(ii) p(-1) = 5(-1) – 4(-1)2 + 3

= – 5x – 4x2 + 3 = -9 + 3 = -6

Thus, the value of 5x – 4x2 + 3 at x = -1 is -6.

(iii) p(2) = 5(2) – 4(2)2 + 3 = 10 – 4(4) + 3

= 10 – 16 + 3 = -3

Thus, the value of 5x – 4x2 + 3 at x = 2 is – 3.

Question 2. Find p (0), p (1) and p (2) for each of the following polynomials.

(i) p(y) = y2 – y +1

(ii) p (t) = 2 +1 + 2t2 -t3

(iii) P (x) = x3

(iv) p (x) = (x-1) (x+1)

Solution:

(i) Given that p(y) = y2 – y +1.

∴ P(0) = (0)2 – 0 + 1 = 0 – 0 + 1 = 1

p(1) = (1)2 – 1 + 1 = 1 – 1 + 1 = 1

p(2) = (2)2 – 2 + 1 = 4 – 2 + 1 = 3

(ii) Given that p(t) = 2 +1 + 2t2 -t3

∴p(0) = 2 + 0 + 2(0)2 – (0)3

= 2 + 0 + 0 – 0=2

P(1) = 2 + 1 + 2(1)2 – (1)3

= 2 + 1 + 2 – 1 = 4

p( 2) = 2 + 2 + 2(2)2 – (2)3

= 2 + 2 + 8 – 8 = 4

(iii) Given that p(x) = x3

∴ p(0) = (0)3 = 0, p(1) = (1)3 = 1

p(2) = (2)3 = 8

(iv) Given that p(x) = (x – 1)(x + 1)

∴ p(0) = (0 – 1)(0 + 1) = (-1)(1) = -1

p(1) = (1 – 1)(1 +1) = (0)(2) = 0

P(2) = (2 – 1)(2 + 1) = (1)(3) = 3

Question 3. Verify whether the following are zeroes of the polynomial, indicated against them.

(i) p(x) = 3x + 1,x = –

(ii) p (x) = 5x – π, x =

(iii) p (x) = x2 – 1, x = x – 1

(iv) p (x) = (x + 1) (x – 2), x = – 1,2

(v) p (x) = x2, x = 0

(vi) p (x) = 1x + m, x = –

(vii) P (x) = 3x2 – 1, x = – ,

(viii) p (x) = 2x + 1, x =

Solution:

(i) We have , p(x) = 3x + 1

(ii) We have, p(x) = 5x – π

∴

(iii) We have, p(x) = x2 – 1

∴ p(1) = (1)2 – 1 = 1 – 1=0

Since, p(1) = 0, so x = 1 is a zero of x2 -1.

Also, p(-1) = (-1)2 -1 = 1 – 1 = 0

Since p(-1) = 0, so, x = -1, is also a zero of x2 – 1.

(iv) We have, p(x) = (x + 1)(x – 2)

∴ p(-1) = (-1 +1) (-1 – 2) = (0)(- 3) = 0

Since, p(-1) = 0, so, x = -1 is a zero of (x + 1)(x – 2).

Also, p( 2) = (2 + 1)(2 – 2) = (3)(0) = 0

Since, p(2) = 0, so, x = 2 is also a zero of (x + 1)(x – 2).

(v) We have, p(x) = x2

∴ p(o) = (0)2 = 0

Since, p(0) = 0, so, x = 0 is a zero of x2.

(vi) We have, p(x) = lx + m

(vii) We have, p(x) = 3x2 – 1

(viii) We have, p(x) = 2x + 1

∴

Since, ≠ 0, so, x = is not a zero of 2x + 1.

Question 4. Find the zero of the polynomial in each of the following cases

(i) p(x)=x+5

(ii) p (x) = x – 5

(iii) p (x) = 2x + 5

(iv) p (x) = 3x – 2

(v) p (x) = 3x

(vi) p (x)= ax, a≠0

(vii) p (x) = cx + d, c ≠ 0 where c and d are real numbers.

Solution:

(i) We have, p(x) = x + 5. Since, p(x) = 0

⇒ x + 5 = 0

⇒ x = -5.

Thus, zero of x + 5 is -5.

(ii) We have, p(x) = x – 5.

Since, p(x) = 0 ⇒ x – 5 = 0 ⇒ x = -5

Thus, zero of x – 5 is 5.

(iii) We have, p(x) = 2x + 5. Since, p(x) = 0

⇒ 2x + 5 =0

⇒ 2x = -5

⇒ x =

Thus, zero of 2x + 5 is .

(iv) We have, p(x) = 3x – 2. Since, p(x) = 0

⇒ 3x – 2 = 0

⇒ 3x = 2

⇒ x =

Thus, zero of 3x – 2 is

(v) We have, p(x) = 3x. Since, p(x) = 0

⇒ 3x = 0 ⇒ x = 0

Thus, zero of 3x is 0.

(vi) We have, p(x) = ax, a ≠ 0.

Since, p(x) = 0 => ax = 0 => x-0

Thus, zero of ax is 0.

(vii) We have, p(x) = cx + d. Since, p(x) = 0

⇒ cx + d = 0 ⇒ cx = -d ⇒

Thus, zero of cx + d is