NCERT Solutions Class 9 Maths Chapter 2 Polynomials Exercise 2.2

Exercise 2.2 of Chapter 2 - Polynomials, teaches you how to find the value of a polynomial by substituting the given value of the variable. This concept will help understand the calculation of polynomials better. This understanding is necessary for both school exams as well as the future topics in algebra.

The questions in this exercise are based on the most recent NCERT syllabus and are structured according to the CBSE exam structure. By practicing these solutions, you will increase your proficiency in substitution based questions and this chapter overall.

To aid in your learning, we have provided simple NCERT Solutions for Exercise 2.2. The followed solutions are step-by-step solutions to keep you from making mistakes and to enhance your problem-solving skills. You can download the free PDF and practice at your pace!

1.0Download NCERT Solutions of Class 9 Maths Chapter 2 Polynomials Exercise 2.2: Free PDF

The NCERT Solutions for Class 9 Maths Chapter 2 helps you learn how to put values into polynomials and solve them easily. Avail ALLEN’s curated solutions to help students to study better and score well in exams. Download the free PDF of Exercise 2.2 from Class 9 maths Chapter 2 from below:

2.0NCERT Exercise Solutions Class 9 Chapter 2- Polynomials: All Exercises 

3.0NCERT Class 9 Maths Chapter 2 Exercise 2.2 : Detailed Solutions

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

(i) x=0

(ii) x=−1

(iii) x=2

Sol. Let f(x)=5x−4x²+3

(i) Value of f(x) at x=0 = f(0) = 5(0)−4(0)²+3 = 0−0+3 = 3

(ii) Value of f(x) at x=−1 = f(−1) = 5(−1)−4(−1)²+3 = −5−4(1)+3 = −5−4+3 = −9+3 = −6

(iii) Value of f(x) at x=2 = f(2) = 5(2)−4(2)²+3 = 10−4(4)+3 = 10−16+3 = −6+3 = −3

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+t+2t²−t³

(iii) p(x)=x³

(iv) p(x)=(x−1)(x+1)

Sol. (i) p(y)=y²−y+1

p(0)=(0)²−(0)+1=1

p(1)=(1)²−(1)+1=1−1+1=1

p(2)=(2)²−(2)+1=4−2+1=3.

(ii) p(t)=2+t+2t²−t³

p(0)=2+0+2(0)²−(0)³ = 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) p(x)=x³

p(0)=(0)³ = 0

p(1)=(1)³ = 1

p(2)=(2)³ = 8

(iv) 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

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)=x²−1, x=1,−1

(iv) p(x)=(x+1)(x−2), x=−1,2

(v) p(x)=x², x=0

(vi) p(x)=ℓx+m, x=−m/ℓ

(vii) p(x)=3x²−1, x=−1/√3, 2/√3

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

Sol. (i) p(x)=3x+1, x=−1/3

p(−1/3) = 3(−1/3)+1 = −1+1 = 0.

Therefore, −1/3 is a zero of p(x).

(ii) p(x)=5x−π, x=4/5

p(4/5) = 5(4/5)−π = 4−π.

Since 4−π ≠ 0, 4/5 is not a zero of p(x).

(iii) p(x)=x²−1, x=1,−1

p(1) = (1)²−1 = 1−1 = 0.

p(−1) = (−1)²−1 = 1−1 = 0.

Therefore, 1 and −1 are zeroes of p(x).

(iv) p(x)=(x+1)(x−2), x=−1,2

p(−1) = (−1+1)(−1−2) = (0)(−3) = 0.

p(2) = (2+1)(2−2) = (3)(0) = 0.

Therefore, −1 and 2 are zeroes of p(x).

(v) p(x)=x², x=0

p(0) = (0)² = 0.

Therefore, 0 is a zero of p(x).

(vi) p(x)=ℓx+m, x=−m/ℓ

p(−m/ℓ) = ℓ(−m/ℓ)+m = −m+m = 0.

Therefore, −m/ℓ is a zero of p(x).

(vii) p(x)=3x²−1, x=−1/√3, 2/√3

p(−1/√3) = 3(−1/√3)²−1 = 3(1/3)−1 = 1−1 = 0.

p(2/√3) = 3(2/√3)²−1 = 3(4/3)−1 = 4−1 = 3.

Since 3≠0, 2/√3 is not a zero of p(x).

So, −1/√3 is a zero of p(x) and 2/√3 is not a zero of p(x).

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

p(1/2) = 2(1/2)+1 = 1+1 = 2.

Since 2≠0, 1/2 is not a zero of p(x).

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, c,d are real numbers.

Sol. (i) p(x)=x+5

To find the zero, set p(x)=0 => x+5=0 => x=−5.

Therefore, −5 is the zero of the polynomial p(x).

(ii) p(x)=x−5

To find the zero, set p(x)=0 => x−5=0 => x=5.

Therefore, 5 is the zero of the polynomial p(x).

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

To find the zero, set p(x)=0 => 2x+5=0 => 2x=−5 => x=−5/2.

Therefore, −5/2 is the zero of the polynomial p(x).

(iv) p(x)=3x−2

To find the zero, set p(x)=0 => 3x−2=0 => 3x=2 => x=2/3.

Therefore, 2/3 is the zero of the polynomial p(x).

(v) p(x)=3x

To find the zero, set p(x)=0 => 3x=0 => x=0.

Therefore, 0 is the zero of the polynomial p(x).

(vi) p(x)=ax, a≠0

To find the zero, set p(x)=0 => ax=0. Since a≠0, then x=0.

Therefore, 0 is the zero of p(x).

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

To find the zero, set p(x)=0 => cx+d=0 => cx=−d => x=−d/c.

Therefore, −d/c is the zero of the polynomial p(x).

4.0Key Features and Benefits Class 9 Maths Chapter 2 Number System: Exercise 2.2

• Exercise 2.2 focuses on evaluating polynomials by substituting values of the variable.
• The questions in this exercise are formulated as per the latest NCERT syllabus and CBSE exam pattern.
• The exercise builds a foundation for advanced questions related to algebra in higher classes.
• Practicing NCERT solutions helps you understand each step clearly and helps avoid silly mistakes.
• The solved answers improve your confidence and speed while solving similar CBSE questions.
• Regular practice with these solutions also supports preparation for competitive exams.