Class 10 Maths Test Paper Chapter 2 Polynomials Paper 2 | CBSE Board | Detailed Solutions & Marking Scheme
VEDANT SKILL ASSESSMENT SERIES
ACADEMIC YEAR 2026-27
Class: X | Subject: Mathematics
Chapter: 2 - Polynomials (Easy Level)
Maximum Marks: 30 | Time Allowed: 1 Hour
GENERAL INSTRUCTIONS
Read all questions carefully before answering.
Misreading the question and then blaming the paper will not increase your marks.Show proper steps wherever required.
“Sir, answer toh yahi aana tha” is not an accepted mathematical method.Write neatly and clearly.
If your handwriting requires a decoder machine, checking may become an adventure.Manage your time wisely.
Spending 45 minutes on one question and calling the rest “optional” is not a strategy.If you do not know the answer, you may cry silently.
Loud crying, emotional speeches, and negotiations for hints are strictly prohibited.
SECTION A (7 MARKS)
Q1. If the graph of a polynomial intersects the x-axis at exactly three distinct points, then the number of zeroes of the polynomial is: [1]
A. 1
B. 2
C. 3
D. 0
Q2. The zeroes of the quadratic polynomial p(x) = x² − 9 are: [1]
A. 3, 3
B. 3, −3
C. −3, −3
D. 9, −9
Q3. If α and β are the zeroes of the quadratic polynomial f(x) = x² + 7x + 10, then the value of α + β is: [1]
A. 7
B. −7
C. 10
D. −10
Q4. If the product of the zeroes of a quadratic polynomial ax² + bx + c is given by c/a, then the product of the zeroes of 2x² − 5x + 8 is: [1]
A. −5/2
B. 5/2
C. 4
D. −4
Q5. A quadratic polynomial whose sum of zeroes is 0 and one zero is 4 is given by: [1]
A. x² − 16
B. x² + 16
C. x² − 4
D. x² + 4
Direction for Q6-Q7:
For the following questions, two statements are given - one labeled Assertion (A) and the other labeled Reason (R). Select the correct answer from the options below:
A. Both A and R are true, R is the correct explanation of A
B. Both A and R are true, R is not the correct explanation of A
C. A is true, R is false
D. A is false, R is true
Q6. [1]
Assertion (A): The absolute maximum number of zeroes that a linear polynomial can have is one.
Reason (R): A polynomial of degree n can have at most n zeroes.
Q7. [1]
Assertion (A): The sum of the zeroes of the polynomial x² − 4x + 4 is 4.
Reason (R): The sum of zeroes of a quadratic polynomial is given by b/a.
SECTION B (8 MARKS)
Q8. Find the zeroes of the quadratic polynomial f(x) = x² + 5x + 6 and verify the relationship between its zeroes and coefficients. [2]
Q9. Find a quadratic polynomial, the sum and product of whose zeroes are −3 and 2 respectively. [2]
Q10. If α and β are the zeroes of the quadratic polynomial p(x) = x² − 6x + 8, without finding the zeroes, find the value of: [2]
αβ + (α + β)
Q11. Draw a rough sketch of the graph of the polynomial y = x². State whether it opens upwards or downwards and mention its vertex coordinates. [2]
SECTION C (6 MARKS)
Q12. If one zero of the quadratic polynomial f(x) = 3x² + kx − 4 is 2, find the value of k. Also, find the other zero of the polynomial. [3]
Q13. If α and β are the zeroes of the quadratic polynomial f(x) = 2x² − 7x + 3, calculate the value of:
(1/α) + (1/β) [3]
SECTION D (5 MARKS)
Q14. If α and β are the zeroes of the quadratic polynomial x² − 5x + 4, find the values of the following expressions: [5]
(i) α² + β²
(ii) α²β + αβ²
SECTION E (4 MARKS)
Q15. Case Study Based Question [4]
During an activities session in a school, a sports teacher sets up a path for a running drill. The path is shaped like a parabola opening upwards. The path is modeled by the quadratic polynomial:
p(x) = x² − 2x − 3
where the x-axis represents the main reference line on the playground floor.
(i) Find the points (zeroes) where the running path meets the reference ground line (x-axis). [1]
(ii) Write down the sum and product of the zeroes for the given polynomial directly using its coefficients. [1]
(iii) Find the value of p(3) and interpret what it means regarding the position of a student running on the path at that point. [2]
— End of Paper —
VEDANT SKILL ASSESSMENT SERIES
ACADEMIC YEAR 2026-27
Class: X | Subject: Mathematics
Chapter: 2 - Polynomials (Easy Level)
ANSWER KEY WITH MARKING SCHEME
SECTION A (7 MARKS)
Q1. Correct Answer: C. 3 [1]
Marking Scheme:
Correct option selected = 1 mark
Q2. Correct Answer: B. 3, −3 [1]
Explanation:
x² − 9 = 0
(x − 3)(x + 3) = 0
Zeroes = 3, −3
Marking Scheme:
Correct option selected = 1 mark
Q3. Correct Answer: B. −7 [1]
For quadratic polynomial ax² + bx + c,
Sum of zeroes = −b/a
Here, a = 1 and b = 7
α + β = −7/1 = −7
Marking Scheme:
Correct option selected = 1 mark
Q4. Correct Answer: C. 4 [1]
Product of zeroes = c/a
= 8/2 = 4
Marking Scheme:
Correct option selected = 1 mark
Q5. Correct Answer: A. x² − 16 [1]
Sum of zeroes = 0
One zero = 4
Other zero = −4
Polynomial:
(x − 4)(x + 4)
= x² − 16
Marking Scheme:
Correct option selected = 1 mark
Q6. Correct Answer: A [1]
Assertion (A): True
A linear polynomial has degree 1, so it can have at most one zero.
Reason (R): True
A polynomial of degree n can have at most n zeroes.
R correctly explains A.
Marking Scheme:
Correct option selected = 1 mark
Q7. Correct Answer: C [1]
Assertion (A): True
For x² − 4x + 4,
Sum of zeroes = −b/a
= −(−4)/1
= 4
Reason (R): False
Correct formula:
Sum of zeroes = −b/a
(not b/a)
Marking Scheme:
Correct option selected = 1 mark
SECTION B (8 MARKS)
Q8. Find the zeroes of f(x) = x² + 5x + 6 and verify relationship between zeroes and coefficients. [2]
x² + 5x + 6 = 0
(x + 2)(x + 3) = 0
Zeroes = −2, −3
Verification:
α + β = (−2) + (−3) = −5
Using formula:
−b/a = −5/1 = −5 ✓
αβ = (−2)(−3) = 6
Using formula:
c/a = 6/1 = 6 ✓
Marking Scheme:
Finding correct zeroes = 1 mark
Correct verification = 1 mark
Q9. Find a quadratic polynomial whose sum and product of zeroes are −3 and 2 respectively. [2]
Required polynomial:
x² − (sum)x + (product)
= x² − (−3)x + 2
= x² + 3x + 2
Marking Scheme:
Correct formula used = 1 mark
Correct polynomial = 1 mark
Q10. Find αβ + (α + β) for p(x) = x² − 6x + 8. [2]
For x² − 6x + 8,
α + β = −(−6)/1 = 6
αβ = 8/1 = 8
Required value:
αβ + (α + β)
= 8 + 6
= 14
Marking Scheme:
Finding α + β and αβ = 1 mark
Final answer = 1 mark
Q11. Sketch y = x² and state nature and vertex. [2]
Graph shape: Parabola
Opens: Upwards
Vertex: (0, 0)
Marking Scheme:
Correct rough sketch = 1 mark
Nature and vertex stated correctly = 1 mark
SECTION C (6 MARKS)
Q12. One zero of 3x² + kx − 4 is 2. Find k and other zero. [3]
Given polynomial:
3x² + kx − 4
Since x = 2 is a zero,
3(2)² + k(2) − 4 = 0
12 + 2k − 4 = 0
8 + 2k = 0
2k = −8
k = −4
Polynomial becomes:
3x² − 4x − 4
Product of zeroes = c/a
= −4/3
Let other zero = β
2β = −4/3
β = −2/3
Marking Scheme:
Substitution of x = 2 = 1 mark
Finding k = 1 mark
Finding other zero = 1 mark
Q13. Find (1/α) + (1/β) for 2x² − 7x + 3. [3]
For 2x² − 7x + 3,
α + β = −(−7)/2
= 7/2
αβ = 3/2
Now,
(1/α) + (1/β)
= (α + β)/αβ
= (7/2)/(3/2)
= 7/3
Answer = 7/3
Marking Scheme:
Finding α + β and αβ = 1 mark
Correct formula used = 1 mark
Final answer = 1 mark
SECTION D (5 MARKS)
Q14. If α and β are zeroes of x² − 5x + 4, find: [5]
Given:
α + β = 5
αβ = 4
(i) α² + β²
Formula:
α² + β² = (α + β)² − 2αβ
= 5² − 2(4)
= 25 − 8
= 17
(ii) α²β + αβ²
= αβ(α + β)
= 4(5)
= 20
Marking Scheme:
Finding α + β and αβ = 1 mark
Formula for part (i) = 1 mark
Correct answer for part (i) = 1 mark
Formula for part (ii) = 1 mark
Correct answer for part (ii) = 1 mark
SECTION E (4 MARKS)
Q15. Case Study Based Question [4]
Given:
p(x) = x² − 2x − 3
(i) Find zeroes [1]
x² − 2x − 3 = 0
(x − 3)(x + 1) = 0
Zeroes = 3, −1
(ii) Sum and product of zeroes [1]
Sum of zeroes:
−b/a = −(−2)/1 = 2
Product of zeroes:
c/a = −3
(iii) Find p(3) and interpret [2]
p(3) = (3)² − 2(3) − 3
= 9 − 6 − 3
= 0
Interpretation:
At x = 3, the student is exactly on the reference ground line (x-axis), because the polynomial value is 0.
Marking Scheme:
Correct value of p(3) = 1 mark
Correct interpretation = 1 mark
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