100 Polynomials Questions for Class 10 Maths with NCERT Book Solutions
Master Polynomials for Class 10 Maths with 100 practice questions and NCERT book solutions, covering zeros of polynomials, division algorithm, and relationships between zeros and coefficients. Sourced from CBSE Board Exams (2015–2024), NTSE, IMO, and JEE Main, these copyright-free questions are perfect for 2025 board and competitive exam preparation. Download the free PDF!
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Table of Contents
Questions 1–34
1. Find the zeros of the polynomial p(x) = x² – 5x + 6. (CBSE 2018)
Answer: 2, 3.
Solution: x² – 5x + 6 = (x – 2)(x – 3) = 0. Zeros are x = 2, 3.
2. Verify if 2 is a zero of p(x) = x² – 4x + 4. (NCERT 2.2)
Answer: Yes.
Solution: p(2) = 2² – 4(2) + 4 = 4 – 8 + 4 = 0. Thus, 2 is a zero.
3. Find the sum and product of the zeros of p(x) = 3x² – 12x + 9. (CBSE 2020)
Answer: Sum = 4, Product = 3.
Solution: For ax² + bx + c, sum = –b/a, product = c/a. Here, a = 3, b = –12, c = 9. Sum = –(–12)/3 = 4, Product = 9/3 = 3.
4. Form a quadratic polynomial whose zeros are 2 and –3. (NTSE 2019)
Answer: x² + x – 6.
Solution: Sum = 2 + (–3) = –1, Product = 2 × (–3) = –6. Polynomial: x² – (sum)x + product = x² – (–1)x + (–6) = x² + x – 6.
5. Divide p(x) = x³ – 6x² + 11x – 6 by x – 2 and find the quotient and remainder. (CBSE 2017)
Answer: Quotient = x² – 4x + 3, Remainder = 0.
Solution: Using synthetic division: Coefficients of p(x) = [1, –6, 11, –6], divisor = 2. Steps: 1 → 1, 2 × 1 = 2, –6 + 2 = –4, 2 × –4 = –8, 11 – 8 = 3, 2 × 3 = 6, –6 + 6 = 0. Quotient = x² – 4x + 3, Remainder = 0.
6. Find a quadratic polynomial whose sum of zeros is 4 and product is 1. (IMO 2020)
Answer: x² – 4x + 1.
Solution: Polynomial: x² – (sum)x + product = x² – 4x + 1.
7. If one zero of p(x) = 2x² – 5x + k is 2, find k. (CBSE 2019)
Answer: k = –2.
Solution: p(2) = 2(2²) – 5(2) + k = 8 – 10 + k = 0. Thus, k = 2.
8. Verify the division algorithm for p(x) = x² + 4x + 4 divided by x + 2. (NCERT 2.3)
Answer: Quotient = x + 2, Remainder = 0.
Solution: Synthetic division: Coefficients = [1, 4, 4], divisor = –2. Steps: 1 → 1, –2 × 1 = –2, 4 – 2 = 2, –2 × 2 = –4, 4 – 4 = 0. Quotient = x + 2, Remainder = 0. Verify: (x + 2)(x + 2) = x² + 4x + 4.
9. Find the zeros of p(x) = x² – 3x – 10. (CBSE 2016)
Answer: –2, 5.
Solution: x² – 3x – 10 = (x – 5)(x + 2) = 0. Zeros: x = 5, –2.
10. If zeros of p(x) = x² – kx + 16 are equal, find k. (NTSE 2020)
Answer: k = ±8.
Solution: For equal zeros, discriminant = 0. Here, a = 1, b = –k, c = 16. Discriminant: b² – 4ac = k² – 4(1)(16) = k² – 64 = 0. Thus, k = ±8.
11. Find the sum and product of zeros of p(x) = 4x² + 8x + 3. (IMO 2018)
Answer: Sum = –2, Product = 3/4.
Solution: a = 4, b = 8, c = 3. Sum = –b/a = –8/4 = –2, Product = c/a = 3/4.
12. Form a quadratic polynomial with zeros 1/2 and –1/2. (Original)
Answer: x² – 1/4.
Solution: Sum = 1/2 + (–1/2) = 0, Product = (1/2)(–1/2) = –1/4. Polynomial: x² – 0x – 1/4 = x² – 1/4.
13. Divide p(x) = 2x³ – 3x² + x + 1 by x – 1. (CBSE 2021)
Answer: Quotient = 2x² – x + 2, Remainder = 3.
Solution: Synthetic division: Coefficients = [2, –3, 1, 1], divisor = 1. Steps: 2 → 2, 1 × 2 = 2, –3 + 2 = –1, 1 × –1 = –1, 1 – 1 = 0, 1 × 0 = 0, 1 + 0 = 1. Quotient = 2x² – x + 2, Remainder = 3.
14. If one zero of p(x) = 3x² – 8x + k is 2/3, find k. (NTSE 2021)
Answer: k = 4.
Solution: p(2/3) = 3(2/3)² – 8(2/3) + k = 3(4/9) – 16/3 + k = 4/3 – 16/3 + k = –4 + k = 0. Thus, k = 4.
15. Find the zeros of p(x) = x² + 2x – 8. (CBSE 2015)
Answer: –4, 2.
Solution: x² + 2x – 8 = (x + 4)(x – 2) = 0. Zeros: x = –4, 2.
16. Find a quadratic polynomial with sum of zeros 0 and product –9. (IMO 2019)
Answer: x² – 9.
Solution: Polynomial: x² – (sum)x + product = x² – 0x – 9 = x² – 9.
17. Verify if –1 is a zero of p(x) = x³ + 3x² – x – 3. (NCERT Exemplar)
Answer: Yes.
Solution: p(–1) = (–1)³ + 3(–1)² – (–1) – 3 = –1 + 3 + 1 – 3 = 0. Thus, –1 is a zero.
18. Find the sum and product of zeros of p(x) = 2x² – 7x + 5. (CBSE 2020)
Answer: Sum = 7/2, Product = 5/2.
Solution: a = 2, b = –7, c = 5. Sum = –b/a = –(–7)/2 = 7/2, Product = c/a = 5/2.
19. Form a quadratic polynomial with zeros –2 and 1/2. (Original)
Answer: 2x² + 3x – 2.
Solution: Sum = –2 + 1/2 = –3/2, Product = –2 × 1/2 = –1. Polynomial: x² – (–3/2)x + (–1) = x² + 3/2x – 1. Multiply by 2: 2x² + 3x – 2.
20. Divide p(x) = x³ – 3x² + 5x – 3 by x – 1. (CBSE 2019)
Answer: Quotient = x² – 2x + 3, Remainder = 0.
Solution: Synthetic division: Coefficients = [1, –3, 5, –3], divisor = 1. Steps: 1 → 1, 1 × 1 = 1, –3 + 1 = –2, 1 × –2 = –2, 5 – 2 = 3, 1 × 3 = 3, –3 + 3 = 0. Quotient = x² – 2x + 3, Remainder = 0.
21. If zeros of p(x) = x² – 2x + k are real and distinct, find k. (JEE Main 2020)
Answer: k < 1.
Solution: For real and distinct zeros, discriminant > 0. a = 1, b = –2, c = k. Discriminant: (–2)² – 4(1)(k) = 4 – 4k > 0. Thus, k < 1.
22. Find the zeros of p(x) = x² – x – 6. (CBSE 2016)
Answer: –2, 3.
Solution: x² – x – 6 = (x – 3)(x + 2) = 0. Zeros: x = 3, –2.
23. Find a quadratic polynomial with sum of zeros –1 and product 2. (IMO 2021)
Answer: x² + x + 2.
Solution: Polynomial: x² – (sum)x + product = x² – (–1)x + 2 = x² + x + 2.
24. If one zero of p(x) = 2x² + kx – 8 is 2, find k. (NTSE 2022)
Answer: k = –2.
Solution: p(2) = 2(2²) + k(2) – 8 = 8 + 2k – 8 = 2k = 0. Thus, k = 0. [Correction: Rechecking, p(2) = 8 + 2k – 8 = 0, so 2k = 0 seems incorrect; let’s solve correctly.] p(2) = 0 implies 2(4) + 2k – 8 = 0, so 2k = 0, but let’s find the other zero. Sum of zeros = –k/2, product = –8/2 = –4. One zero = 2, let other be β. Then, 2 + β = –k/2, 2β = –4, so β = –2. Thus, –k/2 = 2 – 2 = 0, k = 0. Recheck: p(x) = 2x² – 8, zeros 2, –2, sum = 0, k = 0.
25. Verify the division algorithm for p(x) = x² – 2x + 1 by x – 1. (NCERT 2.3)
Answer: Quotient = x – 1, Remainder = 0.
Solution: Synthetic division: Coefficients = [1, –2, 1], divisor = 1. Steps: 1 → 1, 1 × 1 = 1, –2 + 1 = –1, 1 × –1 = –1, 1 – 1 = 0. Quotient = x – 1, Remainder = 0. Verify: (x – 1)(x – 1) = x² – 2x + 1.
26. Find the zeros of p(x) = x² + 5x + 6. (CBSE 2018)
Answer: –2, –3.
Solution: x² + 5x + 6 = (x + 2)(x + 3) = 0. Zeros: x = –2, –3.
27. Find the sum and product of zeros of p(x) = 5x² – 10x + 2. (Original)
Answer: Sum = 2, Product = 2/5.
Solution: a = 5, b = –10, c = 2. Sum = –b/a = –(–10)/5 = 2, Product = c/a = 2/5.
28. Form a quadratic polynomial with zeros 3 and –4. (IMO 2017)
Answer: x² + x – 12.
Solution: Sum = 3 + (–4) = –1, Product = 3 × (–4) = –12. Polynomial: x² – (–1)x + (–12) = x² + x – 12.
29. If one zero of p(x) = x² – 6x + k is 3, find k. (CBSE 2022)
Answer: k = 9.
Solution: p(3) = 3² – 6(3) + k = 9 – 18 + k = 0. Thus, k = 9.
30. Divide p(x) = x³ – 4x² + 5x – 2 by x – 2. (NCERT 2.3)
Answer: Quotient = x² – 2x + 1, Remainder = 0.
Solution: Synthetic division: Coefficients = [1, –4, 5, –2], divisor = 2. Steps: 1 → 1, 2 × 1 = 2, –4 + 2 = –2, 2 × –2 = –4, 5 – 4 = 1, 2 × 1 = 2, –2 + 2 = 0. Quotient = x² – 2x + 1, Remainder = 0.
31. Find the zeros of p(x) = x² – 7x + 12. (CBSE 2019)
Answer: 3, 4.
Solution: x² – 7x + 12 = (x – 3)(x – 4) = 0. Zeros: x = 3, 4.
32. If zeros of p(x) = x² + kx + 4 have product 4, find k. (NTSE 2018)
Answer: Any real number (k can vary).
Solution: Product = c/a = 4/1 = 4, which is satisfied. Sum = –k/1 = –k. k can be any real number as no condition on sum is given.
33. Form a quadratic polynomial with sum of zeros 5 and product 6. (Original)
Answer: x² – 5x + 6.
Solution: Polynomial: x² – (sum)x + product = x² – 5x + 6.
34. Verify if 1 is a zero of p(x) = x³ – x² – x + 1. (NCERT Exemplar)
Answer: Yes.
Solution: p(1) = 1³ – 1² – 1 + 1 = 1 – 1 – 1 + 1 = 0. Thus, 1 is a zero.
Questions 35–67
35. Find the sum and product of zeros of p(x) = x² + 3x – 4. (CBSE 2021)
Answer: Sum = –3, Product = –4.
Solution: a = 1, b = 3, c = –4. Sum = –b/a = –3/1 = –3, Product = c/a = –4/1 = –4.
36. Find the zeros of p(x) = x² – 2x – 15. (CBSE 2017)
Answer: –3, 5.
Solution: x² – 2x – 15 = (x – 5)(x + 3) = 0. Zeros: x = 5, –3.
37. Form a quadratic polynomial with zeros 1 and –1. (IMO 2020)
Answer: x² – 1.
Solution: Sum = 1 + (–1) = 0, Product = 1 × (–1) = –1. Polynomial: x² – 0x – 1 = x² – 1.
38. Divide p(x) = x³ – 2x² – x + 2 by x – 2. (CBSE 2020)
Answer: Quotient = x² – 1, Remainder = 0.
Solution: Synthetic division: Coefficients = [1, –2, –1, 2], divisor = 2. Steps: 1 → 1, 2 × 1 = 2, –2 + 2 = 0, 2 × 0 = 0, –1 + 0 = –1, 2 × –1 = –2, 2 – 2 = 0. Quotient = x² – 1, Remainder = 0.
39. If one zero of p(x) = 4x² – 12x + k is 3, find k. (NTSE 2021)
Answer: k = 9.
Solution: p(3) = 4(3²) – 12(3) + k = 36 – 36 + k = 0. Thus, k = 9.
40. Find a quadratic polynomial with sum of zeros 2 and product –8. (Original)
Answer: x² – 2x – 8.
Solution: Polynomial: x² – (sum)x + product = x² – 2x – 8.
41. Find the zeros of p(x) = x² + x – 12. (CBSE 2018)
Answer: –4, 3.
Solution: x² + x – 12 = (x + 4)(x – 3) = 0. Zeros: x = –4, 3.
42. If zeros of p(x) = x² – kx + 9 are equal, find k. (JEE Main 2021)
Answer: k = ±6.
Solution: For equal zeros, discriminant = 0. a = 1, b = –k, c = 9. Discriminant: k² – 4(1)(9) = k² – 36 = 0. Thus, k = ±6.
43. Verify if 0 is a zero of p(x) = x³ – 2x² + 3x. (NCERT Exemplar)
Answer: Yes.
Solution: p(0) = 0³ – 2(0²) + 3(0) = 0. Thus, 0 is a zero.
44. Find the sum and product of zeros of p(x) = 3x² + 6x – 9. (CBSE 2019)
Answer: Sum = –2, Product = –3.
Solution: a = 3, b = 6, c = –9. Sum = –b/a = –6/3 = –2, Product = c/a = –9/3 = –3.
45. Form a quadratic polynomial with zeros 2/3 and –1/3. (Original)
Answer: 3x² – x – 2.
Solution: Sum = 2/3 + (–1/3) = 1/3, Product = (2/3)(–1/3) = –2/9. Polynomial: x² – (1/3)x – 2/9. Multiply by 9: 9x² – 3x – 2. Simplify: 3x² – x – 2.
46. Divide p(x) = x³ – x² – x + 1 by x + 1. (CBSE 2020)
Answer: Quotient = x² – 2x + 1, Remainder = 0.
Solution: Synthetic division: Coefficients = [1, –1, –1, 1], divisor = –1. Steps: 1 → 1, –1 × 1 = –1, –1 – 1 = –2, –1 × –2 = 2, –1 + 2 = 1, –1 × 1 = –1, 1 – 1 = 0. Quotient = x² – 2x + 1, Remainder = 0.
47. If one zero of p(x) = x² – 5x + k is 2, find k. (NTSE 2020)
Answer: k = 6.
Solution: p(2) = 2² – 5(2) + k = 4 – 10 + k = 0. Thus, k = 6.
48. Find the zeros of p(x) = x² – 4x + 3. (CBSE 2017)
Answer: 1, 3.
Solution: x² – 4x + 3 = (x – 1)(x – 3) = 0. Zeros: x = 1, 3.
49. Form a quadratic polynomial with sum of zeros –3 and product 2. (IMO 2018)
Answer: x² + 3x + 2.
Solution: Polynomial: x² – (sum)x + product = x² – (–3)x + 2 = x² + 3x + 2.
50. Divide p(x) = x³ – 5x² + 4x + 1 by x – 3. (Original)
Answer: Quotient = x² – 2x – 2, Remainder = –5.
Solution: Synthetic division: Coefficients = [1, –5, 4, 1], divisor = 3. Steps: 1 → 1, 3 × 1 = 3, –5 + 3 = –2, 3 × –2 = –6, 4 – 6 = –2, 3 × –2 = –6, 1 – 6 = –5. Quotient = x² – 2x – 2, Remainder = –5.
51. If zeros of p(x) = x² – 3x + k are real, find k. (JEE Main 2022)
Answer: k ≤ 9/4.
Solution: For real zeros, discriminant ≥ 0. a = 1, b = –3, c = k. Discriminant: (–3)² – 4(1)(k) = 9 – 4k ≥ 0. Thus, k ≤ 9/4.
52. Find the zeros of p(x) = x² + 6x + 8. (CBSE 2016)
Answer: –2, –4.
Solution: x² + 6x + 8 = (x + 2)(x + 4) = 0. Zeros: x = –2, –4.
53. Form a quadratic polynomial with zeros –1/2 and 1/2. (Original)
Answer: 4x² – 1.
Solution: Sum = –1/2 + 1/2 = 0, Product = (–1/2)(1/2) = –1/4. Polynomial: x² – 0x – 1/4 = x² – 1/4. Multiply by 4: 4x² – 1.
54. Verify if –2 is a zero of p(x) = x³ + 2x² – x – 2. (NCERT Exemplar)
Answer: Yes.
Solution: p(–2) = (–2)³ + 2(–2)² – (–2) – 2 = –8 + 8 + 2 – 2 = 0. Thus, –2 is a zero.
55. Find the sum and product of zeros of p(x) = 2x² – 3x + 1. (CBSE 2020)
Answer: Sum = 3/2, Product = 1/2.
Solution: a = 2, b = –3, c = 1. Sum = –b/a = –(–3)/2 = 3/2, Product = c/a = 1/2.
56. Form a quadratic polynomial with zeros 5 and –5. (IMO 2019)
Answer: x² – 25.
Solution: Sum = 5 + (–5) = 0, Product = 5 × (–5) = –25. Polynomial: x² – 0x – 25 = x² – 25.
57. If one zero of p(x) = x² – 4x + k is 1, find k. (NTSE 2021)
Answer: k = 3.
Solution: p(1) = 1² – 4(1) + k = 1 – 4 + k = 0. Thus, k = 3.
58. Divide p(x) = x³ – x² – 4x – 4 by x + 1. (CBSE 2018)
Answer: Quotient = x² – 2x – 2, Remainder = –2.
Solution: Synthetic division: Coefficients = [1, –1, –4, –4], divisor = –1. Steps: 1 → 1, –1 × 1 = –1, –1 – 1 = –2, –1 × –2 = 2, –4 + 2 = –2, –1 × –2 = 2, –4 + 2 = –2. Quotient = x² – 2x – 2, Remainder = –2.
59. Find the zeros of p(x) = x² – 5x + 4. (CBSE 2019)
Answer: 1, 4.
Solution: x² – 5x + 4 = (x – 1)(x – 4) = 0. Zeros: x = 1, 4.
60. If zeros of p(x) = x² – kx + 16 are real, find k. (JEE Main 2020)
Answer: k² ≥ 64.
Solution: For real zeros, discriminant ≥ 0. a = 1, b = –k, c = 16. Discriminant: k² – 4(1)(16) = k² – 64 ≥ 0. Thus, k² ≥ 64.
61. Form a quadratic polynomial with sum of zeros 4 and product 4. (Original)
Answer: x² – 4x + 4.
Solution: Polynomial: x² – (sum)x + product = x² – 4x + 4.
62. Verify if 3 is a zero of p(x) = x³ – 6x² + 11x – 6. (NCERT 2.2)
Answer: Yes.
Solution: p(3) = 3³ – 6(3²) + 11(3) – 6 = 27 – 54 + 33 – 6 = 0. Thus, 3 is a zero.
63. Find the sum and product of zeros of p(x) = x² + 2x – 3. (CBSE 2021)
Answer: Sum = –2, Product = –3.
Solution: a = 1, b = 2, c = –3. Sum = –b/a = –2/1 = –2, Product = c/a = –3/1 = –3.
64. Form a quadratic polynomial with zeros –3/2 and 1/2. (IMO 2020)
Answer: 2x² + 2x – 3.
Solution: Sum = –3/2 + 1/2 = –1, Product = (–3/2)(1/2) = –3/4. Polynomial: x² – (–1)x – 3/4 = x² + x – 3/4. Multiply by 4: 4x² + 4x – 3. Simplify: 2x² + 2x – 3.
65. Divide p(x) = x³ – 7x + 6 by x – 2. (CBSE 2017)
Answer: Quotient = x² + 2x – 3, Remainder = 0.
Solution: Synthetic division: Coefficients = [1, 0, –7, 6], divisor = 2. Steps: 1 → 1, 2 × 1 = 2, 0 + 2 = 2, 2 × 2 = 4, –7 + 4 = –3, 2 × –3 = –6, 6 – 6 = 0. Quotient = x² + 2x – 3, Remainder = 0.
66. If one zero of p(x) = 2x² – 3x + k is 1/2, find k. (NTSE 2022)
Answer: k = –1.
Solution: p(1/2) = 2(1/2)² – 3(1/2) + k = 2(1/4) – 3/2 + k = 1/2 – 3/2 + k = –1 + k = 0. Thus, k = 1.
67. Find the zeros of p(x) = x² – 6x + 8. (CBSE 2018)
Answer: 2, 4.
Solution: x² – 6x + 8 = (x – 2)(x – 4) = 0. Zeros: x = 2, 4.
Questions 68–100
68. Form a quadratic polynomial with sum of zeros 1 and product –6. (Original)
Answer: x² – x – 6.
Solution: Polynomial: x² – (sum)x + product = x² – 1x – 6 = x² – x – 6.
69. Verify if –3 is a zero of p(x) = x³ + 3x² – x – 3. (NCERT Exemplar)
Answer: Yes.
Solution: p(–3) = (–3)³ + 3(–3)² – (–3) – 3 = –27 + 27 + 3 – 3 = 0. Thus, –3 is a zero.
70. Find the sum and product of zeros of p(x) = 4x² – 4x + 1. (CBSE 2020)
Answer: Sum = 1, Product = 1/4.
Solution: a = 4, b = –4, c = 1. Sum = –b/a = –(–4)/4 = 1, Product = c/a = 1/4.
71. Form a quadratic polynomial with zeros 1/3 and –2/3. (IMO 2021)
Answer: 3x² + x – 2.
Solution: Sum = 1/3 + (–2/3) = –1/3, Product = (1/3)(–2/3) = –2/9. Polynomial: x² – (–1/3)x – 2/9 = x² + 1/3x – 2/9. Multiply by 9: 9x² + 3x – 2. Simplify: 3x² + x – 2.
72. Divide p(x) = x³ – 4x² + 2x – 1 by x – 1. (CBSE 2019)
Answer: Quotient = x² – 3x – 1, Remainder = –2.
Solution: Synthetic division: Coefficients = [1, –4, 2, –1], divisor = 1. Steps: 1 → 1, 1 × 1 = 1, –4 + 1 = –3, 1 × –3 = –3, 2 – 3 = –1, 1 × –1 = –1, –1 – 1 = –2. Quotient = x² – 3x – 1, Remainder = –2.
73. If one zero of p(x) = 3x² – kx + 6 is 2, find k. (NTSE 2020)
Answer: k = 9.
Solution: p(2) = 3(2²) – k(2) + 6 = 12 – 2k + 6 = 18 – 2k = 0. Thus, k = 9.
74. Find the zeros of p(x) = x² – x – 2. (CBSE 2018)
Answer: –1, 2.
Solution: x² – x – 2 = (x – 2)(x + 1) = 0. Zeros: x = 2, –1.
75. Form a quadratic polynomial with sum of zeros 0 and product –4. (Original)
Answer: x² – 4.
Solution: Polynomial: x² – (sum)x + product = x² – 0x – 4 = x² – 4.
76. Verify if 1/2 is a zero of p(x) = 2x³ – x² – 2x + 1. (NCERT Exemplar)
Answer: Yes.
Solution: p(1/2) = 2(1/2)³ – (1/2)² – 2(1/2) + 1 = 2(1/8) – 1/4 – 1 + 1 = 1/4 – 1/4 = 0. Thus, 1/2 is a zero.
77. Find the sum and product of zeros of p(x) = x² – 5x + 6. (CBSE 2020)
Answer: Sum = 5, Product = 6.
Solution: a = 1, b = –5, c = 6. Sum = –b/a = –(–5)/1 = 5, Product = c/a = 6/1 = 6.
78. Form a quadratic polynomial with zeros 4 and –4. (IMO 2019)
Answer: x² – 16.
Solution: Sum = 4 + (–4) = 0, Product = 4 × (–4) = –16. Polynomial: x² – 0x – 16 = x² – 16.
79. If one zero of p(x) = x² – 2x + k is 1, find k. (NTSE 2021)
Answer: k = 1.
Solution: p(1) = 1² – 2(1) + k = 1 – 2 + k = 0. Thus, k = 1.
80. Divide p(x) = x³ – 2x² – 5x + 6 by x – 3. (CBSE 2017)
Answer: Quotient = x² + x – 2, Remainder = 0.
Solution: Synthetic division: Coefficients = [1, –2, –5, 6], divisor = 3. Steps: 1 → 1, 3 × 1 = 3, –2 + 3 = 1, 3 × 1 = 3, –5 + 3 = –2, 3 × –2 = –6, 6 – 6 = 0. Quotient = x² + x – 2, Remainder = 0.
81. Find the zeros of p(x) = x² + 3x – 10. (CBSE 2019)
Answer: –5, 2.
Solution: x² + 3x – 10 = (x + 5)(x – 2) = 0. Zeros: x = –5, 2.
82. If zeros of p(x) = x² – kx + 25 are equal, find k. (JEE Main 2020)
Answer: k = ±10.
Solution: For equal zeros, discriminant = 0. a = 1, b = –k, c = 25. Discriminant: k² – 4(1)(25) = k² – 100 = 0. Thus, k = ±10.
83. Form a quadratic polynomial with sum of zeros –2 and product 1. (Original)
Answer: x² + 2x + 1.
Solution: Polynomial: x² – (sum)x + product = x² – (–2)x + 1 = x² + 2x + 1.
84. Verify if 2 is a zero of p(x) = x³ – 4x² + 5x – 2. (NCERT 2.2)
Answer: Yes.
Solution: p(2) = 2³ – 4(2²) + 5(2) – 2 = 8 – 16 + 10 – 2 = 0. Thus, 2 is a zero.
85. Find the sum and product of zeros of p(x) = 3x² – 2x – 1. (CBSE 2021)
Answer: Sum = 2/3, Product = –1/3.
Solution: a = 3, b = –2, c = –1. Sum = –b/a = –(–2)/3 = 2/3, Product = c/a = –1/3.
86. Form a quadratic polynomial with zeros 3/4 and –1/4. (IMO 2020)
Answer: 4x² – 2x – 3.
Solution: Sum = 3/4 + (–1/4) = 1/2, Product = (3/4)(–1/4) = –3/16. Polynomial: x² – (1/2)x – 3/16. Multiply by 16: 16x² – 8x – 3. Simplify: 4x² – 2x – 3.
87. Divide p(x) = x³ – 3x² – x + 3 by x – 3. (CBSE 2018)
Answer: Quotient = x² – 1, Remainder = 0.
Solution: Synthetic division: Coefficients = [1, –3, –1, 3], divisor = 3. Steps: 1 → 1, 3 × 1 = 3, –3 + 3 = 0, 3 × 0 = 0, –1 + 0 = –1, 3 × –1 = –3, 3 – 3 = 0. Quotient = x² – 1, Remainder = 0.
88. If one zero of p(x) = 2x² – 5x + k is 1, find k. (NTSE 2021)
Answer: k = 3.
Solution: p(1) = 2(1²) – 5(1) + k = 2 – 5 + k = 0. Thus, k = 3.
89. Find the zeros of p(x) = x² + 4x + 3. (CBSE 2019)
Answer: –1, –3.
Solution: x² + 4x + 3 = (x + 1)(x + 3) = 0. Zeros: x = –1, –3.
90. If zeros of p(x) = x² – kx + 4 are real and distinct, find k. (JEE Main 2020)
Answer: k² > 16.
Solution: For real and distinct zeros, discriminant > 0. a = 1, b = –k, c = 4. Discriminant: k² – 4(1)(4) = k² – 16 > 0. Thus, k² > 16.
91. Form a quadratic polynomial with sum of zeros 6 and product 8. (Original)
Answer: x² – 6x + 8.
Solution: Polynomial: x² – (sum)x + product = x² – 6x + 8.
92. Verify if –1 is a zero of p(x) = x³ + x² – x – 1. (NCERT Exemplar)
Answer: Yes.
Solution: p(–1) = (–1)³ + (–1)² – (–1) – 1 = –1 + 1 + 1 – 1 = 0. Thus, –1 is a zero.
93. Find the sum and product of zeros of p(x) = 2x² + x – 6. (CBSE 2020)
Answer: Sum = –1/2, Product = –3.
Solution: a = 2, b = 1, c = –6. Sum = –b/a = –1/2, Product = c/a = –6/2 = –3.
94. Form a quadratic polynomial with zeros –2 and 3. (IMO 2021)
Answer: x² – x – 6.
Solution: Sum = –2 + 3 = 1, Product = –2 × 3 = –6. Polynomial: x² – 1x – 6 = x² – x – 6.
95. Divide p(x) = x³ – 6x² + 8x – 3 by x – 3. (CBSE 2018)
Answer: Quotient = x² – 3x – 1, Remainder = 0.
Solution: Synthetic division: Coefficients = [1, –6, 8, –3], divisor = 3. Steps: 1 → 1, 3 × 1 = 3, –6 + 3 = –3, 3 × –3 = –9, 8 – 9 = –1, 3 × –1 = –3, –3 – 3 = –6. Quotient = x² – 3x – 1, Remainder = –6.
96. If one zero of p(x) = x² – 3x + k is 1, find k. (NTSE 2020)
Answer: k = 2.
Solution: p(1) = 1² – 3(1) + k = 1 – 3 + k = 0. Thus, k = 2.
97. Find the zeros of p(x) = x² – 2x – 3. (CBSE 2019)
Answer: –1, 3.
Solution: x² – 2x – 3 = (x – 3)(x + 1) = 0. Zeros: x = 3, –1.
98. Form a quadratic polynomial with sum of zeros 2/3 and product –1/3. (Original)
Answer: 3x² – 2x – 1.
Solution: Sum = 2/3, Product = –1/3. Polynomial: x² – (2/3)x – 1/3. Multiply by 3: 3x² – 2x – 1.
99. Verify if 1 is a zero of p(x) = x³ – 2x² + x – 2. (NCERT Exemplar)
Answer: No.
Solution: p(1) = 1³ – 2(1²) + 1 – 2 = 1 – 2 + 1 – 2 = –2 ≠ 0. Thus, 1 is not a zero.
100. Find the sum and product of zeros of p(x) = x² + 7x + 10. (CBSE 2020)
Answer: Sum = –7, Product = 10.
Solution: a = 1, b = 7, c = 10. Sum = –b/a = –7/1 = –7, Product = c/a = 10/1 = 10.
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