**NCERT Solutions for Class
9 Maths Chapter 2 Polynomials Ex 2.2**

NCERT Solutions for Class
9 Maths Chapter 2 Polynomials Ex 2.2 are the part of NCERT Solutions for Class
9 Maths. Here you can find the NCERT Solutions for Class 9 Maths Chapter 2 Polynomials
Ex 2.2.

**NCERT Solutions for Class 9 Maths Chapter 2 Number Systems Ex 2.1****NCERT Solutions for Class 9 Maths Chapter 2 Number Systems Ex 2.2****NCERT Solutions for Class 9 Maths Chapter 2 Number Systems Ex 2.3****NCERT Solutions for Class 9 Maths Chapter 2 Number Systems Ex 2.4****NCERT Solutions for Class 9 Maths Chapter 2 Number Systems Ex 2.5**

**Ex 2.2 Class 9 Maths Question 1.
**Find the value of the polynomial 5x
– 4x

^{2}+ 3 at:

**(i)**x = 0

**(ii)**x = –1

**(iii)**x = 2

**Solution:****
**Let p(x) = 5x – 4x

^{2}+ 3

**(i)**Putting x = 0, we get, p(0) = 5(0) – 4(0)

^{2}+ 3 = 0 – 0 + 3 = 3

Thus, the value of 5x – 4x

^{2}+ 3 at x = 0 is 3.

**(ii)**Putting x = –1, we get, p(–1) = 5(–1) – 4(–1)

^{2}+ 3 = – 5 – 4 + 3 = –9 + 3 = –6

Thus, the value of 5x – 4x

^{2}+ 3 at x = –1 is –6.

**(iii)**Putting x = 2, we get, p(2) = 5(2) – 4(2)

^{2}+ 3 = 10 – 4(4) + 3 = 10 – 16 + 3 = –3

Thus, the value of 5x – 4x

^{2}+ 3 at x = 2 is –3.

**Ex
2.2 Class 9 Maths Question 2.**

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

**(i) **p(y)
= y^{2} – y + 1 **(ii)** p(t)
= 2 + t + 2t^{2} – t^{3}

**(iii)**
p(x) = x^{3
}**(iv)** p(x) = (x – 1) (x + 1)

**Solution:****
(i)** Given that p(y) = y

^{2}– 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 + t + 2t^{2 }– t^{3}

∴ 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) = x^{3}

∴ 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

**Ex
2.2 Class 9 Maths Question 3.
**Verify whether the following are
zeroes of the polynomial, indicated against them.

**(i)**p(x) = 3x + 1, x = –1/3

**(ii)**p(x) = 5x – Ï€, x = 4/5

**(iii)**p(x) = x

^{2}– 1, x = x – 1

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

**(v)**p(x) = x

^{2}, x = 0

**(vi)**p(x) = lx + m, x = –m/l

**(vii)**P(x) = 3x

^{2}– 1, x = –1/√3, 2/√3

**(viii)**p(x) = 2x + 1, x = 12

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

p(-1/3)
= 3(-1/3) + 1 = -1 + 1 = 0

Since, p(-1/3) = 0, therefore, x = -1/3 is a zero of 3x + 1.

**(ii)** We have, p(x) = 5x – Ï€

∴ p(4/5) = 5(4/5) – Ï€ = 4 – Ï€

Therefore,
x = 4/5 is not a zero of 5x – Ï€.

**(iii)** We have, p(x) = x^{2} – 1

∴ p(1) = (1)^{2} – 1 = 1 – 1 = 0

Since, p(1) = 0, so x = 1 is a zero of x^{2} – 1.

Also, p(-1) = (-1)^{2} – 1 = 1 – 1 = 0

Since p(-1) = 0, so, x = -1 is also a zero of x^{2} – 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) = x^{2}

∴ p(0) = (0)^{2} = 0

Since, p(0) = 0, so, x = 0 is a zero of x^{2}.

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

∴ p(-m/l) = l(-m/l) + m = -m + m = 0

Since, p(-m/l) = 0,
so, x = -m/l is a zero of lx + m.

**(vii)**
We have, p(x) = 3x^{2} – 1

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

∴ p(1/2) = 2(1/2) + 1 = 1 + 1 = 2

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

**Ex 2.2 Class 9 Maths 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 = −5/2

Thus, zero of 2x + 5 is −5/2.

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

⇒ 3x – 2 = 0

⇒ 3x = 2

⇒ x = 2/3

Thus, zero of 3x – 2 is 2/3.

**(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 ⇒ x = −d/c

Thus, zero of cx + d is −d/c.

**NCERT Solutions for Maths Class 10**