Looking for accurate SEBA Class 9 Maths Exercise 2.1 (Polynomials) textual solutions for the 2026–27 academic year? Access complete step-by-step solutions covering identifying polynomials in one variable, determining coefficients, finding the degree of polynomials, and classifying algebraic expressions as linear, quadratic, or cubic polynomials to score top marks in your annual exams.
EXERCISE 2.1
1. Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.
(i) 4x² – 3x + 7
Answer: It is a polynomial in one variable (x).
Reason: The exponent of x in each term is a whole number.
(ii) y² + √2
Answer: It is a polynomial in one variable (y).
Reason: The exponent of y is a whole number (2).
(iii) 3√t + t√2
Answer: It is not a polynomial.
Reason: The exponent of variable t in 3√t is 1/2, which is not a whole number.
(iv) y + 2/y
Answer: It is not a polynomial.
Reason: The exponent of variable y in 2/y is -1, which is not a whole number.
(v) x¹⁰ + y³ + t⁵⁰
Answer: It is not a polynomial in one variable.
Reason: It contains three different variables (x, y, and t).
2. Write the coefficients of x² in each of the following:
(i) 2 + x² + x
The coefficient of x² is 1.
(ii) 2 – x² + x³
The coefficient of x² is -1.
(iii) (π/2)x² + x
The coefficient of x² is π/2.
(iv) √2x – 1
The coefficient of x² is 0.
(v) (2x – 3)(x² – 3x + 1)
= 2x(x² – 3x + 1) – 3(x² – 3x + 1)
= 2x³ – 6x² + 2x – 3x² + 9x – 3
= 2x³ – 9x² + 11x – 3
The coefficient of x² is -9.
SEBA Class 9 Maths Exercise 2.1 Solutions
3. Give one example each of a binomial of highest power 35, and of a monomial of highest power 100.
Answer: Example of a binomial of highest power 35: x³⁵ + 7
Example of a monomial of highest power 100: 5y¹⁰⁰
4. Write the highest power of the variable in each of the following polynomials:
(i) 5x³ + 4x² + 7x
Highest power = 3
(ii) 4 – y²
Highest power = 2
(iii) 5t – √7
Highest power = 1
(iv) 3
Highest power = 0
Class 9 Maths Chapter 2 Exercise 2.1 Question Answer SEBA 2026-27
5. Classify the following as linear, quadratic and cubic polynomials:
(i) x² + x
Answer: Quadratic polynomial
(ii) x – x³
Answer: Cubic polynomial
(iii) y + y² + 4
Answer: Quadratic polynomial
(iv) 1 + x
Answer: Linear polynomial
(v) 3t
Answer: Linear polynomial
(vi) r²
Answer: Quadratic polynomial
(vii) 7x³
Answer: Cubic polynomial
Polynomials Class 9 SEBA Ex 2.1 Solutions
6. Two statements are given below.
Statement-1 : The power of the zero polynomial is not defined.
Statement-2 : The power of the non zero constant polynomial is zero.
Choose the correct option:
(a) Statement 1 and statement 2 both are true
(b) Statement 1 and statement 2 both are false
(c) Statement 1 is true but statement 2 is false.
(d) Statement 1 is false but statement 2 is true.
Answer: Statement-1: The power of the zero polynomial is not defined. (True)
Statement-2: The power of the non zero constant polynomial is zero. (True)
Answer: (a) Statement 1 and statement 2 both are true
7. Which of the following are constant polynomials.
(i) 5
(ii) x
(iii) 0
(iv) 5x
Choose the correct option:
(a) (i) and (ii)
(b) (iii) and (iv)
(c) (i) and (iii)
(d) (i) and (iv)
Answer: 5 and 0 are constant polynomials.
Answer: (c) (i) and (iii)
SEBA Class 9 Maths Chapter 2 Exercise 2.1 Textbook Answers
8. Match column-I with column-II
Column-I:
(A) Zero of the zero polynomial is
(B) √2 is a polynomial of power
(C) Power of the zero polynomial is
Column-II:
(P) Zero
(Q) not defined
(R) any real number
Choose the correct alternative:
(a) A -> P, B -> Q, C -> R
(b) A -> R, B -> P, C -> Q
(c) A -> R, B -> Q, C -> P
(d) A -> P, B -> R, C -> Q
Answer: (A) Zero of the zero polynomial is any real number -> (R)
(B) √2 is a polynomial of power Zero -> (P)
(C) Power of the zero polynomial is not defined -> (Q)
Matching order: A -> R, B -> P, C -> Q
Answer: (b) A -> R, B -> P, C -> Q
9. If (3x – 1)⁷ = a₇x⁷ + a₆x⁶ + a₅x⁵ + a₄x⁴ + a₃x³ + a₂x² + a₁x + a₀, then a₇ + a₆ + a₅ + a₄ + a₃ + a₂ + a₁ + a₀ is:
(a) 0
(b) 128
(c) 1
(d) 64
[Hint: The sum of the co-efficients of a polynomial p(x) can be obtained by putting x = 1]
Given:
P(x) = (3x – 1)⁷ = a₇x⁷ + a₆x⁶ + a₅x⁵ + a₄x⁴ + a₃x³ + a₂x² + a₁x + a₀
Putting x = 1 on both sides:
(3(1) – 1)⁷ = a₇(1)⁷ + a₆(1)⁶ + a₅(1)⁵ + a₄(1)⁴ + a₃(1)³ + a₂(1)² + a₁(1) + a₀
(3 – 1)⁷ = a₇ + a₆ + a₅ + a₄ + a₃ + a₂ + a₁ + a₀
(2)⁷ = a₇ + a₆ + a₅ + a₄ + a₃ + a₂ + a₁ + a₀
128 = a₇ + a₆ + a₅ + a₄ + a₃ + a₂ + a₁ + a₀
Answer: (b) 128
Class 9 Maths Exercise 2.1 Degree of Polynomial
📐 Updated Solutions Notice: This page features complete, step-by-step SEBA Class 9 Mathematics Chapter 2 (Polynomials) Exercise 2.1 Textual Solutions updated for the 2026–27 academic session. All algebraic classifications, coefficient identification steps, and degree determination rules strictly follow the latest revised Assam Board (SEBA) curriculum.
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