CBSE Class 10 Maths Ch 2 Polynomials Question Bank with Solutions

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Class 10 Maths · Chapter 2 · Question Bank + Solutions

CBSE · NCERT

MATHEMATICS – CHAPTER 2 : POLYNOMIALS

Chapter-wise Question Bank (CBSE Board + Competitive · Moderate to HOTS Level)

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Class : XSubject : MathematicsName : ..................

Questions : 89Session : 2026–27Total Marks : 151

General Instructions

1. This question bank is based on NCERT Class X Mathematics, Chapter 2 (Polynomials), and is set in the CBSE board pattern with competitive (HOTS) questions mixed in.
2. Section A: Q.1 to Q.50 are MCQs of 1 mark each (50 marks). Section B: Q.51 to Q.60 are Assertion–Reason questions of 1 mark each (10 marks).
3. Section C: Q.61 to Q.70, Very Short Answer, 2 marks each (20 marks). Section D: Q.71 to Q.80, Short Answer, 3 marks each (30 marks).
4. Section E: Q.81 to Q.84, Case-Based, 4 marks each (16 marks). Section F: Q.85 to Q.89, Long Answer, 5 marks each (25 marks).
5. Show all steps for Sections C to F. Complete solutions are given after the questions.

SECTION – A (Multiple Choice Questions · 1 Mark each)

Q.1 – Q.50 Choose the correct option.50 × 1 = 50

1. The degree of the polynomial 5x³ – 2x⁴ + 7 is –

(a) 3(b) 7(c) 4(d) 2

2. The zero of the polynomial p(x) = 3x – 6 is –

(a) –2(b) 3(c) 2(d) 6

3. A linear polynomial has exactly how many zeroes?

(a) 0(b) 2(c) 1(d) 3

4. A quadratic polynomial can have at most how many zeroes?

(a) 1(b) 3(c) 4(d) 2

5. If the graph of y = p(x) cuts the x-axis at exactly 3 points, the number of zeroes of p(x) is –

(a) 1(b) 2(c) 0(d) 3

6. The graph of y = ax² + bx + c opens upwards when –

(a) a < 0(b) b > 0(c) c > 0(d) a > 0

7. If the graph of a quadratic polynomial does not intersect the x-axis, the polynomial has –

(a) no real zero(b) one real zero(c) two real zeroes(d) three real zeroes

8. If the graph of a quadratic polynomial touches the x-axis at exactly one point, then it has –

(a) two equal real zeroes (one repeated zero)(b) two distinct zeroes(c) no zero(d) three zeroes

9. The sum of the zeroes of the polynomial x² – 5x + 6 is –

(a) –5(b) 6(c) 5(d) –6

10. The product of the zeroes of 2x² – 7x + 3 is –

(a) 3/2(b) 7/2(c) –3/2(d) 2/3

11. A zero of the polynomial x² – 3x – 4 is –

(a) 2(b) 3(c) 4(d) –4

12. A quadratic polynomial whose zeroes are –2 and –3 is –

(a) x² – 5x + 6(b) x² + 5x + 6(c) x² + 5x – 6(d) x² – x – 6

13. A quadratic polynomial whose sum and product of zeroes are 3 and –2 respectively is –

(a) x² + 3x – 2(b) x² – 3x – 2(c) x² – 3x + 2(d) x² + 3x + 2

14. If α and β are the zeroes of x² – 7x + 10, then α² + β² equals –

(a) 49(b) 29(c) 9(d) 39

15. If α and β are the zeroes of x² – 5x + 6, then 1/α + 1/β is –

(a) 6/5(b) 5/6(c) –5/6(d) 1/6

16. If α and β are the zeroes of x² – 7x + 12, then |α – β| equals –

(a) 1(b) 7(c) 5(d) 12

17. If 2 is a zero of x² – 5x + k, the value of k is –

(a) –6(b) 10(c) 4(d) 6

18. The zeroes of x² + 4x + 3 are –

(a) 1 and 3(b) –1 and –3(c) 1 and –3(d) –1 and 3

19. The sum of the zeroes of 4x² – 8x + 3 is –

(a) –2(b) 3/4(c) 2(d) –3/4

20. If the sum of the zeroes of kx² + 2x + 3k is equal to their product, then k equals –

(a) 2/3(b) 3/2(c) –3/2(d) –2/3

21. If α and β are the zeroes of 2x² – 5x + 3, then α + β + αβ equals –

(a) 4(b) 5/2(c) 3/2(d) 1

22. A quadratic polynomial whose zeroes are 2 + √3 and 2 – √3 is –

(a) x² + 4x + 1(b) x² – 4x – 1(c) x² – 2x + 1(d) x² – 4x + 1

23. The zeroes of x² – 2 are –

(a) 2 and –2(b) √2 and 2(c) 1 and –2(d) √2 and –√2

24. The number of real zeroes of x² + 1 is –

(a) 1(b) 2(c) 3(d) 0

25. If –1 is a zero of x² + 3x + k, the value of k is –

(a) 2(b) –2(c) 4(d) –4

26. The division algorithm for polynomials states that if p(x) and g(x) (g(x) ≠ 0) are polynomials, then p(x) = g(x)·q(x) + r(x), where –

(a) r(x) = 0 or degree r(x) < degree g(x)(b) degree r(x) > degree g(x)(c) degree r(x) = degree g(x)(d) degree q(x) = degree g(x)

27. When x³ – 3x² + 5x – 3 is divided by x² – 2, the remainder is –

(a) 5x – 3(b) x – 3(c) 7x + 9(d) 7x – 9

28. When a polynomial of degree 5 is divided by a polynomial of degree 2, the degree of the quotient is –

(a) 2(b) 3(c) 5(d) 7

29. The remainder when x³ + 2x² – 5x + 6 is divided by x – 1 is –

(a) 6(b) 2(c) 0(d) 4

30. The sum of the zeroes of the cubic polynomial x³ – 6x² + 11x – 6 is –

(a) 6(b) –6(c) 11(d) –11

31. The product of the zeroes of x³ – 6x² + 11x – 6 is –

(a) –6(b) 11(c) 6(d) 1

32. For the cubic polynomial x³ – 6x² + 11x – 6, the sum of the products of its zeroes taken two at a time is –

(a) 6(b) –11(c) 11(d) –6

33. The maximum number of zeroes a cubic polynomial can have is –

(a) 1(b) 2(c) 3(d) 4

34. How many quadratic polynomials have zeroes 2 and 5?

(a) exactly one(b) infinitely many(c) exactly two(d) none

35. The zeroes of the polynomial 4x² – 4x + 1 are –

(a) real and distinct(b) not real(c) real and equal (1/2 and 1/2)(d) –1/2 and –1/2

36. If α and β are the zeroes of x² + px + q, then α + β equals –

(a) –p(b) p(c) q(d) –q

37. If the zeroes of x² + px + q are 2 and –3, then (p, q) is –

(a) (1, –6)(b) (–1, –6)(c) (1, 6)(d) (–1, 6)

38. If one zero of 3x² + 8x + k is the reciprocal of the other, then k equals –

(a) 1/3(b) 3(c) –3(d) 8

39. A quadratic polynomial whose zeroes are –3 and 4 is –

(a) x² – x – 12(b) x² + x – 12(c) x² – x + 12(d) x² + 7x – 12

40. If 2 and –3 are the zeroes of x² + (a + 1)x + b, then the values of a and b are –

(a) a = 0, b = –6(b) a = 1, b = –6(c) a = 0, b = 6(d) a = –1, b = 6

41. If α and β are the zeroes of x² – px + q, then α²β + αβ² equals –

(a) p + q(b) p/q(c) pq(d) q/p

42. If α and β are the zeroes of 6x² + x – 2, then α/β + β/α equals –

(a) 25/12(b) –25/12(c) –12/25(d) 13/12

43. If α and β are the zeroes of x² – 6x + k and 3α + 2β = 20, then k equals –

(a) –16(b) 16(c) 8(d) –8

44. If α and β are the zeroes of x² – x – 2, the quadratic polynomial with zeroes 2α and 2β is –

(a) x² – 2x + 8(b) x² + 2x – 8(c) x² – 4x – 4(d) x² – 2x – 8

45. If α and β are the zeroes of x² – 5x + 6, the polynomial whose zeroes are 1/α and 1/β is –

(a) x² – 5x + 6(b) 6x² + 5x + 1(c) x² – 6x + 5(d) 6x² – 5x + 1

46. The number of zeroes of the polynomial x³ – x is –

(a) 1(b) 3(c) 2(d) 0

47. Given that 2 is a zero of x³ – 4x² + x + 6, the other two zeroes are –

(a) –2 and 3(b) 3 and –1(c) 1 and –3(d) 6 and 1

48. If α and β are the zeroes of x² – 3x + 1, then α² + β² equals –

(a) 9(b) 7(c) 5(d) 11

49. The graph of y = (x – 1)(x – 3) cuts the x-axis at the points –

(a) (–1, 0) and (–3, 0)(b) (0, 1) and (0, 3)(c) (1, 0) and (3, 0)(d) (1, 0) and (–3, 0)

50. If the product of the zeroes of 2x² – 7x + k is 4, then k equals –

(a) 4(b) 8(c) 2(d) –8

SECTION – B (Assertion (A) – Reason (R) · 1 Mark each)

Q.51 – Q.60 Choose: (a) Both A and R true, R explains A; (b) Both true, R does not explain A; (c) A true, R false; (d) A false, R true.10 × 1 = 10

51. Assertion (A): The polynomial x² + 1 has no real zero.
Reason (R): The graph of y = x² + 1 does not intersect the x-axis.

52. Assertion (A): A quadratic polynomial can have at most two zeroes.
Reason (R): A polynomial of degree n has at most n zeroes.

53. Assertion (A): The zeroes of x² – 3x + 2 are 1 and 2.
Reason (R): A quadratic polynomial can have at most two zeroes.

54. Assertion (A): The polynomial x² – 4x + 4 has two equal zeroes.
Reason (R): The graph of x² – 4x + 4 cuts the x-axis at two distinct points.

55. Assertion (A): The polynomial x² + 5 has real zeroes.
Reason (R): x² + 5 ≥ 5 for every real number x.

56. Assertion (A): The sum of the zeroes of 2x² – 8x + 6 is 4.
Reason (R): For ax² + bx + c, the sum of the zeroes is –b/a.

57. Assertion (A): A quadratic polynomial can have three zeroes.
Reason (R): The graph of a quadratic polynomial is a parabola, which can cut the x-axis at most twice.

58. Assertion (A): The cubic polynomial x³ – x has three real zeroes.
Reason (R): The graph of every cubic polynomial cuts the x-axis at exactly three points.

59. Assertion (A): In the division algorithm p(x) = g(x)q(x) + r(x), the degree of r(x) is less than the degree of g(x) (or r(x) = 0).
Reason (R): The division process is stopped when the degree of the remainder becomes less than the degree of the divisor.

60. Assertion (A): If p(x) = g(x)q(x) + r(x) with r(x) = 0, then g(x) is a factor of p(x).
Reason (R): The degree of g(x) is less than or equal to the degree of p(x).

SECTION – C (Very Short Answer · 2 Marks each)

61. Find the zeroes of the polynomial x² – 3x – 4 and verify the relationship between the zeroes and the coefficients.(2)

62. Find a quadratic polynomial, the sum and product of whose zeroes are –3 and 2 respectively.(2)

63. Find the sum and the product of the zeroes of 3x² – 5x – 2.(2)

64. Find the zeroes of x² + 7x + 10 and verify the relationship between the zeroes and coefficients.(2)

65. If α and β are the zeroes of x² – 5x + k such that α – β = 1, find k.(2)

66. Find a quadratic polynomial whose zeroes are 1/4 and –1.(2)

67. Divide 2x² + 3x + 1 by x + 2 and write the quotient and the remainder.(2)

68. Find the zero(s) of the polynomial p(x) = x² – 6x + 9. How many times does its graph meet the x-axis?(2)

69. If α and β are the zeroes of x² – 7x + 10, find the value of α² + β².(2)

70. If 2 is a zero of both 2x² + ax + 14 and a(x² + x) – b, find a and b.(2)

SECTION – D (Short Answer · 3 Marks each)

71. Find the zeroes of 6x² – 3 – 7x and verify the relationship between the zeroes and the coefficients.(3)

72. Find a quadratic polynomial whose sum and product of zeroes are √2 and 1/3 respectively.(3)

73. If α and β are the zeroes of 3x² – 5x – 2, find (i) α² + β² and (ii) 1/α + 1/β.(3)

74. Divide 3x³ + x² + 2x + 5 by 1 + 2x + x² and find the quotient and the remainder. Verify the division algorithm.(3)

75. On dividing x³ – 3x² + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and –2x + 4 respectively. Find g(x).(3)

76. If α and β are the zeroes of x² – 6x + k and 3α + 2β = 20, find the value of k.(3)

77. Verify that 3, –1 and –1/3 are the zeroes of the cubic polynomial 3x³ – 5x² – 11x – 3 and verify the relationship between the zeroes and the coefficients.(3)

78. If α and β are the zeroes of x² – x – 2, find a quadratic polynomial whose zeroes are 2α and 2β.(3)

79. If α and β are the zeroes of x² – 5x + 6, form a quadratic polynomial whose zeroes are 1/α and 1/β.(3)

80. If one zero of the polynomial 3x² + 8x + k is the reciprocal of the other, find k and the zeroes.(3)

SECTION – E (Case-Based Questions · 4 Marks each)

81. Read the passage and answer the questions that follow.(4)

A ball is thrown upwards from a point 5 m above the ground. Its height h (in metres) after t seconds is given by the polynomial h(t) = –t² + 4t + 5.

(i) At what time does the ball hit the ground? (Find the zeroes of h(t).)(1)

(ii) Which zero is rejected, and why? So at what time does the ball land?(1)

(iii) Find the maximum height reached by the ball and the time at which it is reached.(2)

82. Read the passage and answer the questions that follow.(4)

The graph of a cubic polynomial p(x) cuts the x-axis at exactly three points: x = –2, x = 1 and x = 3.

(i) How many zeroes does p(x) have? Name them.(1)

(ii) Write a cubic polynomial having these zeroes (leading coefficient 1).(1)

(iii) For the polynomial in (ii), find the sum of the zeroes, the sum of the products taken two at a time, and the product of the zeroes, and verify with the coefficients.(2)

83. Read the passage and answer the questions that follow.(4)

A farmer has a rectangular field whose area (in square metres) is given by the polynomial x² + 7x + 12, and whose length is (x + 4) metres.

(i) Find the breadth of the field.(1)

(ii) Find the zeroes of the area polynomial.(1)

(iii) If x = 6, find the perimeter and the area of the field.(2)

84. Read the passage and answer the questions that follow.(4)

In a class activity, the quadratic polynomial p(x) = x² – (k + 6)x + 2(2k – 1) is given. The teacher tells the students that the sum of its zeroes is half of their product.

(i) Write the sum and the product of the zeroes of p(x) in terms of k.(1)

(ii) Form the equation using the teacher's condition.(1)

(iii) Find k and the polynomial p(x). Verify the condition.(2)

SECTION – F (Long Answer · 5 Marks each)

85. Divide p(x) = x⁴ – 3x² + 4x + 5 by g(x) = x² + 1 – x and find the quotient and the remainder. Verify the division algorithm.(5)

86. Given that 3 is a zero of the polynomial x³ – 5x² – 2x + 24, find all its zeroes. Also verify the relationship between the zeroes and the coefficients.(5)

87. If α and β are the zeroes of the polynomial kx² + 4x + 4 such that α² + β² = 24, find the value(s) of k.(5)

88. If α and β are the zeroes of the polynomial 2x² – 4x + 5, find the value of (i) α² + β², (ii) (α – β)², (iii) 1/α + 1/β, (iv) α³ + β³.(5)

89. Given that √2 and –√2 are two zeroes of the polynomial 2x⁴ – 3x³ – 3x² + 6x – 2, find all the other zeroes.(5)

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CBSE · NCERT

MATHEMATICS – CHAPTER 2 : COMPLETE SOLUTIONS

Polynomials · Answer Key with Step-by-Step Solutions

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Class : XSubject : MathematicsTotal Marks : 151

SECTION – A (MCQ Solutions)

Quick Answer Key (Q.1 – Q.50)

1. (c)	2. (c)	3. (c)	4. (d)	5. (d)	6. (d)	7. (a)	8. (a)	9. (c)	10. (a)
11. (c)	12. (b)	13. (b)	14. (b)	15. (b)	16. (a)	17. (d)	18. (b)	19. (c)	20. (d)
21. (a)	22. (d)	23. (d)	24. (d)	25. (a)	26. (a)	27. (d)	28. (b)	29. (d)	30. (a)
31. (c)	32. (c)	33. (c)	34. (b)	35. (c)	36. (a)	37. (a)	38. (b)	39. (a)	40. (a)
41. (c)	42. (b)	43. (a)	44. (d)	45. (d)	46. (b)	47. (b)	48. (b)	49. (c)	50. (b)

Solutions with Explanation

1. Ans. (c) 4

The highest power of x is 4 (from –2x⁴), so the degree is 4.

2. Ans. (c) 2

p(x) = 0 ⇒ 3x = 6 ⇒ x = 2.

3. Ans. (c) 1

ax + b = 0 gives x = –b/a, a single zero.

4. Ans. (d) 2

A polynomial of degree n has at most n zeroes; for n = 2 the maximum is 2.

5. Ans. (d) 3

The zeroes of p(x) are the x-coordinates of the points where its graph meets the x-axis.

6. Ans. (d) a > 0

The parabola opens upwards if a > 0 and downwards if a < 0.

7. Ans. (a) no real zero

No point of the graph lies on the x-axis, so p(x) = 0 has no real solution.

8. Ans. (a) two equal real zeroes (one repeated zero)

The parabola just touches the axis at its vertex: the zero is repeated (discriminant = 0).

9. Ans. (c) 5

Sum = –b/a = –(–5)/1 = 5. (The zeroes are 2 and 3.)

10. Ans. (a) 3/2

Product = c/a = 3/2.

11. Ans. (c) 4

p(4) = 16 – 12 – 4 = 0. (The zeroes are 4 and –1.)

12. Ans. (b) x² + 5x + 6

Sum = –5, product = 6 ⇒ x² – (sum)x + product = x² + 5x + 6.

13. Ans. (b) x² – 3x – 2

x² – (sum)x + (product) = x² – 3x – 2. (Any non-zero multiple k of it also works.)

14. Ans. (b) 29

α + β = 7, αβ = 10. α² + β² = (α + β)² – 2αβ = 49 – 20 = 29.

15. Ans. (b) 5/6

1/α + 1/β = (α + β)/(αβ) = 5/6.

16. Ans. (a) 1

(α – β)² = (α + β)² – 4αβ = 49 – 48 = 1. (The zeroes are 3 and 4.)

17. Ans. (d) 6

p(2) = 4 – 10 + k = 0 ⇒ k = 6.

18. Ans. (b) –1 and –3

x² + 4x + 3 = (x + 1)(x + 3) ⇒ x = –1, –3.

19. Ans. (c) 2

Sum = –b/a = 8/4 = 2.

20. Ans. (d) –2/3

Sum = –2/k; product = 3k/k = 3. So –2/k = 3 ⇒ k = –2/3.

21. Ans. (a) 4

α + β = 5/2, αβ = 3/2 ⇒ α + β + αβ = 4.

22. Ans. (d) x² – 4x + 1

Sum = 4, product = 4 – 3 = 1 ⇒ x² – 4x + 1.

23. Ans. (d) √2 and –√2

x² = 2 ⇒ x = ±√2.

24. Ans. (d) 0

x² + 1 ≥ 1 for all real x, so it never equals 0.

25. Ans. (a) 2

p(–1) = 1 – 3 + k = 0 ⇒ k = 2.

26. Ans. (a) r(x) = 0 or degree r(x) < degree g(x)

The division stops when the remainder is 0 or its degree is less than that of the divisor.

27. Ans. (d) 7x – 9

Quotient x – 3: (x – 3)(x² – 2) = x³ – 3x² – 2x + 6. Remainder = (x³ – 3x² + 5x – 3) – (x³ – 3x² – 2x + 6) = 7x – 9.

28. Ans. (b) 3

deg q = deg p – deg g = 5 – 2 = 3.

29. Ans. (d) 4

By the remainder theorem the remainder is p(1) = 1 + 2 – 5 + 6 = 4.

30. Ans. (a) 6

For ax³ + bx² + cx + d, sum = –b/a = 6. (Zeroes: 1, 2, 3.)

31. Ans. (c) 6

Product = –d/a = –(–6)/1 = 6 (1 × 2 × 3 = 6).

32. Ans. (c) 11

αβ + βγ + γα = c/a = 11 (check: 2 + 6 + 3 = 11).

33. Ans. (c) 3

A polynomial of degree n has at most n zeroes.

34. Ans. (b) infinitely many

Every polynomial k(x² – 7x + 10), where k ≠ 0 is a real number, has the zeroes 2 and 5.

35. Ans. (c) real and equal (1/2 and 1/2)

4x² – 4x + 1 = (2x – 1)², so the zero 1/2 is repeated.

36. Ans. (a) –p

Sum of zeroes = –(coefficient of x)/(coefficient of x²) = –p.

37. Ans. (a) (1, –6)

–p = 2 + (–3) = –1 ⇒ p = 1; q = 2 × (–3) = –6.

38. Ans. (b) 3

Product of zeroes = k/3 = α × (1/α) = 1 ⇒ k = 3.

39. Ans. (a) x² – x – 12

Sum = 1, product = –12 ⇒ x² – x – 12.

40. Ans. (a) a = 0, b = –6

Sum = –(a + 1) = 2 – 3 = –1 ⇒ a = 0; product = b = 2 × (–3) = –6.

41. Ans. (c) pq

α²β + αβ² = αβ(α + β) = q × p = pq.

42. Ans. (b) –25/12

α + β = –1/6, αβ = –1/3. α² + β² = 1/36 + 2/3 = 25/36. So α/β + β/α = (25/36) ÷ (–1/3) = –25/12.

43. Ans. (a) –16

α + β = 6 ⇒ 3α + 2β = 2(α + β) + α = 12 + α = 20 ⇒ α = 8, β = –2. k = αβ = –16.

44. Ans. (d) x² – 2x – 8

α + β = 1, αβ = –2. New sum = 2(α + β) = 2, new product = 4αβ = –8 ⇒ x² – 2x – 8. (The zeroes 2, –1 become 4, –2.)

45. Ans. (d) 6x² – 5x + 1

Sum = (α + β)/αβ = 5/6, product = 1/6 ⇒ x² – (5/6)x + 1/6, i.e. 6x² – 5x + 1.

46. Ans. (b) 3

x³ – x = x(x – 1)(x + 1); the zeroes are 0, 1 and –1.

47. Ans. (b) 3 and –1

Dividing by (x – 2) gives x² – 2x – 3 = (x – 3)(x + 1); the zeroes are 3 and –1. (Check: product = 2 × 3 × (–1) = –6 = –d/a.)

48. Ans. (b) 7

(α + β)² – 2αβ = 9 – 2 = 7.

49. Ans. (c) (1, 0) and (3, 0)

The zeroes are x = 1 and x = 3, so the graph meets the x-axis at (1, 0) and (3, 0).

50. Ans. (b) 8

Product = k/2 = 4 ⇒ k = 8.

SECTION – B (Assertion – Reason)

51. Ans. (a)

Both true; the zeroes of a polynomial are the x-coordinates of the points where its graph meets the x-axis, and there are none.

52. Ans. (a)

Both true; R is the general result of which A is the case n = 2.

53. Ans. (b)

Both true, but R does not explain why the zeroes are 1 and 2 (x² – 3x + 2 = (x – 1)(x – 2)).

54. Ans. (c)

A is true ((x – 2)², zero 2 repeated). R is false: the graph only touches the x-axis at (2, 0).

55. Ans. (d)

A is false: x² + 5 is never 0. R is true.

56. Ans. (a)

Both true; sum = –(–8)/2 = 4.

57. Ans. (d)

A is false. R is true.

58. Ans. (c)

A is true (0, 1, –1). R is false: a cubic cuts the x-axis at 1, 2 or 3 points (e.g. x³ + x has only one zero).

59. Ans. (a)

Both true; R is the reason A holds.

60. Ans. (b)

Both true, but R is not the reason g(x) is a factor.

SECTION – C (2 Marks each)

61. Ans.

x² – 3x – 4 = (x – 4)(x + 1) ⇒ zeroes 4 and –1. [1] Sum = 3 = –(–3)/1 ✓; product = –4 = –4/1 ✓. [1]

62. Ans.

Polynomial = k[x² – (sum)x + product]. [1] With k = 1: x² – (–3)x + 2 = x² + 3x + 2. [1]

63. Ans.

Here a = 3, b = –5, c = –2. [½] Sum = –b/a = 5/3. [½] Product = c/a = –2/3. [1]

64. Ans.

x² + 7x + 10 = (x + 2)(x + 5) ⇒ zeroes –2 and –5. [1] Sum = –7 = –7/1 ✓; product = 10 = 10/1 ✓. [1]

65. Ans.

α + β = 5 and α – β = 1 ⇒ α = 3, β = 2. [1] k = αβ = 6. [1]

66. Ans.

Sum = 1/4 – 1 = –3/4; product = (1/4)(–1) = –1/4. [1] Polynomial: x² + (3/4)x – 1/4, or 4x² + 3x – 1. [1]

67. Ans.

2x² + 3x + 1 = (x + 2)(2x – 1) + 3. [1] Quotient = 2x – 1, remainder = 3. [1]

68. Ans.

x² – 6x + 9 = (x – 3)² ⇒ the only (repeated) zero is x = 3. [1] The graph touches the x-axis at exactly one point, (3, 0). [1]

69. Ans.

α + β = 7 and αβ = 10. [1] α² + β² = (α + β)² – 2αβ = 49 – 20 = 29. [1]

70. Ans.

2x² + ax + 14 at x = 2: 8 + 2a + 14 = 0 ⇒ a = –11. [1] a(x² + x) – b at x = 2: a(4 + 2) – b = 0 ⇒ b = 6a = –66. [1]

SECTION – D (3 Marks each)

71. Ans.

6x² – 7x – 3 = 6x² – 9x + 2x – 3 = (2x – 3)(3x + 1). [1] Zeroes: x = 3/2 and x = –1/3. [½] Sum = 3/2 – 1/3 = 7/6 = –(–7)/6 ✓. [½] Product = (3/2)(–1/3) = –1/2 = –3/6 ✓. [1]

72. Ans.

Polynomial = k[x² – (sum)x + product] = k[x² – √2 x + 1/3]. [2] Taking k = 3: 3x² – 3√2 x + 1. [1]

73. Ans.

α + β = 5/3 and αβ = –2/3. [1] (i) α² + β² = (5/3)² – 2(–2/3) = 25/9 + 4/3 = 37/9. [1] (ii) 1/α + 1/β = (α + β)/(αβ) = (5/3) ÷ (–2/3) = –5/2. [1]

74. Ans.

Divisor g(x) = x² + 2x + 1. 3x³ ÷ x² = 3x: subtract 3x(x² + 2x + 1) = 3x³ + 6x² + 3x ⇒ –5x² – x + 5. [1] –5x² ÷ x² = –5: subtract –5(x² + 2x + 1) = –5x² – 10x – 5 ⇒ 9x + 10 (degree 1 < 2, stop). [1] Quotient = 3x – 5, remainder = 9x + 10. Check: (x² + 2x + 1)(3x – 5) + 9x + 10 = 3x³ + x² – 7x – 5 + 9x + 10 = 3x³ + x² + 2x + 5 ✓. [1]

75. Ans.

By the division algorithm p(x) = g(x)q(x) + r(x) ⇒ g(x)q(x) = p(x) – r(x). [1] p(x) – r(x) = x³ – 3x² + x + 2 + 2x – 4 = x³ – 3x² + 3x – 2. [1] g(x) = (x³ – 3x² + 3x – 2) ÷ (x – 2) = x² – x + 1. [1]

76. Ans.

α + β = 6 …(i); 3α + 2β = 20 …(ii). [1] (ii) – 2×(i): α = 20 – 12 = 8; so β = 6 – 8 = –2. [1] k = αβ = 8 × (–2) = –16. [1]

77. Ans.

p(3) = 81 – 45 – 33 – 3 = 0; p(–1) = –3 – 5 + 11 – 3 = 0; p(–1/3) = –1/9 – 5/9 + 11/3 – 3 = 0 ✓. [1] Sum = 3 – 1 – 1/3 = 5/3 = –b/a ✓. [½] Sum of products in pairs = –3 + 1/3 – 1 = –11/3 = c/a ✓. [1] Product = 3 × (–1) × (–1/3) = 1 = –d/a = 3/3 ✓. [½]

78. Ans.

α + β = 1 and αβ = –2. [1] New sum = 2α + 2β = 2; new product = 4αβ = –8. [1] Polynomial: x² – 2x – 8. [1]

79. Ans.

α + β = 5 and αβ = 6. [1] Sum = 1/α + 1/β = (α + β)/αβ = 5/6; product = 1/(αβ) = 1/6. [1] Polynomial: x² – (5/6)x + 1/6, i.e. 6x² – 5x + 1. [1]

80. Ans.

Let the zeroes be α and 1/α. Product = k/3 = α × (1/α) = 1 ⇒ k = 3. [1] The polynomial is 3x² + 8x + 3. x = [–8 ± √(64 – 36)]/6 = (–8 ± 2√7)/6. [1] Zeroes: (–4 + √7)/3 and (–4 – √7)/3 (product = (16 – 7)/9 = 1 ✓). [1]

SECTION – E (Case-Based · 4 Marks each)

81. Ans.

(i) h(t) = 0 ⇒ t² – 4t – 5 = 0 ⇒ (t – 5)(t + 1) = 0 ⇒ t = 5 or t = –1. The zeroes are 5 and –1. [1]

(ii) t = –1 is rejected because time cannot be negative. The ball lands at t = 5 seconds. [1]

(iii) h(t) = –(t² – 4t) + 5 = –(t – 2)² + 9. [1] The maximum value is 9 m, at t = 2 s (midway between the zeroes –1 and 5, i.e. t = –b/2a). [1] [2]

82. Ans.

(i) Three zeroes: –2, 1 and 3. [1]

(ii) p(x) = (x + 2)(x – 1)(x – 3) = (x² + x – 2)(x – 3) = x³ – 2x² – 5x + 6. [1]

(iii) Sum = –2 + 1 + 3 = 2 = –b/a = 2 ✓. Pair-products: (–2)(1) + (1)(3) + (3)(–2) = –2 + 3 – 6 = –5 = c/a ✓. [1] Product = (–2)(1)(3) = –6 = –d/a = –6 ✓. [1] [2]

83. Ans.

(i) Breadth = area ÷ length = (x² + 7x + 12) ÷ (x + 4). x² + 7x + 12 = (x + 3)(x + 4) ⇒ breadth = (x + 3) m. [1]

(ii) (x + 3)(x + 4) = 0 ⇒ x = –3 and –4. [1]

(iii) Length = 10 m, breadth = 9 m. [½] Perimeter = 2(10 + 9) = 38 m. [1] Area = 10 × 9 = 90 m² (= 36 + 42 + 12 ✓). [½] [2]

84. Ans.

(i) Sum = k + 6; product = 2(2k – 1) = 4k – 2. [1]

(ii) Sum = ½ × product ⇒ k + 6 = ½(4k – 2) = 2k – 1. [1]

(iii) k + 6 = 2k – 1 ⇒ k = 7. [1] p(x) = x² – 13x + 26. Sum = 13; product = 26; 13 = ½ × 26 ✓. [1] [2]

SECTION – F (Long Answer · 5 Marks each)

85. Ans.

Write g(x) = x² – x + 1 and p(x) = x⁴ + 0x³ – 3x² + 4x + 5. [½]
Step 1: x⁴ ÷ x² = x²; subtract x²(x² – x + 1) = x⁴ – x³ + x² ⇒ x³ – 4x² + 4x + 5. [1]
Step 2: x³ ÷ x² = x; subtract x(x² – x + 1) = x³ – x² + x ⇒ –3x² + 3x + 5. [1]
Step 3: –3x² ÷ x² = –3; subtract –3(x² – x + 1) = –3x² + 3x – 3 ⇒ 8. Degree 0 < 2, so stop. [1]
Quotient q(x) = x² + x – 3, remainder r(x) = 8. [½]
Verification: g(x)q(x) + r(x) = (x² – x + 1)(x² + x – 3) + 8 = x⁴ – 3x² + 4x – 3 + 8 = x⁴ – 3x² + 4x + 5 = p(x) ✓. [1]

86. Ans.

Since 3 is a zero, (x – 3) is a factor. [½] Dividing: x³ – 5x² – 2x + 24 = (x – 3)(x² – 2x – 8). [1½] x² – 2x – 8 = (x – 4)(x + 2). [1] The zeroes are 3, 4 and –2. [½]
Verification: Sum = 3 + 4 – 2 = 5 = –(–5)/1 ✓. [½] Pair-products: 12 – 8 – 6 = –2 = c/a ✓. [½] Product = 3 × 4 × (–2) = –24 = –24/1 ✓. [½]

87. Ans.

α + β = –4/k and αβ = 4/k. [1] α² + β² = (α + β)² – 2αβ = 16/k² – 8/k. [1] Set = 24: 16 – 8k = 24k² ⇒ 24k² + 8k – 16 = 0 ⇒ 3k² + k – 2 = 0. [1] (3k – 2)(k + 1) = 0 ⇒ k = 2/3 or k = –1. [1] Check: k = 2/3 gives sum –6, product 6, α² + β² = 36 – 12 = 24 ✓; k = –1 gives sum 4, product –4, 16 + 8 = 24 ✓. [1]

88. Ans.

α + β = 4/2 = 2 and αβ = 5/2. [1]
(i) α² + β² = (α + β)² – 2αβ = 4 – 5 = –1. [1]
(ii) (α – β)² = (α + β)² – 4αβ = 4 – 10 = –6. [1]
(iii) 1/α + 1/β = (α + β)/αβ = 2 ÷ (5/2) = 4/5. [1]
(iv) α³ + β³ = (α + β)³ – 3αβ(α + β) = 8 – 3 × (5/2) × 2 = 8 – 15 = –7. [1]

89. Ans.

(x – √2)(x + √2) = x² – 2 is a factor. [1] Divide 2x⁴ – 3x³ – 3x² + 6x – 2 by x² – 2: 2x² × (x² – 2) = 2x⁴ – 4x² ⇒ –3x³ + x² + 6x – 2; –3x × (x² – 2) = –3x³ + 6x ⇒ x² – 2; 1 × (x² – 2) ⇒ 0. [2] Quotient = 2x² – 3x + 1 = (2x – 1)(x – 1). [1] Its zeroes are 1/2 and 1. So all the zeroes are √2, –√2, 1/2 and 1. [1]

QUICK REVISION · KEY POINTS (CBSE / COMPETITIVE)

1. Degree = highest power of x. A polynomial of degree n has at most n zeroes; linear – 1, quadratic – at most 2, cubic – at most 3.
2. Zeroes of p(x) are the x-coordinates of the points where the graph y = p(x) meets the x-axis. Parabola: opens up if a > 0, down if a < 0; no x-intercept – no real zero; touches the axis – equal zeroes.
3. Quadratic ax² + bx + c: sum of zeroes α + β = –b/a; product αβ = c/a. Polynomial from zeroes: k[x² – (α + β)x + αβ].
4. Useful identities: α² + β² = (α + β)² – 2αβ; (α – β)² = (α + β)² – 4αβ; 1/α + 1/β = (α + β)/αβ; α³ + β³ = (α + β)³ – 3αβ(α + β).
5. Cubic ax³ + bx² + cx + d: α + β + γ = –b/a; αβ + βγ + γα = c/a; αβγ = –d/a.
6. Division algorithm: p(x) = g(x)·q(x) + r(x), where r(x) = 0 or deg r < deg g; deg q = deg p – deg g. Hence g(x)·q(x) = p(x) – r(x).
7. If a zero (or a factor) is given, divide to get the other factor, then factorise the quotient to find the remaining zeroes.
8. Shortcut: for zeroes α, β of a quadratic, the polynomial with zeroes kα, kβ has sum k(α + β) and product k²αβ; with zeroes 1/α, 1/β it has sum (α + β)/αβ and product 1/αβ.

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