Polynomials Chapter Test — Class 10 Maths Practice Test (with Answers)

Gurpreet Singh · Every answer on this page is computed and verified before publishing. About the team

A quick practice test on Polynomials to check how ready you are for the chapter. Attempt it fully on paper first — give yourself about 45 minutes — then reveal the answer key at the bottom to mark yourself. Total: 20 marks. Need a refresher first? Read the Polynomials notes.

Section A — Objective (5 × 1 mark)

1. If one zero of the quadratic polynomial x² − 7x + k is 3, then the value of k is:
   (a) 10   (b) 12   (c) −12   (d) 4
2. The sum and the product of the zeroes of 5x² − 11x + 2 are respectively:
   (a) −11/5, 2/5   (b) 11/5, 2/5   (c) 11/5, −2/5   (d) 2/5, 11/5
3. A quadratic polynomial whose zeroes are 3 + √2 and 3 − √2 is:
   (a) x² − 6x + 7   (b) x² + 6x + 7   (c) x² − 6x − 7   (d) x² − 7x + 6
4. If the zeroes of the quadratic polynomial x² + (a + 1)x + b are 2 and −3, then:
   (a) a = −7, b = −1   (b) a = 5, b = −1   (c) a = 2, b = −6   (d) a = 0, b = −6
5. Assertion (A): The polynomial q(x) = x² + 4 has no real zeroes.
   Reason (R): For every real number x, x² ≥ 0, so x² + 4 ≥ 4 > 0.
   (a) Both A and R are true and R is the correct explanation of A.   (b) Both A and R are true but R is not the correct explanation of A.   (c) A is true, R is false.   (d) A is false, R is true.

Section B — Short Answer (5 × 2 marks)

6. Find the zeroes of the quadratic polynomial p(x) = 6x² − 5x − 6, and verify the relationship between the zeroes and the coefficients.
7. Find a quadratic polynomial the sum of whose zeroes is −3/2 and the product of whose zeroes is −5.
8. If α and β are the zeroes of p(x) = x² − 8x + 5, find the value of 1/α + 1/β.
9. Without drawing the graph, state how many times the graph of p(x) = x² − 10x + 25 meets the x-axis. Give a reason for your answer.
10. If one zero of the polynomial p(x) = x² + (m − 2)x − 49 is the negative of the other, find m and hence the zeroes.

Section C — Long Answer (1 × 5 marks)

11. The daily profit (in hundreds of rupees) of a small stationery stall on selling x dozen notebooks is modelled by P(x) = −x² + 11x − 24, where x ≥ 0.
    (i) Find the values of x for which the profit is zero. (2)
    (ii) Verify the relationship between the zeroes and the coefficients of P(x). (2)
    (iii) Write a quadratic polynomial whose zeroes are each 2 more than the zeroes of P(x). (1)

Show the Full Answer Key

Section A

1. (b) 12. If 3 is a zero, p(3) = 0 ⇒ 9 − 21 + k = 0 ⇒ k = 12. (The zeroes are then 3 and 4.)

2. (b) 11/5 and 2/5. Here a = 5, b = −11, c = 2. Sum = −b/a = 11/5; product = c/a = 2/5.

3. (a) x² − 6x + 7. Sum = (3 + √2) + (3 − √2) = 6; product = 3² − (√2)² = 9 − 2 = 7. So the polynomial is x² − (sum)x + product = x² − 6x + 7.

4. (d) a = 0, b = −6. Sum of zeroes = 2 + (−3) = −1 = −(a + 1) ⇒ a + 1 = 1 ⇒ a = 0. Product = 2 × (−3) = −6 = b. Check: x² + x − 6 is 0 at x = 2 and at x = −3.

5. (a). x² is never negative, so x² + 4 is at least 4 for every real x and can never equal 0 — hence no real zeroes. R is true and is exactly the reason A is true.

Section B

6. Split the middle term: 6x² − 5x − 6 = 6x² − 9x + 4x − 6 = 3x(2x − 3) + 2(2x − 3) = (2x − 3)(3x + 2).
Zeroes: x = 3/2 and x = −2/3.
Verification: sum = 3/2 − 2/3 = 5/6, and −b/a = −(−5)/6 = 5/6 ✓. Product = (3/2)(−2/3) = −1, and c/a = −6/6 = −1 ✓.

7. A polynomial with given sum S and product P of zeroes is k[x² − Sx + P]. Here S = −3/2, P = −5, so x² + (3/2)x − 5. Taking k = 2 to clear the fraction: 2x² + 3x − 10. (Any non-zero multiple is equally correct.) Check: −b/a = −3/2 ✓, c/a = −10/2 = −5 ✓.

8. α + β = 8 and αβ = 5. So 1/α + 1/β = (α + β)/(αβ) = 8/5.

9. Exactly once. x² − 10x + 25 = (x − 5)², so both zeroes are 5 — they coincide. A repeated zero means the graph touches the x-axis at (5, 0) instead of crossing it, so it meets the axis at one point only.

10. Let the zeroes be α and −α. Their sum is 0, and sum = −(m − 2), so −(m − 2) = 0 ⇒ m = 2. The polynomial becomes x² − 49 = (x − 7)(x + 7), so the zeroes are 7 and −7. (Consistent with product = −49.)

Section C

11. (i) Profit is zero when P(x) = 0: −x² + 11x − 24 = 0 ⇒ x² − 11x + 24 = 0 ⇒ (x − 3)(x − 8) = 0 ⇒ x = 3 or x = 8 dozen notebooks. Both satisfy x ≥ 0, so both are admissible.

(ii) For P(x) = −x² + 11x − 24: a = −1, b = 11, c = −24.
Sum of zeroes = 3 + 8 = 11, and −b/a = −11/(−1) = 11 ✓.
Product of zeroes = 3 × 8 = 24, and c/a = −24/(−1) = 24 ✓.

(iii) The new zeroes are 3 + 2 = 5 and 8 + 2 = 10. Sum = 15, product = 50, so the polynomial is x² − 15x + 50.

Score it honestly, then go back to only the questions you got wrong — fixing two mistakes properly today beats re-reading the whole chapter tomorrow.