CBSE Class 10 · Chapter 2 · Algebra · 2026-27

Polynomials Class 10: MCQ and case study questions

15 multiple-choice questions and 2 case-based questions on the current (rationalised) syllabus, in the style of the board paper's Sections A and E. Try each one first; the answer, a one-line reason and a worked solution open on a tap.

15 MCQs (1 mark each)
2 case studies (4 marks each)
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Multiple-choice questions

Tap an option to check it; the answer and a worked solution open. Section A of the board paper has 18 MCQs of 1 mark each, spread over all chapters.

Q1

·1 mark·Multiple choiceZeroes of a polynomial

The zeroes of the polynomial $x^{2} - 9 x + 14$ are

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Answer: (b)

2

and

7

Why: Split −9x as −2x − 7x: x² − 9x + 14 = (x − 2)(x − 7).

$x^{2} - 9 x + 14 = (x - 2) (x - 7)$ , which is $0$ when $x = 2$ or $x = 7$ .

Q2

·1 mark·Multiple choiceRelationship between zeroes and coefficients

The sum and the product of the zeroes of $4 x^{2} + 7 x - 2$ are respectively

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Answer: (c)

- \frac{7}{4}

and

- \frac{1}{2}

Why: Sum = −b/a and product = c/a with a = 4, b = 7, c = −2.

Sum $= - \frac{b}{a} = - \frac{7}{4}$ ; product $= \frac{c}{a} = - \frac{2}{4} = - \frac{1}{2}$ . (The zeroes are $\frac{1}{4}$ and $- 2$ .)

Q3

·1 mark·Multiple choiceRelationship between zeroes and coefficients

A quadratic polynomial whose zeroes are $3 + \sqrt{2}$ and $3 - \sqrt{2}$ is

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Answer: (a)

x^{2} - 6 x + 7

Why: Sum of zeroes = 6, product = 9 − 2 = 7, so x² − 6x + 7.

Sum $= 6$ , product $= (3 + \sqrt{2}) (3 - \sqrt{2}) = 9 - 2 = 7$ . Polynomial: $x^{2} - (sum) x + product = x^{2} - 6 x + 7$ .

Q4

·1 mark·Multiple choiceZeroes of a polynomial

If $- 4$ is a zero of $x^{2} + k x - 20$ , then $k$ equals

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Answer: (d)

- 1

Why: Substitute x = −4 and set the value to 0.

$(- 4)^{2} + k (- 4) - 20 = 0 \Rightarrow 16 - 4 k - 20 = 0 \Rightarrow k = - 1$ . (The other zero is $5$ .)

Q5

·1 mark·Multiple choiceGeometrical meaning of zeroes

The graph of a quadratic polynomial $y = p (x)$ touches the $x$ -axis at exactly one point and does not cross it. The number of distinct zeroes of $p (x)$ is

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Answer: (b)

1

Why: The zeroes are the x-coordinates of the points where the graph meets the x-axis.

Each point where the graph meets the $x$ -axis gives one zero. One point of contact gives exactly one zero (e.g. $y = (x - 2)^{2}$ has only the zero $2$ ).

Q6

·1 mark·Multiple choiceRelationship between zeroes and coefficients

If $α$ and $β$ are the zeroes of $x^{2} - 6 x + 5$ , then $α^{2} β + α β^{2}$ equals

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Answer: (c)

30

Why: α²β + αβ² = αβ(α + β) = 5 × 6.

$α + β = 6$ , $α β = 5$ . So $α^{2} β + α β^{2} = α β (α + β) = 5 \times 6 = 30$ .

Q7

·1 mark·Multiple choiceRelationship between zeroes and coefficients

The zeroes of $x^{2} - 12 x + k$ are in the ratio $1 : 5$ . The value of $k$ is

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Answer: (a)

20

Why: Let the zeroes be a and 5a: 6a = 12 gives a = 2, so k = 2 × 10.

Zeroes $a$ and $5 a$ : $a + 5 a = 12 \Rightarrow a = 2$ . Zeroes $2$ and $10$ ; $k =$ product $= 20$ .

Q8

·1 mark·Multiple choiceRelationship between zeroes and coefficients

If one zero of $5 x^{2} + 13 x + k$ is the reciprocal of the other, then $k$ equals

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Answer: (d)

5

Why: Zeroes α and 1/α have product 1, and product = k/5.

Product of zeroes $= α \cdot \frac{1}{α} = 1 = \frac{k}{5}$ , so $k = 5$ . (Then $5 x^{2} + 13 x + 5$ has real zeroes since $169 - 100 > 0$ .)

Q9

·1 mark·Multiple choiceZeroes of a polynomial

The zeroes of $x^{2} - 3$ are

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Answer: (c)

\sqrt{3}

and

- \sqrt{3}

Why: x² = 3 gives x = ±√3.

$x^{2} - 3 = (x - \sqrt{3}) (x + \sqrt{3})$ , so the zeroes are $\sqrt{3}$ and $- \sqrt{3}$ .

Q10

·1 mark·Multiple choiceRelationship between zeroes and coefficients

For the quadratic polynomial $a x^{2} + b x + c$ ( $a \neq 0$ ), if $c = 0$ then

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Answer: (b) one zero is

0

Why: c = 0 makes the product of zeroes c/a = 0, so one zero is 0.

With $c = 0$ : $a x^{2} + b x = x (a x + b)$ , whose zeroes are $0$ and $- \frac{b}{a}$ . So one zero is $0$ (both only if also $b = 0$ ).

Q11

·1 mark·Multiple choiceGeometrical meaning of zeroes

For some $b > 0$ , the graph of $y = x^{2} + b x + 9$ touches the $x$ -axis at exactly one point. That point is

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Answer: (a)

(- 3, 0)

Why: Touching at one point means one repeated zero: x² + bx + 9 = (x + 3)² with b = 6.

One point of contact means the polynomial is a perfect square: $x^{2} + b x + 9 = (x + 3)^{2}$ or $(x - 3)^{2}$ . With $b > 0$ , it is $(x + 3)^{2}$ ( $b = 6$ ), whose only zero is $- 3$ . The graph touches the $x$ -axis at $(- 3, 0)$ . ( $(0, 9)$ is where it meets the $y$ -axis.)

Q12

·1 mark·Multiple choiceRelationship between zeroes and coefficients

A quadratic polynomial whose zeroes have sum $- 4$ and product $- 21$ is

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Answer: (d)

x^{2} + 4 x - 21

Why: Use x² − (sum)x + (product).

$x^{2} - (- 4) x + (- 21) = x^{2} + 4 x - 21$ . Check: $(x + 7) (x - 3)$ has zeroes $- 7, 3$ with sum $- 4$ , product $- 21$ .

Q13

·1 mark·Multiple choiceRelationship between zeroes and coefficients

If $α$ and $β$ are the zeroes of $2 x^{2} - 7 x + 3$ , then $(α - β)^{2}$ equals

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Answer: (a)

\frac{25}{4}

Why: (α − β)² = (α + β)² − 4αβ = 49/4 − 6.

$α + β = \frac{7}{2}$ , $α β = \frac{3}{2}$ . $(α - β)^{2} = \frac{49}{4} - 4 \times \frac{3}{2} = \frac{49}{4} - 6 = \frac{25}{4}$ .

Q14

·1 mark·Multiple choiceRelationship between zeroes and coefficients

The zeroes of $x^{2} - 5 x + 2$ are $α$ and $β$ . The value of $\frac{1}{α^{2}} + \frac{1}{β^{2}}$ is

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Answer: (c)

\frac{21}{4}

Why: 1/α² + 1/β² = (α² + β²)/(αβ)² = ((α + β)² − 2αβ)/(αβ)².

$α + β = 5$ , $α β = 2$ . $α^{2} + β^{2} = 25 - 4 = 21$ , so $\frac{1}{α^{2}} + \frac{1}{β^{2}} = \frac{α^{2} + β^{2}}{(α β)^{2}} = \frac{21}{4}$ .

Q15

·1 mark·Multiple choiceRelationship between zeroes and coefficients

A quadratic polynomial $p (x)$ has zeroes $\frac{1}{2}$ and $- 3$ , and $p (0) = - 6$ . Then $p (x)$ is

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Answer: (b)

4 x^{2} + 10 x - 6

Why: p(x) = k(x − 1/2)(x + 3); p(0) = −3k/2 = −6 gives k = 4.

$p (x) = k (x - \frac{1}{2}) (x + 3)$ ; $p (0) = - \frac{3}{2} k = - 6 \Rightarrow k = 4$ . So $p (x) = (2 x - 1) (2 x + 6) = 4 x^{2} + 10 x - 6$ .

Case study questions

Section E of the board paper has three case-based questions of 4 marks: a real-life passage, then parts of 1, 1 and 2 marks, with a choice (OR) on the 2-mark part. All Class 10 case study questions by chapter.

Case study 1: Bakery profit (4 marks)

The owner of a home bakery models her weekly profit (in hundreds of rupees) from selling $x$ dozen cupcakes by the quadratic polynomial $P (x) = - x^{2} + 14 x - 40$ . The graph of $y = P (x)$ is a parabola opening downwards; she makes a profit only where the graph lies above the $x$ -axis.

(i) Find the zeroes of

P (x)

. [1 mark]

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Answer:

4

and

10

- x^{2} + 14 x - 40 = - (x - 4) (x - 10)

, so the zeroes are

4

and

10

. A1

(ii) Verify that the sum of the zeroes equals

- \frac{b}{a}

for

P (x)

. [1 mark]

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Answer:

4 + 10 = 14 = - \frac{14}{- 1}

Sum

= 4 + 10 = 14

and

- \frac{b}{a} = - \frac{14}{- 1} = 14

. A1

(iii) Find the profit when she sells

7

dozen cupcakes, and state for which whole numbers of dozens between the zeroes she makes a profit. [2 marks]

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Answer: ₹900;

x = 5, 6, 7, 8, 9

P (7) = - 49 + 98 - 40 = 9

, i.e. ₹

900

A1. The graph is above the

x

-axis between the zeroes, so

x = 5, 6, 7, 8, 9

. A1

OR Find a quadratic polynomial whose zeroes are each

2

more than the zeroes of

P (x)

. [2 marks]

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Answer:

x^{2} - 18 x + 72

New zeroes

6

and

12

M1: sum

18

, product

72

, so

x^{2} - 18 x + 72

. A1

Case study 2: Kitchen garden (4 marks)

Meera plans a rectangular kitchen garden on her terrace. Its length is $(x + 5)$ m and its width is $(x - 2)$ m, so its area is $A (x) = (x + 5) (x - 2) = x^{2} + 3 x - 10$ square metres.

(i) Without finding the zeroes, write the sum and the product of the zeroes of

A (x)

. [1 mark]

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Answer: Sum

- 3

, product

- 10

Sum

= - \frac{3}{1} = - 3

, product

= \frac{- 10}{1} = - 10

. A1

(ii) Find the zeroes of

A (x)

. [1 mark]

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Answer:

2

and

- 5

(x + 5) (x - 2) = 0 \Rightarrow x = - 5

x = 2

. A1

(iii) If

α

and

β

are the zeroes of

A (x)

, find

α^{2} + β^{2}

. [2 marks]

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Answer:

29

α^{2} + β^{2} = (α + β)^{2} - 2 α β

= 9 + 20 = 29

. A1

OR Find a quadratic polynomial whose zeroes are

2 α

and

2 β

, where

α, β

are the zeroes of

A (x)

. [2 marks]

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Answer:

x^{2} + 6 x - 40

Sum

= 2 (α + β) = - 6

, product

= 4 α β = - 40

M1; polynomial

x^{2} + 6 x - 40

. A1

Next steps for Polynomials

This free set is separate from the chapter's question bank. On the Polynomials chapter page the revision notes and the first step of the Route to 95 are free for everyone. CBSE Pro adds all 41 questions in the chapter bank (short and long answers, assertion–reason and more case studies) with their full step-marking schemes, the Route to 95 checkpoints with your progress saved, Skill Builders, the full common-mistakes library and AI marking of your written working.

Polynomials: important questions What Pro includes

Original questions written by CBSE Math Revision for the CBSE 2026-27 syllabus; not taken from NCERT, NCERT Exemplar or CBSE papers. Every answer was checked twice, the second time by an independent reviewer. CBSE Math Revision is independent and not affiliated with CBSE or NCERT. Spotted a slip? Tell us and it goes in the corrections log.