SEBA Class 9 Maths Polynomials MCQs (2026–27) – Assam Eduverse
The SEBA Class 9 Maths Polynomials MCQs (2026–27) are prepared according to the latest ASSEB syllabus and the updated board exam pattern. These SEBA Class 9 Maths Polynomials MCQs include conceptual objective questions, formula-based MCQs, and exam-oriented practice sets designed to strengthen algebraic understanding.
Prepared by subject experts of Assam Eduverse, these seba class 9 maths polynomials mcqs focus on important topics such as types of polynomials, degree of a polynomial, coefficients, value of a polynomial, and algebraic expressions. Practicing these polynomials mcqs class 9 seba and assam board class 9 maths objective questions improves conceptual clarity and problem-solving skills.
Regular revision of these ASSEB class 9 maths important MCQs ensures strong preparation for the 2026–27 board examination and helps students perform better in objective-type mathematics questions.
SEBA Class 9 Maths Polynomials MCQs – ASSEB 2026–27 Board Exam Practice
Table of Contents
Q1. In 2 + x + x² the coefficient of x² is:
(a) 2
(b) 1
(c) –2
(d) –1
Answer: (b) 1
The coefficient of x² means the number written in front of x².
In 2 + x + x², the term containing x² is 1×x².
So, the coefficient of x² is 1.
Q2. In 2 – x² + x³ the coefficient of x² is:
(a) 2
(b) 1
(c) –2
(d) –1
Answer: (d) –1
The term containing x² is –x².
–x² means –1 × x².
So, the coefficient of x² is –1.
Q3. In (πx²)/2 + x + 10, the coefficient of x² is:
(a) π/2
(b) 1
(c) –π/2
(d) –1
Answer: (a) π/2
The term containing x² is (πx²)/2.
This can be written as (π/2) × x².
So, the coefficient of x² is π/2.
Q4. The degree of 5t – 7 is:
(a) 0
(b) 1
(c) 2
(d) 3
Answer: (b) 1
The degree of a polynomial is the highest power of the variable.
In 5t – 7, the highest power of t is 1.
So, the degree is 1.
Q5. The degree of 4 – y² is:
(a) 0
(b) 1
(c) 2
(d) 3
Answer: (c) 2
The highest power of y in 4 – y² is 2.
So, the degree of the polynomial is 2.
Q6. The degree of 3 is:
(a) 0
(b) 1
(c) 2
(d) 3
Answer: (a) 0
3 is a constant term.
The degree of a non-zero constant is 0.
So, the degree of 3 is 0.
Q7. The value of p(x) = 5x – 4x² + 3 for x = 0 is:
(a) 3
(b) 2
(c) –3
(d) –2
Answer: (a) 3
Substitute x = 0 in p(x):
p(0) = 5(0) – 4(0)² + 3
= 0 – 0 + 3
= 3
Q8. The value of p(x) = 5x – 4x² + 3 for x = –1 is:
(a) 6
(b) –6
(c) 3
(d) –3
Answer: (b) –6
Substitute x = –1:
p(–1) = 5(–1) – 4(–1)² + 3
= –5 – 4(1) + 3
= –5 – 4 + 3
= –6
Q9. The value of p(x) = (x – 1)(x + 1) for p(1) is:
(a) 1
(b) 0
(c) 2
(d) –2
Answer: (b) 0
Substitute x = 1:
p(1) = (1 – 1)(1 + 1)
= (0)(2)
= 0
Q10. The value of p(t) = 2 + t + 2t² – t³ for p(0) is:
(a) 1
(b) 2
(c) –1
(d) 3
Answer: (b) 2
Substitute t = 0:
p(0) = 2 + 0 + 2(0)² – (0)³
= 2 + 0 + 0 – 0
= 2
Q11. The value of p(t) = 2 + t + 2t² – t³ for p(2) is:
(a) 4
(b) –4
(c) 6
(d) 7
Answer: (a) 4
Substitute t = 2:
p(2) = 2 + 2 + 2(2)² – (2)³
= 2 + 2 + 8 – 8
= 4
Q12. The value of p(y) = y² – y + 1 for p(0) is:
(a) –1
(b) 3
(c) –2
(d) 1
Answer: (d) 1
Substitute y = 0:
p(0) = (0)² – 0 + 1
= 0 – 0 + 1
= 1
Q13. The zero of p(x) = 2x – 7 is:
(a) 7/2
(b) 2/7
(c) –2/7
(d) –7/2
Answer: (a) 7/2
To find the zero, put p(x) = 0.
2x – 7 = 0
2x = 7
x = 7/2
So, the zero is 7/2.
Q14. The zero of p(x) = 9x + 4 is:
(a) 4/9
(b) 9/4
(c) –4/9
(d) –9/4
Answer: (c) –4/9
Put p(x) = 0.
9x + 4 = 0
9x = –4
x = –4/9
So, the zero is –4/9.
Q15. Which are the zeroes of p(x) = x² – 1:
(a) 1, –1
(b) –1, 2
(c) –2, 2
(d) –3, 3
Answer: (a) 1, –1
x² – 1 = 0
(x – 1)(x + 1) = 0
So, x = 1 or x = –1.
Therefore, the zeroes are 1 and –1.
Q16. Which are the zeroes of p(x) = (x – 1)(x – 2):
(a) 1, –2
(b) –1, 2
(c) 1, 2
(d) –1, –2
Answer: (c) 1, 2
Set each factor equal to zero:
x – 1 = 0 ⇒ x = 1
x – 2 = 0 ⇒ x = 2
So, the zeroes are 1 and 2.
Q17. Which one of the following is the zero of p(x) = lx + m:
(a) m/l
(b) l/m
(c) –m/l
(d) –l/m
Answer: (c) –m/l
Put lx + m = 0
lx = –m
x = –m/l
So, the zero is –m/l.
Q18. Which one of the following is the zero of p(x) = 5x – π :
(a) –4π/5
(b) 1/5 π
(c) 4π/5
(d) none of these
Answer: (d) none of these
Set 5x – π = 0
5x = π
x = π/5
Since π/5 is not given in the options,
the correct answer is none of these.
Q19. On dividing x³ + 3x² + 3x + 1 by x we get remainder:
(a) 1
(b) 0
(c) –1
(d) 2
Answer: (a) 1
By Remainder Theorem, remainder = p(0).
p(0) = 0³ + 3(0)² + 3(0) + 1
= 1
So, the remainder is 1.
Q20. On dividing x³ + 3x² + 3x + 1 by x + π we get remainder:
(a) –π³ + 3π² – 3π + 1
(b) π³ – 3π² + 3π + 1
(c) –π³ – 3π² – 3π – 1
(d) –π³ + 3π² – 3π – 1
Answer: (a) –π³ + 3π² – 3π + 1
By Remainder Theorem, remainder = p(–π).
p(–π) = (–π)³ + 3(–π)² + 3(–π) + 1
= –π³ + 3π² – 3π + 1
So, the remainder is –π³ + 3π² – 3π + 1.
Q21. On dividing x³ + 3x² + 3x + 1 by 5 + 2x we get remainder:
(a) 8/27
(b) 27/8
(c) –27/8
(d) –8/27
Answer: (c) –27/8
5 + 2x = 0 ⇒ x = –5/2
By Remainder Theorem, remainder = p(–5/2).
p(x) = (x + 1)³
p(–5/2) = (–5/2 + 1)³ = (–3/2)³ = –27/8.
Q22. If x – 2 is a factor of x³ – 3x + 5a then the value of a is:
(a) 1
(b) –1
(c) 2/5
(d) –2/5
Answer: (d) –2/5
If x – 2 is a factor, then p(2) = 0.
p(2) = 2³ – 3(2) + 5a = 8 – 6 + 5a = 2 + 5a
2 + 5a = 0
5a = –2
a = –2/5.
Q23. (x + 8)(x – 10) in the expanded form is:
(a) x² – 8x – 80
(b) x² – 2x – 80
(c) x² + 2x + 80
(d) x² – 2x + 80
Answer: (b) x² – 2x – 80
(x + 8)(x – 10)
= x(x – 10) + 8(x – 10)
= x² – 10x + 8x – 80
= x² – 2x – 80
Q24. The value of 95 × 96 is:
(a) 9020
(b) 9120
(c) 9320
(d) 9340
Answer: (b) 9120
95 × 96 = (100 – 5)(100 – 4)
= 10000 – 900 + 20
= 9120
Q25. The value of 104 × 96 is:
(a) 9984
(b) 9624
(c) 9980
(d) 9986
Answer: (a) 9984
104 × 96 = (100 + 4)(100 – 4)
= 100² – 4²
= 10000 – 16
= 9984
Q26. Without actual calculating the cubes the value of 28³ + (–15)³ + (–13)³ is:
(a) 16380
(b) –16380
(c) 15380
(d) –15380
Answer: (a) 16380
Since 28 + (–15) + (–13) = 0,
Using identity:
a³ + b³ + c³ = 3abc (when a + b + c = 0)
= 3 × 28 × (–15) × (–13)
= 3 × 28 × 195
= 16380
Q27. If x – 2 is a factor of x³ – 2ax² + ax – 1 then the value of a is:
(a) 7/6
(b) –7/6
(c) 6/7
(d) –6/7
Answer: (a) 7/6
If x – 2 is a factor, then p(2) = 0.
p(2) = 8 – 8a + 2a – 1
= 7 – 6a
7 – 6a = 0
6a = 7
a = 7/6
Q28. If x + 2 is a factor of x³ + 2ax² + ax – 1 then the value of a is:
(a) 2/3
(b) 3/5
(c) 3/2
(d) 1/2
Answer: (c) 3/2
If x + 2 is a factor, then p(–2) = 0.
p(–2) = –8 + 8a – 2a – 1
= –9 + 6a
–9 + 6a = 0
6a = 9
a = 3/2
Q29. If x + y + z = 0 then x³ + y³ + z³ is equal to:
(a) 3xyz
(b) –3xyz
(c) xy
(d) –2xy
Answer: (a) 3xyz
Using identity:
x³ + y³ + z³ – 3xyz = (x + y + z)(...)
Since x + y + z = 0,
x³ + y³ + z³ = 3xyz.
Q30. The factors of 2x² – 7x + 3 are:
(a) (x – 3)(2x – 1)
(b) (x + 3)(2x + 1)
(c) (x – 3)(2x + 1)
(d) (x + 3)(2x – 1)
Answer: (a) (x – 3)(2x – 1)
2x² – 7x + 3
= 2x² – 6x – x + 3
= 2x(x – 3) – 1(x – 3)
= (x – 3)(2x – 1)
Q31. The factors of 6x² + 5x – 6 are:
(a) (2x – 3)(3x – 2)
(b) (2x – 3)(3x + 2)
(c) (2x + 3)(3x – 2)
(d) (2x + 3)(3x + 2)
Answer: (c) (2x + 3)(3x – 2)
6x² + 5x – 6
= 6x² + 9x – 4x – 6
= 3x(2x + 3) – 2(2x + 3)
= (2x + 3)(3x – 2)
Q32. The factors of 3x² – x – 4 are:
(a) (3x – 4)(x – 1)
(b) (3x – 4)(x + 1)
(c) (3x + 4)(x – 1)
(d) (3x + 4)(x + 1)
Answer: (b) (3x – 4)(x + 1)
3x² – x – 4
= 3x² + 3x – 4x – 4
= 3x(x + 1) – 4(x + 1)
= (3x – 4)(x + 1)
Q33. The factors of 12x² – 7x + 1 are:
(a) (4x – 1)(3x – 1)
(b) (4x – 1)(3x + 1)
(c) (4x + 1)(3x – 1)
(d) (4x + 1)(3x + 1)
Answer: (a) (4x – 1)(3x – 1)
12x² – 7x + 1
= 12x² – 4x – 3x + 1
= 4x(3x – 1) – 1(3x – 1)
= (4x – 1)(3x – 1)
Q34. The factors of x³ – 2x² – x + 2 are:
(a) (x – 1)(x – 1)(x – 5)
(b) (x + 1)(x + 1)(x + 5)
(c) (x + 1)(x – 1)(x + 5)
(d) (x + 1)(x – 1)(x – 2)
Answer: (d) (x + 1)(x – 1)(x – 2)
x³ – 2x² – x + 2
= x²(x – 2) – 1(x – 2)
= (x² – 1)(x – 2)
= (x + 1)(x – 1)(x – 2)
Q35. Which of the following is not a polynomial?
(a) x² + √2x + 3
(b) x² + √2x + 6
(c) x³ + 3x² – 3
(d) 6x + 4
Answer: (b) x² + √2x + 6
In a polynomial, the powers of x must be whole numbers.
√2x means x is multiplied by √2, which is allowed.
But if the term is √(2x), then power is 1/2 which is not allowed.
Hence option (b) is not a polynomial.
Q36. The degree of the polynomial 3x³ – x⁴ + 5x + 3 is:
(a) –4
(b) 4
(c) 1
(d) 3
Answer: (b) 4
The highest power of x is 4 (from –x⁴).
So, the degree is 4.
Q37. Zero of the polynomial p(x) = a²x, a ≠ 0 is:
(a) x = 0
(b) x = 1
(c) x = –1
(d) a = 0
Answer: (a) x = 0
p(x) = a²x.
Since a ≠ 0, a² ≠ 0.
So, a²x = 0 ⇒ x = 0.
Q38. Which of the following is a term of a polynomial?
(a) 2x
(b) 3/x
(c) x√x
(d) √x
Answer: (a) 2x
A polynomial term must have whole number powers of x.
2x has power 1 (allowed).
3/x = 3x⁻¹ (not allowed).
x√x = x^(3/2) (not allowed).
√x = x^(1/2) (not allowed).
Q39. If p(x) = 5x² – 3x + 7, then p(1) equals:
(a) –10
(b) 9
(c) –9
(d) 10
Answer: (b) 9
p(1) = 5(1)² – 3(1) + 7
= 5 – 3 + 7
= 9
Q40. Factorisation of x³ + 1 is:
(a) (x + 1)(x² – x + 1)
(b) (x + 1)(x² + x + 1)
(c) (x + 1)(x² – x – 1)
(d) (x + 1)(x² + 1)
Answer: (a) (x + 1)(x² – x + 1)
Using identity:
a³ + b³ = (a + b)(a² – ab + b²)
x³ + 1³ = (x + 1)(x² – x + 1)
Q41. If x + y + 2 = 0, then x³ + y³ + 8 equals:
(a) (x + y + 2)³
(b) 0
(c) 6xy
(d) –6xy
Answer: (c) 6xy
x + y + 2 = 0 ⇒ x + y = –2.
Using identity:
x³ + y³ + 2³ – 3xy(2) = (x + y + 2)(...)
Since x + y + 2 = 0,
x³ + y³ + 8 = 6xy.
Q42. If x = 2 is a zero of the polynomial 2x² + 3x – p, then the value of p is:
(a) –4
(b) 0
(c) 8
(d) 14
Answer: (d) 14
Since 2 is a zero, p(2) = 0.
2(2)² + 3(2) – p = 0
8 + 6 – p = 0
14 – p = 0
p = 14
SEBA Class 9 Maths Polynomials MCQs (2026–27) Important Objective Questions
The SEBA Class 9 Maths Polynomials MCQs provided here are prepared according to the latest ASSEB syllabus 2026–27. These SEBA Class 9 Maths Polynomials MCQs include conceptual objective questions and exam-oriented practice sets designed for board preparation.
Students preparing for the board examination should regularly practice SEBA Class 9 Maths Polynomials MCQs. These seba class 9 maths polynomials mcqs cover important topics such as algebraic expressions, types of polynomials, degree of polynomials, coefficients, and evaluating the value of a polynomial.
The polynomials mcqs class 9 seba provided here are prepared by subject experts to ensure alignment with the latest board exam pattern. These ASSEB class 9 maths important MCQs help students understand polynomial concepts and algebraic operations clearly.
Regular revision using SEBA Class 9 Maths Polynomials MCQs along with assam board class 9 maths objective questions improves conceptual clarity, algebraic skills, and board exam performance.
Frequently Asked Questions (FAQ)
1. Are these SEBA Class 9 Maths Polynomials MCQs based on the latest syllabus?
Yes, these SEBA Class 9 Maths Polynomials MCQs follow the latest ASSEB syllabus for the 2026–27 academic session.
2. Are these seba class 9 maths polynomials mcqs useful for board exam preparation?
Yes, these seba class 9 maths polynomials mcqs are designed according to the latest board exam pattern.
3. Do polynomials mcqs class 9 seba include conceptual algebra questions?
Yes, polynomials mcqs class 9 seba include questions related to types, degree, and evaluation of polynomials.
4. Who prepared these ASSEB class 9 maths important MCQs?
These ASSEB class 9 maths important MCQs are prepared by subject experts of Assam Eduverse.
5. Are assam board class 9 maths objective questions from Polynomials frequently asked?
Yes, objective questions from Polynomials are commonly asked in Assam Board Class 9 Mathematics examinations.
6. Which important topics are included in Polynomials MCQs?
Topics include types of polynomials, degree of polynomial, coefficients, algebraic expressions, and value of polynomials.
7. Can practising SEBA Class 9 Maths Polynomials MCQs improve exam scores?
Yes, regular practice improves conceptual clarity, algebraic understanding, and accuracy in the board examination.
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