Download NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.2 – Polynomials. This Exercise contains 4 questions, for which detailed answers have been provided in this note. In case you are looking at studying the remaining Exercise for Class 9 for Maths NCERT solutions for Chapter 2 or other Chapters, you can click the link at the end of this Note.

### NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.2 – Polynomials

**NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.2 – Polynomials**

**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

Then value of polynomial at

**(i) ** x = 0 is given by

p(0) = 5(0) – 4(0)^{2} + 3 = 0 – 0 + 3 = 3

**(ii)** x = –1 is given by

p(-1) = 5(-1) – 4(-1)^{2 }+ 3 = – 5 – 4 + 3 = -6

**(iii) ** x = 2 is given by

P(2) = 5(2) – 4(2)^{2 }+ 3 = 10 – 16 + 3 = -3

**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:**

To find p(x) at x = a , put x = a in the equation and evaluate the expression obtained as follows:

**(i) Given p(y) = y ^{2} – y + 1 **Therefore,

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 p(t) = 2 + t + 2t ^{2} – t^{3 }**Therefore,

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 p(x) = x ^{3 }**Therefore,

p(0) = 0

^{3 }=

^{ }0

p(1) = 1

^{3 }= 1

p(2) = 8

^{3 }= 8

**(iv) Given , p(x) = (x – 1) (x + 1) **Therefore,

p(0) = (0 – 1)(0 + 1) = -1

p(1) = (1 – 1)(1 + 1) = 0

p(2) = (2 – 1)(2 + 1) = 3

**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 – π, x = (4/5) **

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

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

^{2}– 1, x = 1, –1(iv) p(x) = (x + 1) (x – 2), x = – 1, 2

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

(vi) p(x) = lx + m, x = -m/l

^{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 = 1/2**

**Solution :**

Zero of a polynomial p(x) is a number a such that p(a) = 0.

**(i) p(x) = 3x + 1, x = ****(-1/3)**

We get,

p**(1/3)** = 3 x (-1/3) + 1

= -1 + 1 = 0

Hence, x = (-1/3) is a zero of the given polynomial.

**(ii) p(x) = 5x – π, x = 4/5**

We get,

p(4/5) = 5 x (4/5) – π

= 4 – π ≈ 0.8584 ≠ 0

Hence, x = (4/5) is not a zero of the given polynomial.

**(iii) p(x) = x ^{2} – 1, x = 1, –1**

We get,

p(1) = 1

^{2}– 1 = 1 – 1 = 0

p(-1) = (-1)

^{2}– 1 = 1 – 1 = 0

Hence, x = 1, –1 both are the zeroes of the given polynomial.

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

We get,

p(-1) = (-1 + 1) (-1 – 2) = 0

p(2) = (2 + 1) (2 – 2) = 0

Hence, x = –1 , 2 both are the zeroes of the given polynomial.

**(v) p(x) = x ^{2} , x = 0**

We get,

p(0 ) = 0

^{2 }= 0

Hence, x = 0 is a zero of the given polynomial.

**(vi) p(x) = lx + m, x = (-m/l)**

We get

p(-m/l ) = l(-m/l) + m

= -m + m = 0

Hence, x = -m/l is a zero of the given polynomial.

**(vii) p(x) = 3x ^{2} – 1, x = **

**−(1/√3), (2/√3)**

We get,

p[−(1/√3)] = 3(−1/√3)

^{2}– 1

= 3 x (1/3) – 1 = 1 – 1 = 0

p(2/√3) = 3(2/√3)

^{2}– 1

= 4 – 1 = 3

Hence, x = −(1/√3) is a zero of the given polynomial, but x = 2/√3 is not the zero of the given polynomial.

**(viii) p(x) = 2x + 1, x = ½**

We get,

p(½ ) = 2 x (1/2) + 1

= 1 + 1= 2

Hence, x = (1/2) is not a zero of the given polynomial.

**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

(v) p(x) = 3x

(vi) p(x) = ax, a ≠ 0

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

** ****Solution:**

Solving the equation p(x) = 0 will give us the zero of the given polynomial.

**(i) p(x) = x + 5**

Solving p(x) = 0 we get ,

x + 5 = 0

Or, x = -5

Thus we obtain x = -5 as the zero of the given polynomial.

**(ii) p(x) = x – 5**

Solving p(x) = 0 we get ,

x – 5 = 0

Or, x = 5

Thus we obtain x = 5 as the zero of the given polynomial.

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

Solving p(x) = 0 we get ,

2x + 5 = 0

Or, x = (-5/2)

Thus we obtain x = (-5/2) as the zero of the given polynomial.

**(iv) p(x) = 3x – 2**

Solving p(x) = 0 we get ,

3x – 2 = 0

Or, x = (2/3)

Thus we obtain x = 2/3 as the zero of the given polynomial.

**(v) p(x) = 3x**

Solving p(x) = 0 we get ,

3x = 0

Or, x = 0

Thus we obtain x = 0 as the zero of the given polynomial.

**(vi) p(x) = ax, a ≠ 0**

Solving p(x) = 0 we get ,

ax = 0

Or, x = 0

Thus we obtain x = 0 as the zero of the given polynomial.

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

Solving p(x) = 0 we get ,

cx + d = 0

Or, x = (-d/c)

Thus we obtain x = (-d/c) as the zero of the given polynomial.

**NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.2 – Polynomials, has been designed by the NCERT to test the knowledge of the student on the topic – Zeroes of a Polynomial**

**The next Exercise for** **NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.3 – Number System can be accessed by** clicking here**.**

**Download NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.2 – Polynomials**