Class 10 Mathematics: Polynomials important questions

Polynomials is a chapter of Class 10 Mathematics for the CBSE. These 11 questions are marked Very Important, Important or Practice, and each answer has been checked by our team. Use them to revise, not as a prediction of what will come in the exam.

1. Practice · 1 marks

   1. If α and β are the zeroes of f(x) = px² − 2x + 3p and α + β = αβ, then p is: (a) −2/3 (b) 2/3 (c) 1/3 (d) −1/3

   (b) 2/3 α + β = 2/p and αβ = 3p/p = 3. So 2/p = 3 and p = 2/3.

2. Important · 3 marks

   2. If the zeroes of x² + px + q are double the zeroes of 2x² − 5x − 3, find p and q.

   For 2x² − 5x − 3: α + β = 5/2, αβ = −3/2. The zeroes of x² + px + q are 2α and 2β: 2(α + β) = −p gives p = −5, and 4αβ = q gives q = −6.

3. Important · 3 marks

   3. If α and β are the zeroes of 5x² + 5x + 1, find (1) α² + β² and (2) α⁻¹ + β⁻¹.

   α + β = −1, αβ = 1/5. (1) α² + β² = (α + β)² − 2αβ = 1 − 2/5 = 3/5. (2) 1/α + 1/β = (α + β)/αβ = −1 ÷ (1/5) = −5.

4. Practice · 1 marks

   4. The graph of a quadratic polynomial p(x) passes through the points (−6, 0), (0, −30), (4, −20) and (6, 0). The zeroes of the polynomial are: (A) −6, 0 (B) 4, 6 (C) −30, −20 (D) −6, 6

   (D) −6, 6 The zeroes are where the graph meets the x-axis (y = 0): (−6, 0) and (6, 0).

5. Practice · 1 marks

   5. A quadratic polynomial having zeroes −√(5/2) and √(5/2) is: (A) x² − 5√2x + 1 (B) 8x² − 20 (C) 15x² − 6 (D) x² − 2√5x − 1

   (B) 8x² − 20 Sum of zeroes = 0 and product = −5/2, so the polynomial is k(x² − 5/2); with k = 8 it is 8x² − 20.

6. Important · 3 marks

   6. If α and β are the zeroes of 6x² − 5x + 1, form a quadratic polynomial whose zeroes are α² and β².

   α + β = 5/6, αβ = 1/6. α² + β² = (α + β)² − 2αβ = 25/36 − 1/3 = 13/36, and α²β² = 1/36. Polynomial: x² − (13/36)x + 1/36, or 36x² − 13x + 1.

7. Practice · 1 marks

   7. If the zeroes of ax² + bx + c (a, c ≠ 0) are equal, then: (A) c and b have opposite signs (B) c and a have opposite signs (C) c and b have the same sign (D) c and a have the same sign

   (D) Equal zeroes need b² = 4ac, so ac = b²/4 > 0, meaning a and c have the same sign.

8. Important · 3 marks

   8. Find the zeroes of 2x² − (1 + 2√2)x + √2 and verify the relationship between the zeroes and the coefficients.

   2x² − x − 2√2x + √2 = x(2x − 1) − √2(2x − 1) = (2x − 1)(x − √2), so the zeroes are 1/2 and √2. Sum = 1/2 + √2 = (1 + 2√2)/2 = −b/a. ✓ Product = √2/2 = c/a. ✓

9. Practice · 1 marks

   9. If f(x) = px² + qx + r, p ≠ 0, and p + r = q, then one of the zeroes of f(x) is: (A) q/p (B) r/p (C) −r/p (D) −q/p

   (C) −r/p Substitute q = p + r: f(x) = px² + (p + r)x + r = (x + 1)(px + r), so the zeroes are −1 and −r/p.

10. Practice · 1 marks

    10. Assertion (A): If the zeroes of (2k − 1)x² + 4x − 3 are reciprocals of each other, then k = −1. Reason (R): If a = c, the zeroes of ax² + bx + c (a ≠ 0) are reciprocals of each other. A. Both A and R are true, and R is the correct explanation of A. B. Both A and R are true, and R is not the correct explanation of A. C. A is true but R is false. D. A is false but R is true.

    A. Both are true, and R explains A: reciprocal zeroes need product c/a = 1, so 2k − 1 = −3 and k = −1.

11. Important · 3 marks

    11. If α and β are the zeroes of 2x² − 8x + 5, find (α + 1/β) × (β + 1/α).

    α + β = 4 and αβ = 5/2. (α + 1/β)(β + 1/α) = [(αβ + 1)/β] × [(αβ + 1)/α] = (αβ + 1)²/(αβ) = (5/2 + 1)² ÷ (5/2) = (49/4) × (2/5) = 49/10.