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Class 10 Maths – Chapter 2: Polynomials (20 Important Questions + 10 FAQs with Solutions)

Class 10 Maths – Chapter 2: Polynomials (20 Important Questions + 10 FAQs with Solutions)

Polynomials is one of the most scoring chapters in Class 10 Mathematics. Students often find it easy to understand but tricky when it comes to applying the concepts in exams. That is why CBSE frequently includes problem-solving and conceptual questions from this chapter in board examinations.

At Fuzy Math Academy, we aim to simplify your preparation by providing hand-picked 20 important CBSE-style questions from Chapter 2 (Polynomials) along with their detailed step-by-step solutions. In addition, you will also find 10 FAQs that clear common doubts and misconceptions about this chapter.

Our unique approach powered by Learning Management System (LMS) and 24/7 AI chat support ensures that you never get stuck. Whether you are revising at midnight or practicing during the day, our support system is always available to guide you.

Importance of Chapter 2: Polynomials in Class 10

Polynomials form the foundation of algebra. They are not only important in Class 10 but also useful in higher studies such as Class 11, 12, engineering, and competitive exams. Some of the key areas covered in this chapter include:

  • Zeros of a polynomial and their relationship with coefficients

  • Division algorithm for polynomials

  • Graphical representation of quadratic and cubic polynomials

  • Application of factor theorem

  • Standard problem-solving techniques

Mastering these concepts ensures you can score well in both 1-mark conceptual questions and 4-mark long-answer questions in the board exams.

Why Practice with Step-by-Step Solutions?

Most students lose marks not because they do not know the answer, but because they make mistakes in writing proper steps. CBSE marking scheme is very particular about stepwise presentation. That is why in this blog we have provided detailed solutions with proper steps for every important question.

You will also find FAQs written in a snippet-friendly style so that they are easy to revise and recall quickly.

What This Blog Contains

  •  20 Important CBSE-style questions from Polynomials with complete stepwise solutions
  • 10 FAQs addressing the most common doubts of students
  •  Detailed explanations in simple language
  • Exam tips for scoring full marks

Benefits of Studying with Fuzy Math Academy

At Fuzy Math Academy, we don’t just provide content — we make learning smarter. Here’s why thousands of students trust us:

  • Affordable Online Coaching (Classes 5–12): Our fees are far lower than traditional coaching, making quality education accessible to all.

  • LMS (Learning Management System): Tracks your progress, shows mistakes, and provides targeted practice.

  • 24/7 AI Chat Support: Ask doubts anytime, get instant answers, and never waste time stuck on a problem.

  • Concept Clarity + Practice: Live classes, recorded lectures, notes, assignments, and mock tests — all in one platform.

How to Use This Blog for Revision

  1. First, go through the concepts in NCERT textbook.

  2. Attempt the 20 important questions given below without looking at solutions.

  3. Check your answers with the step-by-step solutions provided.

  4. Read the 10 FAQs carefully as they cover exam-focused conceptual doubts.

  5. Revise the mistakes using your own notes or Fuzy Math Academy LMS.

By following this approach, you will see noticeable improvement in accuracy and confidence.

Conclusion

Polynomials is a chapter where you can easily score full marks if you practice systematically. The 20 important questions with stepwise solutions and 10 FAQs in this blog are enough to give you strong command over this topic.

If you want to go beyond and get personalized guidance, join Fuzy Math Academy today. With LMS-powered coaching, AI chat support, and experienced teachers, your preparation for Class 10 Mathematics becomes smoother and more effective.

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20 Important Questions on Polynomials – Class 10 (Complete Step-by-Step Solutions)

  1. Q1. Find the zeros of p(x) = 2x² − 5x + 3.

  1. Solution (step-by-step)
    1. Identify coefficients: a = 2, b = −5, c = 3.
    2. Compute discriminant: D = b² − 4ac = (−5)² − 4(2)(3) = 25 − 24 = 1.
    3. Use quadratic formula: x = [−b ± √D] / (2a) = [5 ± 1] / 4.
    4. So x₁ = (5 + 1)/4 = 6/4 = 3/2, and x₂ = (5 − 1)/4 = 4/4 = 1.
    5. Zeros: 1 and 3/2.
  2. Q2. Verify zeros–coefficients relationship for x² − 7x + 12.
    Solution
    1. Factorize: x² − 7x + 12 = (x − 3)(x − 4).
    2. Zeros are 3 and 4. Sum = 3 + 4 = 7; product = 3 × 4 = 12.
    3. For ax² + bx + c with a = 1, b = −7, c = 12: sum = −b/a = 7, product = c/a = 12. Verified.
  3. Q3. Find a quadratic polynomial whose zeros are 2 and −3.
    Solution
    1. Zeros α = 2, β = −3 ⇒ polynomial ∝ (x − α)(x − β) = (x − 2)(x + 3).
    2. Multiply out: (x − 2)(x + 3) = x² + 3x − 2x − 6 = x² + x − 6.
    3. Choose monic polynomial: x² + x − 6.
  4. Q4. Construct a quadratic polynomial whose sum of zeros is 4 and product is −5.
    Solution
    1. For sum S and product P, polynomial is x² − Sx + P.
    2. Here S = 4, P = −5 ⇒ polynomial is x² − 4x − 5.
  5. Q5. For what value of k does x² + 2(k−1)x + (k² − 1) have equal zeros?
    Solution
    1. Equal zeros ⇔ discriminant D = 0.
    2. Here a = 1, b = 2(k − 1), c = k² − 1.
    3. Compute D = b² − 4ac = [2(k − 1)]² − 4(k² − 1) = 4[(k − 1)² − (k² − 1)].
    4. Expand: (k − 1)² − (k² − 1) = (k² − 2k + 1) − k² + 1 = 2 − 2k = 2(1 − k).
    5. So D = 4 × 2(1 − k) = 8(1 − k). Set D = 0 ⇒ 1 − k = 0 ⇒ k = 1.
  6. Q6. One zero is 2 and sum of zeros is 7. Find other zero and polynomial.
    Solution
    1. Let zeros be 2 and β. Then 2 + β = 7 ⇒ β = 5.
    2. Polynomial with zeros 2 and 5: (x − 2)(x − 5) = x² − 7x + 10.
  7. Q7. Form a quadratic with leading coefficient 3, one zero −2, and product of zeros 6.
    Solution
    1. Let zeros be α = −2 and β. Given αβ = 6 ⇒ (−2)β = 6 ⇒ β = −3.
    2. Polynomial (leading coefficient 3): 3(x − (−2))(x − (−3)) = 3(x + 2)(x + 3).
    3. Expand: (x + 2)(x + 3) = x² + 5x + 6 ⇒ multiply by 3 ⇒ 3x² + 15x + 18.
  8. Q8. If p(x) = x² − 6x + a has a zero 2, find a and the other zero.
    Solution
    1. Since 2 is a zero, p(2) = 0 ⇒ 2² − 6·2 + a = 0 ⇒ 4 − 12 + a = 0 ⇒ a = 8.
    2. Now polynomial is x² − 6x + 8. Sum of zeros = 6 ⇒ other zero = 6 − 2 = 4.
  9. Q9. Divide 2x³ + 3x² − 11x − 6 by x − 2. Find quotient and remainder.
    Solution (synthetic division)
    1. Use synthetic division with root 2. Coefficients: 2, 3, −11, −6.
    2. Bring down 2. Multiply by 2 → 4; add to next coefficient: 3 + 4 = 7.
    3. Multiply 7 by 2 → 14; add to −11 ⇒ 3. Multiply 3 by 2 → 6; add to −6 ⇒ 0.
    4. So quotient coefficients are 2, 7, 3 ⇒ quotient = 2x² + 7x + 3. Remainder = 0.
  10. Q10. Factorize x³ − 7x² + 14x − 8, given one zero is 2.
    Solution
    1. Since 2 is a zero, divide polynomial by (x − 2) (synthetic or long division).
    2. Synthetic with 2 and coefficients 1, −7, 14, −8: bring 1; ×2 → 2; add to −7 ⇒ −5; ×2 → −10; add to 14 ⇒ 4; ×2 → 8; add to −8 ⇒ 0.
    3. Quotient is x² − 5x + 4 ⇒ factorize: (x − 1)(x − 4).
    4. Complete factorization: (x − 2)(x − 1)(x − 4).
  11. Q11. Find remainder when x³ − 4x² + 5x − 2 is divided by x − 1.
    Solution
    1. Remainder theorem: remainder = p(1).
    2. Compute p(1) = 1 − 4 + 5 − 2 = 0. So remainder = 0; (x − 1) is a factor.
  12. Q12. Find k such that x = 1 is a zero of p(x) = kx² − 3x + 2.
    Solution
    1. Set p(1) = 0 ⇒ k(1)² − 3(1) + 2 = 0 ⇒ k − 3 + 2 = 0 ⇒ k − 1 = 0 ⇒ k = 1.
  13. Q13. Construct a quadratic polynomial whose sum of zeros is 0 and product is −9.
    Solution
    1. Using x² − Sx + P: here S = 0, P = −9 ⇒ polynomial is x² − 9.
    2. Factor: x² − 9 = (x − 3)(x + 3); zeros are 3 and −3.
  14. Q14. For p(x) = 6x² − x − 2, find sum and product of zeros.
    Solution
    1. For ax² + bx + c: sum = −b/a, product = c/a.
    2. Here a = 6, b = −1, c = −2 ⇒ sum = −(−1)/6 = 1/6; product = (−2)/6 = −1/3.
  15. Q15. Form a quadratic with zeros 2 and 2/3 having integer coefficients.
    Solution
    1. Monic polynomial with these zeros: (x − 2)(x − 2/3) = x² − (8/3)x + 4/3.
    2. Multiply by 3 to clear denominators: 3x² − 8x + 4 (integer coefficients).
  16. Q16. Divide x³ + 4x² + x − 6 by x² + x − 2 (Division Algorithm).
    Solution
    1. Perform polynomial long division:
    2. First term: x³ / x² = x ⇒ multiply divisor by x: x³ + x² − 2x. Subtract from dividend → (x³ + 4x² + x − 6) − (x³ + x² − 2x) = 3x² + 3x − 6.
    3. Next: 3x² / x² = 3 ⇒ multiply divisor by 3: 3x² + 3x − 6. Subtract → remainder 0.
    4. Quotient = x + 3, remainder = 0.
  17. Q17. If (x − 1) is a factor of x³ + kx² + kx + 1, find k and factorize.
    Solution
    1. Since (x − 1) is a factor, p(1) = 0. Compute p(1): 1 + k + k + 1 = 2 + 2k = 0 ⇒ k = −1.
    2. Substitute k = −1: polynomial → x³ − x² − x + 1.
    3. Factor by grouping: x²(x − 1) − 1(x − 1) = (x − 1)(x² − 1) = (x − 1)(x − 1)(x + 1) = (x − 1)²(x + 1).
  18. Q18. Find a monic degree-4 polynomial with zeros ±1 and ±2.
    Solution
    1. Zeros ±1, ±2 ⇒ factors (x − 1)(x + 1)(x − 2)(x + 2) = (x² − 1)(x² − 4).
    2. Multiply: (x² − 1)(x² − 4) = x⁴ − 4x² − x² + 4 = x⁴ − 5x² + 4.
  19. Q19. For 2x² + kx + 3, if sum of zeros is 1, find k.
    Solution
    1. Sum of zeros = −b/a = −k/2. Given = 1 ⇒ −k/2 = 1 ⇒ k = −2.
  20. Q20. If 1 is a zero of x² + kx + 2, find k and the other zero.
    Solution
    1. p(1) = 1 + k + 2 = 0 ⇒ k + 3 = 0 ⇒ k = −3.
    2. Now polynomial is x² − 3x + 2 = (x − 1)(x − 2) ⇒ other zero = 2.

10 FAQs on Polynomials – Class 10 (Complete Explanations)

  1. Q1: What is a polynomial? Give examples.
    Answer (step-by-step)
    1. Definition: A polynomial in x is an expression of the form a_n x^n + a_{n−1} x^{n−1} + ... + a_1 x + a_0 where coefficients a_i are real numbers and n is a non-negative integer.
    2. Requirements: powers are whole numbers (0,1,2,...), coefficients are finite real numbers.
    3. Examples: 3x² − 2x + 1 (polynomial), x^3 − 7 (polynomial). Non-example: 1/x (not a polynomial because x is in denominator), √x (not a polynomial because exponent 1/2).
  2. Q2: How to check if a number a is a zero of p(x)?
    Answer
    1. Substitute x = a in p(x) and compute p(a).
    2. If p(a) = 0, then a is a zero (by definition).
    3. Example: p(x) = x² − 5x + 6, check a = 2 ⇒ p(2) = 4 − 10 + 6 = 0 ⇒ 2 is a zero.
  3. Q3: Relation between zeros and coefficients of a quadratic.
    Answer
    1. Let quadratic be ax² + bx + c and zeros α, β. Then ax² + bx + c = a(x − α)(x − β).
    2. Expand right side: a[x² − (α + β)x + αβ] ⇒ comparing coefficients: −a(α + β) = b and aαβ = c.
    3. Therefore α + β = −b/a and αβ = c/a. Example: x² − 7x + 12 ⇒ α + β = 7, αβ = 12.
  4. Q4: How to form quadratic from given sum S and product P of zeros?
    Answer
    1. If zeros are α and β with α + β = S and αβ = P, quadratic is x² − Sx + P (monic).
    2. Example: S = 4, P = −5 ⇒ x² − 4x − 5.
  5. Q5: What is Factor Theorem and how to use it?
    Answer
    1. Factor Theorem: (x − a) is a factor of polynomial p(x) iff p(a) = 0.
    2. Use: To check if (x − a) divides p(x), compute p(a). If zero ⇒ divide to get quotient; else not a factor.
    3. Example: p(x) = x³ − 7x² + 14x − 8, check a = 2 ⇒ p(2) = 0 ⇒ (x − 2) is a factor.
  6. Q6: What is Division Algorithm for polynomials?
    Answer
    1. Given polynomials p(x) and g(x) ≠ 0, there exist unique q(x) (quotient) and r(x) (remainder) such that p(x) = g(x)·q(x) + r(x), where deg r < deg g.
    2. Procedure: use long division or synthetic division (when divisor is x − a).
    3. Example: Divide x³ + 4x² + x − 6 by x² + x − 2 gives quotient x + 3 and remainder 0 (shown earlier).
  7. Q7: How to decide nature of zeros using discriminant?
    Answer
    1. For ax² + bx + c, discriminant D = b² − 4ac.
    2. If D > 0 → two distinct real zeros; D = 0 → real and equal zeros; D < 0 → two complex (non-real) zeros.
    3. Example: x² − 4x + 3 ⇒ D = 16 − 12 = 4 > 0 ⇒ two distinct real zeros (1 and 3).
  8. Q8: How to get polynomial with integer coefficients when zeros are fractions?
    Answer
    1. Suppose zeros are p/q and r/s (in lowest terms). Form (x − p/q)(x − r/s) and multiply by lcm(q, s)² (or by denominators) to clear fractions.
    2. Example: zeros 1/2 and −3 ⇒ (x − 1/2)(x + 3) = x² + (5/2)x − 3/2. Multiply by 2 ⇒ 2x² + 5x − 3 (integer coefficients).
  9. Q9: How many zeros can a polynomial have?
    Answer
    1. Maximum number of zeros (counting multiplicity and complex roots) equals the degree of the polynomial.
    2. Over reals, some zeros may be non-real. Example: quadratic degree 2 has at most 2 zeros; cubic has at most 3.
    3. Multiplicity: If (x − a)^m divides polynomial, a is a root of multiplicity m (counts m times).
  10. Q10: What is Remainder Theorem? How to use it?
    Answer
    1. Remainder Theorem: When p(x) is divided by (x − a), remainder is p(a).
    2. Use: To find remainder quickly, evaluate p(a) instead of full division.
    3. Example: Remainder of x³ − 4x² + 5x − 2 by (x − 1) is p(1) = 0 (so divisible).

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