Looking for accurate SEBA Class 9 Maths Exercise 2.2 (Polynomials) textual solutions for the 2026–27 academic year? Access complete step-by-step solutions covering evaluating polynomial values p(x) at given variables, verifying zeros of polynomials, and finding zeros of linear polynomials to score top marks in your annual exams.
EXERCISE 2.2
1. Find the value of the polynomial 5x – 4x² + 3 at
(i) x = 0
Answer: Let p(x) = 5x – 4x² + 3
(i) At x = 0:
p(0) = 5(0) – 4(0)² + 3
p(0) = 0 – 0 + 3
p(0) = 3
(ii) x = -1
Answer: At x = -1:
p(-1) = 5(-1) – 4(-1)² + 3
p(-1) = -5 – 4(1) + 3
p(-1) = -5 – 4 + 3
p(-1) = -6
(iii) x = 2
Answer: At x = 2:
p(2) = 5(2) – 4(2)² + 3
p(2) = 10 – 4(4) + 3
p(2) = 10 – 16 + 3
p(2) = -3
SEBA Class 9 Maths Exercise 2.2 Solutions
2. Find p(0), p(1) and p(2) for each of the following polynomials:
(i) p(y) = y² – y + 1
Answer: 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
Answer: 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
Answer: 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)
Answer: 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
Class 9 Maths Chapter 2 Exercise 2.2 Question Answer SEBA 2026-27
3. Verify whether the following are zeroes of the polynomial, indicated against them.
(i) p(x) = 3x + 1, x = -1/3
Answer: p(x) = 3x + 1, x = -1/3
p(-1/3) = 3(-1/3) + 1 = -1 + 1 = 0
Since p(-1/3) = 0, x = -1/3 is a zero of p(x).
(ii) p(x) = 5x – π, x = 4/5
Answer: p(x) = 5x – π, x = 4/5
p(4/5) = 5(4/5) – π = 4 – π ≠ 0
Since p(4/5) ≠ 0, x = 4/5 is not a zero of p(x).
(iii) p(x) = x² – 1, x = 1, -1
Answer: p(x) = x² – 1, x = 1, -1
p(1) = (1)² – 1 = 1 – 1 = 0
p(-1) = (-1)² – 1 = 1 – 1 = 0
Since p(1) = 0 and p(-1) = 0, both x = 1 and x = -1 are zeroes of p(x).
(iv) p(x) = (x + 1)(x – 2), x = -1, 2
Answer: 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
Since p(-1) = 0 and p(2) = 0, both x = -1 and x = 2 are zeroes of p(x).
(v) p(x) = x², x = 0
Answer: p(x) = x², x = 0
p(0) = (0)² = 0
Since p(0) = 0, x = 0 is a zero of p(x).
(vi) p(x) = lx + m, x = -m/l
Answer: p(x) = lx + m, x = -m/l
p(-m/l) = l(-m/l) + m = -m + m = 0
Since p(-m/l) = 0, x = -m/l is a zero of p(x).
(vii) p(x) = 3x² – 1, x = -1/√3, 2/√3
Answer: 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 ≠ 0
Therefore, x = -1/√3 is a zero of p(x), but x = 2/√3 is not.
(viii) p(x) = 2x + 1, x = 1/2
Answer: p(x) = 2x + 1, x = 1/2
p(1/2) = 2(1/2) + 1 = 1 + 1 = 2 ≠ 0
Since p(1/2) ≠ 0, x = 1/2 is not a zero of p(x).
Zeros of Polynomial Class 9 SEBA Ex 2.2 Solutions
4. Find the zero of the polynomial in each of the following cases:
(i) p(x) = x + 5
Answer: To find the zero of p(x), put p(x) = 0:
x + 5 = 0 => x = -5
Zero of p(x) is -5.
(ii) p(x) = x – 5
Answer: x – 5 = 0 => x = 5
Zero of p(x) is 5.
(iii) p(x) = 2x + 5
Answer: 2x + 5 = 0 => 2x = -5 => x = -5/2
Zero of p(x) is -5/2.
(iv) p(x) = 3x – 2
Answer: 3x – 2 = 0 => 3x = 2 => x = 2/3
Zero of p(x) is 2/3.
(v) p(x) = 3x
Answer: 3x = 0 => x = 0
Zero of p(x) is 0.
(vi) p(x) = ax, a ≠ 0
Answer: ax = 0 => x = 0 (since a ≠ 0)
Zero of p(x) is 0.
(vii) p(x) = cx + d, c ≠ 0, c, d are real numbers.
Answer: cx + d = 0 => cx = -d => x = -d/c
Zero of p(x) is -d/c.
Let p(x) = x² – x – 6
SEBA Class 9 Maths Chapter 2 Exercise 2.2 Textbook Answers
5. Which of the following are the zeroes of the polynomial x² – x – 6
(i) 3
(ii) -3
(iii) 2
(iv) -2
Choose the correct alternative—
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (i) and (iv)
(d) (iii) and (iv)
Answer: To find zeroes, put p(x) = 0:
x² – x – 6 = 0
x² – 3x + 2x – 6 = 0
x(x – 3) + 2(x – 3) = 0
(x – 3)(x + 2) = 0
Either x – 3 = 0 => x = 3
Or x + 2 = 0 => x = -2
The zeroes are 3 and -2, which correspond to (i) and (iv).
Answer: (c) (i) and (iv)
Class 9 Maths Exercise 2.2 Find Value of Polynomial
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