SEBA Class 9 Maths Polynomials MCQs (2026–27) – Assam Eduverse
To build a strong foundation in algebra, the SEBA Class 9 Maths Polynomials MCQs are structured as per the latest ASSEB syllabus and aligned with the current academic session’s exam pattern. These SEBA Class 9 Maths Polynomials MCQs feature conceptual, formula-based, and exam-focused questions that help students grasp key algebraic concepts effectively. For more structured practice, students can also explore important maths MCQs chapter-wise.
Curated by subject experts of Assam Eduverse, this set of seba class 9 maths polynomials mcqs covers essential topics like types of polynomials, degree, coefficients, values of polynomials, and algebraic expressions. Practicing polynomials mcqs class 9 seba along with chapterwise MCQs and question answers helps improve both conceptual clarity and problem-solving ability.
With regular practice of these ASSEB class 9 maths important MCQs, students can strengthen their preparation and approach objective-type questions with greater confidence. For complete guidance, students may also refer to Class 9 study materials.
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 – Important Objective Questions
A clear understanding of Polynomials is essential for building a strong base in algebra and higher-level mathematics. Practicing MCQs based on the latest SEBA (ASSEB) syllabus helps students strengthen their concepts while becoming familiar with the pattern of objective questions asked in examinations.
These SEBA Class 9 Maths Polynomials MCQs cover key topics such as algebraic expressions, types and degree of polynomials, coefficients, and evaluation of polynomials. Since these concepts are fundamental to solving algebraic problems, regular practice allows students to develop better clarity and accuracy.
Solving such important objective questions for Class 9 Maths improves logical thinking and helps students understand how algebraic expressions behave under different operations. It also reduces errors in calculations and strengthens problem-solving skills required for exams.
Consistent practice of MCQs enhances speed, accuracy, and confidence, enabling students to solve questions more efficiently during exams. It also supports quick and effective revision, especially before tests and final assessments.
To perform well in school exams and board-based assessments, students should make these MCQs a regular part of their study routine. With proper understanding and continuous practice, mastering polynomials becomes much easier and scoring high becomes more achievable.
FAQs – SEBA Class 9 Maths Polynomials MCQs
1. How many MCQs come from Polynomials in SEBA Class 9 Maths exam?
Usually 3–5 MCQs come from Polynomials out of the 45 MCQs. Focus on identities and zeroes—they are frequently repeated in exams.
2. What are the most important topics for Polynomials MCQs in SEBA Class 9?
Key topics include zeroes of polynomials, factorisation, and algebraic identities. Practice identity-based questions daily—they are easy scoring.
3. Where can I download SEBA Class 9 Polynomials MCQs with answers PDF?
You can download chapter-wise MCQs with answers from Assam Eduverse. Always revise solved examples before attempting PDFs for better accuracy.
4. Are Polynomials MCQs difficult in SEBA Class 9 exams?
No, most MCQs are basic and formula-based. If you remember identities and practice regularly, you can solve them within seconds.
5. How to prepare quickly for Polynomials MCQs before the exam?
Revise formulas, solve previous year MCQs, and practice 20–30 questions daily. Assam Eduverse mock tests can help improve speed and confidence.
6. Do SEBA repeat Polynomials MCQs from previous years?
Yes, similar pattern questions are often repeated. Practice past papers carefully—you may see familiar concepts in the final exam.
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