RS Aggarwal Class 9 Math Chapter 2 Polynomials Exercise 2B Solution

EXERCISE 2B

Question 1:
p(x) = 5 – 4x + 2x²
(i) p(0) = 5 – 4(0) + 2(0)² = 5

(ii) p(3) = 5 – 4(3) + 2(3)²
= 5 – 12 + 18
= 23 – 12 = 11

(iii) p(-2) = 5 – 4(-2) + 2(-2)²
= 5 + 8 + 8 = 21

Question 2:
p(y) = 4 + 3y – y² + 5y³
(i) p(0) = 4 + 3(0) – 0² + 5(0)³
= 4 + 0 – 0 + 0 = 4

(ii) p(2) = 4 + 3(2) – 2² + 5(2)³
= 4 + 6 – 4 + 40
= 10 – 4 + 40 = 46

(iii) p(-1) = 4 + 3(-1) – (-1)² + 5(-1)³
= 4 – 3 – 1 – 5 = -5

Question 3:
f(t) = 4t² – 3t + 6
(i) f(0) = 4(0)² – 3(0) + 6
= 0 – 0 + 6 = 6

(ii) f(4) = 4(4)² – 3(4) + 6
= 64 – 12 + 6 = 58

(iii) f(-5) = 4(-5)² – 3(-5) + 6
= 100 + 15 + 6 = 121

Question 4:
(i) p(x) = 0
⇒ x – 5 = 0
⇒ x = 5
⇒ 5 is the zero of the polynomial p(x).

(ii) q(x) = 0
⇒ x + 4 = 0
⇒ x = -4
⇒ -4 is the zero of the polynomial q(x).

(iii) p(t) = 0
⇒ 2t – 3 = 0
⇒ 2t =3
⇒ t = \(\frac { 3 }{ 2 } \)
⇒ t = \(\frac { 3 }{ 2 } \) is the zero of the polynomial p(t).

(iv) f(x) = 0
⇒ 3x + 1= 0
⇒ 3x = -1
⇒ x = \(\frac { -1 }{ 3 } \)
⇒ x = \(\frac { -1 }{ 3 } \) is the zero of the polynomial f(x).

(v) g(x) = 0
⇒ 5 – 4x = 0
⇒ -4x = -5
⇒ x = \(\frac { 5 }{ 4 } \)
⇒ x = \(\frac { 5 }{ 4 } \) is the zero of the polynomial g(x).

(vi) h(x) = 0
⇒ 6x – 1 = 0
⇒ 6x = 1
⇒ x = \(\frac { 1 }{ 6 } \)
⇒ x = \(\frac { 1 }{ 6 } \) is the zero of the polynomial h(x).

(vii) p(x) = 0
⇒ ax + b = 0
⇒ ax = -b
⇒ x = \(\frac { -b }{ a } \)
⇒ x = \(\frac { -b }{ a } \)is the zero of the polynomial p(x)

(viii) q(x) = 0
⇒ 4x = 0
⇒ x = 0
⇒ 0 is the zero of the polynomial q(x).

(ix) p(x) = 0
⇒ ax = 0
⇒ x = 0
⇒ 0 is the zero of the polynomial p(x).

Question 5:
(i) p(x) = x – 4
Then, p(4) = 4 – 4 = 0
⇒ 4 is a zero of the polynomial p(x).

(ii) p(x) = x – 3
Then, p(-3) = -3 – 3 = -6
⇒ -3 is not a zero of the polynomial p(x).

(iii) p(y) = 2y + 1
Then,
⇒ \(\frac { -1 }{ 2 } \) is a zero of the polynomial p(y).

(iv) p(x) = 2 – 5x
Then,
⇒ \(\frac { 2 }{ 5 } \) is a zero of the polynomial p(x).

(v) p(x) = (x – 1) (x – 2)
Then, p(1) = (1 – 1) (1 – 2) = 0 -1 = 0
⇒ 1 is a zero of the polynomial p(x).
Also, p(2) = (2 – 1)(2 – 2) = 1 0 = 0
⇒ 2 is a zero of the polynomial p(x).
Hence, 1 and 2 are the zeroes of the polynomial p(x).

(vi) p(x) = x² – 3x.
Then, p(0) = 0² – 3(0) = 0
p(3) = (3²) – 3(3) = 9 – 9 = 0
⇒ 0 and 3 are the zeroes of the polynomial p(x).

(vii) p(x) = x² + x – 6
Then, p(2) = 2² + 2 – 6
= 4 + 2 – 6
= 6 – 6 = 0
⇒ 2 is a zero of the polynomial p(x).
Also, p(-3) = (-3)² – 3 – 6
= 9 – 3 – 6 = 0
⇒ -3 is a zero of the polynomial p(x).
Hence, 2 and -3 are the zeroes of the polynomial p(x).