Polynomials Test Class 10 Maths — 40 Marks
Pandey Sir’s BEST Classes
PERIODIC TEST — Polynomials (2026-27) | CLASS 10 | MATHEMATICS
Max Marks: 40 · Time: 1 hour 30 minutes (Time left: 90:00)
| Section | Marks | Questions | Time | Strategy |
|---|---|---|---|---|
| A | 12 | 12 MCQ | 15 min | Quick recall — about 1 minute each |
| B | 8 | 4 Short | 20 min | Write the condition before you solve |
| C | 12 | 4 Long | 30 min | Full steps, then verify the relation |
| D | 8 | 2 Case | 20 min | Read the paragraph, label the parts |
| Review | — | — | 5 min | Cross-check totals and answers |
General Instructions: All questions are compulsory. Section A has 12 MCQs of 1 mark each. Internal choice is given in Sections B, C and D — attempt any ONE of the two options. Write the formula or condition used, then substitute. Draw a rough graph only where asked.
What is this paper? A printable 40-mark Class 10 Maths chapter test on Polynomials with a full per-question marking scheme, internal choice and practice/test modes. It is built to the current CBSE 2026-27 syllabus and is board-pattern practice, not a prediction of any examination paper.
p(x) = ax2 + bx + c
α + β = −ba
αβ = ca
x2 − (α + β)x + αβ
Here α and β are the zeros of the quadratic ax2 + bx + c (a ≠ 0).
Section A — Multiple Choice (12 × 1 = 12 marks)
Choose the single best option.
The zero of the linear polynomial p(x) = 3x − 6 is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Put p(x) = 0
3x − 6 = 0
3x = 6
x = 2
3x = 6
x = 2
Answer: (a) 2
If α and β are the zeros of x2 − 5x + 6, then α + β is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Sum of zeros
Here a = 1, b = −5, c = 6
α + β = −b/a = −(−5)/1 = 5
α + β = −b/a = −(−5)/1 = 5
Answer: (a) 5
The product of the zeros of 2x2 − 7x + 3 is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Product of zeros
a = 2, b = −7, c = 3
αβ = c/a = 3/2
αβ = c/a = 3/2
Answer: (b) 3/2
The graph of a polynomial cuts the x-axis at exactly two points. The number of zeros of the polynomial is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Zeros are the x-intercepts
A zero is a value of x where p(x) = 0, i.e. where the graph meets the x-axis.
Two intersection points → two zeros.
Two intersection points → two zeros.
Answer: (c) 2
A quadratic polynomial whose zeros are 3 and −2 is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Use sum and product
Sum = 3 + (−2) = 1
Product = 3 × (−2) = −6
p(x) = x2 − (sum)x + product
= x2 − x − 6
Product = 3 × (−2) = −6
p(x) = x2 − (sum)x + product
= x2 − x − 6
Answer: (a) x2 − x − 6
If one zero of x2 − 4x + k is 1, then k is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Substitute x = 1
p(1) = 0
1 − 4 + k = 0
k = 3
1 − 4 + k = 0
k = 3
Answer: (a) 3
The zeros of the polynomial x2 − 9 are
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Difference of squares
x2 − 9 = 0
x2 = 9
x = 3 or x = −3
x2 = 9
x = 3 or x = −3
Answer: (b) 3, −3
If α and β are the zeros of x2 + 5x + 6, then αβ is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Product of zeros
a = 1, b = 5, c = 6
αβ = c/a = 6/1 = 6
αβ = c/a = 6/1 = 6
Answer: (a) 6
Which of the following is a linear polynomial?
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Degree of a linear polynomial is 1
(a) degree 2
(b) degree 1 → linear
(c) degree 3
(d) degree 2
(b) degree 1 → linear
(c) degree 3
(d) degree 2
Answer: (b) 5x − 2
If the sum of the zeros of x2 − (k + 2)x + 5 is 7, then k is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Sum of zeros
Sum = −b/a = (k + 2)
(k + 2) = 7
k = 5
(k + 2) = 7
k = 5
Answer: (a) 5
If the graph of a polynomial meets the x-axis at 3 points, then its degree is at least
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Number of zeros ≤ degree
A degree-n polynomial has at most n zeros.
3 zeros need degree ≥ 3.
3 zeros need degree ≥ 3.
Answer: (c) 3
If α and β are the zeros of x2 − 3x + 2, then 1α + 1β is
Marking Scheme: Correct option: 1 (no partial marks).
Total: 1 Mark
Total: 1 Mark
Combine the fractions
α + β = 3, αβ = 2
1α + 1β = α + βαβ
= 3/2
1α + 1β = α + βαβ
= 3/2
Answer: (a) 3/2
Section B — Short Answer (4 × 2 = 8 marks)
Find the zeros of the quadratic polynomial x2 − 2x − 8.
Marking Scheme: Factorising (split the middle term) 1 mark · Zeros stated 1 mark.
Total: 2 Marks
Total: 2 Marks
Split the middle term
x2 − 2x − 8
= x2 − 4x + 2x − 8
= x(x − 4) + 2(x − 4)
= (x − 4)(x + 2)
= x2 − 4x + 2x − 8
= x(x − 4) + 2(x − 4)
= (x − 4)(x + 2)
Put each factor = 0
x − 4 = 0 → x = 4
x + 2 = 0 → x = −2
x + 2 = 0 → x = −2
Answer: x = 4 and x = −2
Find a quadratic polynomial whose sum and product of zeros are 4 and −3 respectively.
Marking Scheme: Correct form of the polynomial 1 mark · Final polynomial 1 mark.
Total: 2 Marks
Total: 2 Marks
Use the standard form
p(x) = x2 − (sum)x + product
= x2 − 4x + (−3)
= x2 − 4x + (−3)
Simplify
p(x) = x2 − 4x − 3
Answer: x2 − 4x − 3
If α and β are the zeros of x2 − 6x + 8, find α2 + β2.
Marking Scheme: Sum and product of zeros 1 mark · Using the identity and answer 1 mark.
Total: 2 Marks
Total: 2 Marks
Sum and product
α + β = 6
αβ = 8
αβ = 8
Identity
α2 + β2 = (α + β)2 − 2αβ
= 62 − 2 × 8
= 36 − 16
= 20
= 62 − 2 × 8
= 36 − 16
= 20
Answer: 20
Answer either Option 1 or Option 2.
Option 1 — Sum and product
Marking Scheme: Correct a, b, c identified 1 mark · Both values 1 mark.
Total: 2 Marks
Total: 2 Marks
Identify the coefficients
For 2x2 − 5x + 3: a = 2, b = −5, c = 3
Apply the relations
α + β = −b/a = 5/2
αβ = c/a = 3/2
αβ = c/a = 3/2
Answer: α + β = 5/2, αβ = 3/2
— OR —
Option 2
If the product of the zeros of x2 − 3x + k is 2, find k and the zeros.
Marking Scheme: Value of k 1 mark · Both zeros 1 mark.
Total: 2 Marks
Total: 2 Marks
Product of zeros
Product = c/a = k/1 = k
k = 2
k = 2
Factorise
x2 − 3x + 2 = (x − 1)(x − 2)
Zeros: 1 and 2
Zeros: 1 and 2
Answer: k = 2; zeros 1 and 2
[Attempt any ONE of the two options of Q16]
Section C — Long Answer (4 × 3 = 12 marks)
Find the zeros of 6x2 − 3 − 7x and verify the relation between the zeros and the coefficients.
Marking Scheme: Correct standard form and factorisation 1 mark · Zeros 1 mark · Verification of sum and product 1 mark.
Total: 3 Marks
Total: 3 Marks
Write in standard form
6x2 − 7x − 3 = 0
Splitting the middle term: 6x2 − 9x + 2x − 3
= 3x(2x − 3) + 1(2x − 3)
= (2x − 3)(3x + 1)
Splitting the middle term: 6x2 − 9x + 2x − 3
= 3x(2x − 3) + 1(2x − 3)
= (2x − 3)(3x + 1)
Zeros
2x − 3 = 0 → x = 3/2
3x + 1 = 0 → x = −1/3
3x + 1 = 0 → x = −1/3
Verify
α = 3/2, β = −1/3
α + β = 3/2 − 1/3 = 7/6
−b/a = −(−7)/6 = 7/6 ✓
αβ = (3/2)×(−1/3) = −1/2
c/a = −3/6 = −1/2 ✓
α + β = 3/2 − 1/3 = 7/6
−b/a = −(−7)/6 = 7/6 ✓
αβ = (3/2)×(−1/3) = −1/2
c/a = −3/6 = −1/2 ✓
Answer: Zeros are 3/2 and −1/3; sum = 7/6, product = −1/2 (verified)
If α and β are the zeros of x2 − 5x + 6, find a quadratic polynomial whose zeros are α + 1 and β + 1.
Marking Scheme: Sum and product of the given zeros 1 mark · New sum and product 1 mark · New polynomial 1 mark.
Total: 3 Marks
Total: 3 Marks
Given zeros
α + β = 5
αβ = 6
αβ = 6
New sum and product
New sum = (α + 1) + (β + 1) = (α + β) + 2 = 7
New product = (α + 1)(β + 1) = αβ + α + β + 1 = 6 + 5 + 1 = 12
New product = (α + 1)(β + 1) = αβ + α + β + 1 = 6 + 5 + 1 = 12
New polynomial
x2 − (new sum)x + new product
= x2 − 7x + 12
= x2 − 7x + 12
Answer: x2 − 7x + 12
Answer either Option 1 or Option 2.
Option 1 — One zero is three times the other
Marking Scheme: Letting the zeros be α and 3α and using the sum 1 mark · Finding α 1 mark · Finding k 1 mark.
Total: 3 Marks
Total: 3 Marks
Let the zeros be α and 3α
For 2x2 − 8x + k: sum = 8/2 = 4
α + 3α = 4
4α = 4 → α = 1
α + 3α = 4
4α = 4 → α = 1
Use the product
Product = α × 3α = 3α2 = 3
c/a = k/2
k/2 = 3 → k = 6
c/a = k/2
k/2 = 3 → k = 6
Check
2x2 − 8x + 6 = 2(x2 − 4x + 3) = 2(x − 1)(x − 3)
Zeros 1 and 3 ✓
Zeros 1 and 3 ✓
Answer: k = 6
— OR —
Option 2
If α and β are the zeros of x2 + 7x + 12, find α2 + β2 − αβ.
Marking Scheme: Sum and product 1 mark · α2 + β2 1 mark · Final value 1 mark.
Total: 3 Marks
Total: 3 Marks
Sum and product
α + β = −7
αβ = 12
αβ = 12
Find α2 + β2
(α + β)2 − 2αβ = (−7)2 − 2(12)
= 49 − 24 = 25
= 49 − 24 = 25
Subtract αβ
25 − 12 = 13
Answer: 13
[Attempt any ONE of the two options of Q19]
Answer either Option 1 or Option 2.
Option 1 — Repeated zero
Marking Scheme: Factorising the polynomial 1 mark · Stating the zero 1 mark · Verification 1 mark.
Total: 3 Marks
Total: 3 Marks
Factorise
4x2 − 4x + 1 = (2x − 1)2
Zero
2x − 1 = 0 → x = 1/2
Both zeros are 1/2 (a repeated zero).
Both zeros are 1/2 (a repeated zero).
Verify
Sum = 1/2 + 1/2 = 1
−b/a = 4/4 = 1 ✓
Product = 1/2 × 1/2 = 1/4
c/a = 1/4 ✓
−b/a = 4/4 = 1 ✓
Product = 1/2 × 1/2 = 1/4
c/a = 1/4 ✓
Answer: Zero 1/2 (repeated); sum = 1, product = 1/4
— OR —
Option 2
If α and β are the zeros of x2 + 4x + 3, find α2β + αβ2.
Marking Scheme: Sum and product 1 mark · Factorising the expression 1 mark · Value 1 mark.
Total: 3 Marks
Total: 3 Marks
Sum and product
α + β = −4
αβ = 3
αβ = 3
Factorise the expression
α2β + αβ2 = αβ(α + β)
Substitute
= 3 × (−4) = −12
Answer: −12
[Attempt any ONE of the two options of Q20]
Section D — Case Study (2 × 4 = 8 marks)
A ball is thrown upward. Its height above the ground, in metres, t seconds after it is thrown is given by h(t) = −t2 + 6t + 7. The ball hits the ground when h(t) = 0.
Marking Scheme: Part (a) 1 mark · Part (b) 1.5 marks · Part (c) 1.5 marks.
Total: 4 Marks
Total: 4 Marks
(a) Factorise h(t)
h(t) = −(t2 − 6t − 7)
t2 − 6t − 7 = (t − 7)(t + 1)
So h(t) = −(t − 7)(t + 1)
t2 − 6t − 7 = (t − 7)(t + 1)
So h(t) = −(t − 7)(t + 1)
(b) Zeros of h(t)
t − 7 = 0 → t = 7
t + 1 = 0 → t = −1
t + 1 = 0 → t = −1
(c) Time to hit the ground
Time cannot be negative, so t = 7 seconds.
Answer: (a) −(t − 7)(t + 1) (b) t = 7, t = −1 (c) 7 seconds
Answer either Option 1 or Option 2.
Option 1 — Break-even sales
A small shop’s daily profit, in hundreds of rupees, after selling x items is modelled by p(x) = x2 − 5x + 6. The shop makes no profit (break-even) when p(x) = 0.
Marking Scheme: Part (a) 1 mark · Part (b) 1.5 marks · Part (c) 1.5 marks.
Total: 4 Marks
Total: 4 Marks
(a) Factorise p(x)
x2 − 5x + 6 = (x − 2)(x − 3)
(b) Break-even values
x − 2 = 0 → x = 2
x − 3 = 0 → x = 3
The shop breaks even at 2 items and at 3 items.
x − 3 = 0 → x = 3
The shop breaks even at 2 items and at 3 items.
(c) Profit at x = 6
p(6) = 36 − 30 + 6 = 12
Profit = ₹1200
Profit = ₹1200
Answer: (a) (x − 2)(x − 3) (b) x = 2 and x = 3 (c) ₹1200
— OR —
Option 2
The area of a rectangular hall is (x2 + 9x + 20) square metres. Its length and breadth are the two linear factors of this polynomial.
Marking Scheme: Part (a) 1 mark · Part (b) 1.5 marks · Part (c) 1.5 marks.
Total: 4 Marks
Total: 4 Marks
(a) Factorise
x2 + 9x + 20 = (x + 4)(x + 5)
(b) Zeros of the polynomial
x + 4 = 0 → x = −4
x + 5 = 0 → x = −5
x + 5 = 0 → x = −5
(c) Dimensions and perimeter when x = 1
Length = 1 + 5 = 6 m, breadth = 1 + 4 = 5 m
Area = 6 × 5 = 30 m2 (also 1 + 9 + 20 = 30 ✓)
Perimeter = 2(6 + 5) = 22 m
Area = 6 × 5 = 30 m2 (also 1 + 9 + 20 = 30 ✓)
Perimeter = 2(6 + 5) = 22 m
Answer: (a) (x + 4)(x + 5) (b) x = −4 and x = −5 (c) 6 m × 5 m, perimeter 22 m
[Attempt any ONE of the two options of Q22]
Self-assessment for Sections B, C and D (28 marks)
The MCQ section is scored for you. For the written questions, tick each one you got right out of the marks shown, and add them up honestly.
Written total: 0 / 28
Keep going with these
Written and checked by Pandey Sir · Pandey Sir’s BEST Classes, Dammaiguda, Hyderabad. Every answer on this paper is worked out and checked before publishing.
This is board-pattern practice for Polynomials, built to the current CBSE Class 10 Maths syllabus 2026-27. It is not a prediction of any examination paper.
Curriculum scope checked against cbseacademic.nic.in. Spotted a mistake? Tell us here.