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!
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:
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).
(Session 2025 - 26)