Polynomials Class 10 Previous Year Questions: 1 Mark (Chapter 2)

1-mark questions on Polynomials

This page has 31 1-mark questions on Polynomials from CBSE Class 10 board papers of 2023, 2024, 2025 and 2026. Of these, 22 appeared only in Standard papers and 9 appeared only in Basic papers. Question formats: 31 MCQs. 14 of them were asked in two or more paper sets.

0 of 31 practised

All 31 Polynomials 1-mark questions

Q1. If the zeroes of the quadratic polynomial \(x^2 + (a + 1)x + b\) are 2 and -3, then

Mark practised

(A) \(a = -7\), \(b = -1\)
(B) \(a = 5\), \(b = -1\)
(C) \(a = 2\), \(b = -6\)
(D) \(a = 0\), \(b = -6\)

Marks: 1
Level: Standard
Asked in: 3 paper sets
Years: 2023
Topic: 2.3 Relationship between Zeroes and Coefficients of a Polynomial
Source: (CBSE 2023 Mathematics Standard, Set 30/6/1, Q17, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(D) a = 0, b = −6, because sum of zeroes = 2 + (−3) = −1 = −(a + 1), so a = 0, and product = 2 × (−3) = −6 = b.
Why? For the quadratic polynomial ax² + bx + c, the sum of the zeroes equals −b/a and the product equals c/a. In this polynomial the coefficient of x² is 1, the coefficient of x is a + 1, and the constant term is b, so the sum formula fixes a and the product formula gives b.

The full 1 mark is for the correct option.

Answer:

(D) a = 0, b = −6

✓ Answer matches the CBSE answer key, 2023 Main, Paper 30/6/1: option (D).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Forgetting the minus sign in the sum formula and treating the sum of the zeroes as a + 1 instead of −(a + 1), which gives the wrong value of a.
Checking only one zero against an option; both 2 and −3 must be zeroes of the polynomial.
Confusing the printed b with the coefficient b in the general form ax² + bx + c; here the printed b is the constant term, so the product of the zeroes equals b.

Why this works

The relation between zeroes and coefficients is the fastest route. For x² + (a + 1)x + b, the coefficient of x² is 1, the coefficient of x is a + 1, and the constant term is b. Since the zeroes are 2 and −3, their sum is −1 and their product is −6. The sum formula gives −(a + 1) = −1, so a = 0. The product formula gives b = −6. The zeroes differ by 5, and indeed (−1 + 5)/2 = 2 and (−1 − 5)/2 = −3. With these values the polynomial is x² + x − 6, which factors as (x − 2)(x + 3). Quick check: substituting x = 2 gives 4 + 2 − 6 = 0, and substituting x = −3 gives 9 − 3 − 6 = 0.

Key idea to remember

Relationship between Zeroes and Coefficients of a Polynomial: see Key ideas used on this page.

Explanation

For a quadratic polynomial ax² + bx + c, if α and β are its zeroes, then α + β = −b/a and αβ = c/a. Here the coefficient of x² is 1, the coefficient of x is a + 1, and the constant term is b. With α = 2 and β = −3, the sum gives −(a + 1) = −1 and the product gives b = −6.

Important formulas

α + β = −b/a · Sum of the zeroes of ax² + bx + c; here the coefficient of x is a + 1.
αβ = c/a · Product of the zeroes; here the constant term is b and the coefficient of x² is 1.

Q2. The number of quadratic polynomials having zeroes -1 and 3 is :

Mark practised

(A) 1
(B) 2
(C) 3
(D) more than 3

Marks: 1
Level: Basic
Asked in: 3 paper sets
Years: 2024
Topic: 2.3 Relationship between Zeroes and Coefficients of a Polynomial
Source: (CBSE 2024 Mathematics Basic, Set 430/4/1, Q7, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(D) more than 3, because k(x² − 2x − 3), for every non-zero real number k, has zeroes −1 and 3, so there are infinitely many.
Why? A quadratic polynomial is not fixed by its zeroes; multiplying by any non-zero constant keeps the same zeroes.

The full 1 mark is for the correct option.

Answer:

(D) more than 3

✓ Answer matches the CBSE answer key, 2024 Main, Paper 430/4/1: option (D).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing A because you take x² − 2x − 3 as the only polynomial, forgetting that k(x² − 2x − 3) works for every non-zero real k.
Choosing C by counting only the values 1, 2 and 3 for k, forgetting that k can be any real number.
Checking (x − 1)(x + 3) as another polynomial, although its zeroes are 1 and −3, not −1 and 3.

Why this works

Zeroes −1 and 3 mean that x + 1 and x − 3 are factors, so the general polynomial is p(x) = k(x + 1)(x − 3). Expanding gives p(x) = k(x² − 2x − 3). Taking k as 1 gives x² − 2x − 3; taking k as 2 gives 2x² − 4x − 6; taking k as 3 gives 3x² − 6x − 9. Since k can be any non-zero real number, there are infinitely many quadratic polynomials. Quick check: putting x = −1 in x² − 2x − 3 gives 1 + 2 − 3 = 0, and putting x = 3 gives 9 − 6 − 3 = 0.

Key idea to remember

Relationship between Zeroes and Coefficients of a Polynomial: see Key ideas used on this page.

Explanation

If α and β are zeroes of a quadratic polynomial, the polynomial can be written as k(x − α)(x − β). With zeroes −1 and 3, the sum is −1 + 3 = 2 and the product is (−1)(3) = −3. So the general polynomial is k(x² − 2x − 3). Since k can be any non-zero real number, the number of such polynomials is more than 3.

Important formulas

p(x) = k(x − α)(x − β) · General quadratic polynomial with zeroes α and β; k is a non-zero real constant.
α + β = −b/a, αβ = c/a · Relationship between zeroes and coefficients of ax² + bx + c.

Q3. Which of the following statements is true for a polynomial \(p(x)\) of degree 3?

Mark practised

(A) \(p(x)\) has at most two distinct zeroes.
(B) \(p(x)\) has at least two distinct zeroes.
(C) \(p(x)\) has exactly three distinct zeroes.
(D) \(p(x)\) has at most three distinct zeroes.

Marks: 1
Level: Standard
Asked in: 3 paper sets
Years: 2025
Topic: 2.2 Geometrical Meaning of the Zeroes of a Polynomial
Source: (CBSE 2025 Mathematics Standard, Set 30/4/1, Q17, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(D) p(x) has at most three distinct zeroes, because a polynomial of degree n has at most n zeroes, so a cubic polynomial can have no more than three zeroes.
Why? A zero of a polynomial is a number k for which p(k) = 0. A polynomial of degree n has at most n zeroes, so a cubic polynomial, whose degree is 3, has at most three zeroes.

The full 1 mark is for the correct option.

Answer:

(D) p(x) has at most three distinct zeroes

✓ Answer matches the CBSE answer key, 2025 Main, Paper 30/4/1: option (D).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing C because you count repeated zeroes as separate; the option says distinct zeroes, so the cubic x³ has only one distinct zero, 0.
Choosing A because you confuse a cubic with a quadratic; the upper limit for a cubic is 3, not 2.
Choosing B because you think a cubic must have at least two zeroes; the rule gives only an upper limit, 'at most', not a lower limit.

Why this works

The degree of a polynomial is the highest power of x in it. A zero of p(x) is a number k for which p(k) = 0. Geometrically, zeroes are the x-coordinates where the graph y = p(x) meets the x-axis. A cubic polynomial has degree 3, and its graph can cut the x-axis at no more than 3 points. So it can have 0, 1, 2 or 3 distinct zeroes, but never 4 or more. For example, x³ − 4x has the three distinct zeroes −2, 0 and 2, while x³ has only one distinct zero, 0. Hence option D is true. The words 'at most' mean 'not more than', so the correct limit is 3, not exactly 3 or at least 2.

Key idea to remember

A polynomial of degree n has at most n zeroes: see Key ideas used on this page.

Explanation

Each distinct zero of a polynomial is the x-coordinate of a point where its graph cuts the x-axis. A polynomial of degree n can cut the x-axis at at most n points. For a cubic polynomial, degree is 3, so the graph can cut it at 0, 1, 2 or 3 points, and the polynomial has at most 3 distinct zeroes. 'At most' allows fewer than 3, so exactly three is not necessary; it also does not force any minimum number.

Important formulas

A polynomial p(x) of degree n has at most n zeroes. · For a cubic polynomial, n = 3, so it has at most 3 distinct zeroes.

Q4. If \(\alpha\) and \(\beta\) are two zeroes of a polynomial \(f(x) = px^2 - 2x + 3p\) and \(\alpha + \beta = \alpha\beta\), then value of \(p\) is :

Mark practised

(A) \(-\frac{2}{3}\)
(B) \(\frac{2}{3}\)
(C) \(\frac{1}{3}\)
(D) \(-\frac{1}{3}\)

Marks: 1
Level: Standard
Asked in: 3 paper sets
Years: 2026
Topic: 2.3 Relationship between Zeroes and Coefficients of a Polynomial
Source: (CBSE 2026 Mathematics Standard, Set 30/2/1, Q5, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 2/3, because for f(x) = px² − 2x + 3p, α + β = 2/p and αβ = 3, so 2/p = 3 gives p = 2/3.
Why? For a quadratic polynomial ax² + bx + c with zeroes α and β, the sum of zeroes is −b/a and the product of zeroes is c/a.

The full 1 mark is for the correct option.

Answer:

(B) 2/3

✓ Answer matches the CBSE answer key, 2026 Main, Paper 30/2/1: option (B).

Open CBSE’s official marking scheme for the 2026 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Using the sum of zeroes as b/a instead of −b/a; for this polynomial the sum would then be −2/p, leading to the wrong option −2/3.
Writing the product as the constant term 3p instead of c/a = 3p/p = 3; then the condition '2/p equals 3p' gives no listed option.

Why this works

Compare f(x) = px² − 2x + 3p with ax² + bx + c. Here a = p, b = −2 and c = 3p. So α + β = −b/a = 2/p, and αβ = c/a = 3p/p = 3. The condition α + β = αβ becomes 2/p = 3. Multiplying by p gives 2 = 3p, so p = 2/3. Quick check: with p = 2/3, the constant term is 3p = 2, so f(x) = (2/3)x² − 2x + 2. Both the sum and the product are 2/(2/3) = 3, so they are equal.

Key idea to remember

Relationship between Zeroes and Coefficients of a Polynomial: see Key ideas used on this page.

Explanation

For a quadratic polynomial ax² + bx + c, if α and β are its zeroes, then α + β = −b/a and αβ = c/a. This solution compares the given polynomial with ax² + bx + c, substitutes the two formulas into the given condition, and solves the resulting linear equation for p.

Important formulas

α + β = −b/a · Sum of the zeroes of ax² + bx + c.
αβ = c/a · Product of the zeroes of ax² + bx + c.

Q5. If α, β are the zeroes of a polynomial p(x) = x² + x - 1, then 1/α + 1/β equals to

Mark practised

(A) 1
(B) 2
(C) -1
(D) -1/2

Marks: 1
Level: Standard
Asked in: 2 paper sets
Years: 2023
Topic: 2.3 Relationship between Zeroes and Coefficients of a Polynomial
Source: (CBSE 2023 Mathematics Standard, Set 30/1/1, Q7, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) 1, because for p(x) = x² + x − 1, α + β = −1 and αβ = −1, so 1/α + 1/β = (α + β)/(αβ) = (−1)/(−1) = 1.
Why? For a quadratic polynomial ax² + bx + c, the sum of the zeroes is −b/a and the product is c/a, so the sum of the reciprocals is the quotient of these two values.

The full 1 mark is for the correct option.

Answer:

(A) 1

✓ Answer matches the CBSE answer key, 2023 Main, Paper 30/1/1: option (A).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Taking the sum of the zeroes as 1 (b/a) instead of −1 (−b/a), which changes the answer to −1.
Reading the constant term of x² + x − 1 as 1 instead of −1, which makes the product of the zeroes 1 instead of −1.
Finding 1/(α + β) instead of adding 1/α and 1/β separately.

Why this works

Compare p(x) = x² + x − 1 with ax² + bx + c: a = 1, b = 1 and c = −1. The NCERT relationship gives α + β = −b/a = −1 and αβ = c/a = −1. To add the reciprocals, use one denominator: 1/α + 1/β = β/(αβ) + α/(αβ) = (α + β)/(αβ). Substituting gives (−1)/(−1) = 1. A quick check is that the two negatives cancel, leaving 1.

Key idea to remember

Relationship between Zeroes and Coefficients of a Polynomial: see Key ideas used on this page.

Explanation

For a quadratic polynomial ax² + bx + c with zeroes α and β, the sum of the zeroes is −b/a and the product is c/a. This solution uses these two relations to rewrite the sum of the reciprocals without finding α and β themselves.

Important formulas

α + β = −b/a · Sum of the zeroes of ax² + bx + c.
αβ = c/a · Product of the zeroes of ax² + bx + c.
1/α + 1/β = (α + β)/(αβ) · Adding reciprocals over a common denominator.

Q6. If \(\alpha\) and \(\beta\) are the zeroes of the quadratic polynomial \(p(x) = x^2 - ax - b\) , then the value of \(\alpha^2 + \beta^2\) is :

Mark practised

(A) \(a^2 - 2b\)
(B) \(a^2 + 2b\)
(C) \(b^2 - 2a\)
(D) \(b^2 + 2a\)

Marks: 1
Level: Standard
Asked in: 2 paper sets
Years: 2023
Topic: 2.3 Relationship between Zeroes and Coefficients of a Polynomial
Source: (CBSE 2023 Mathematics Standard, Set 30/2/1, Q16, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) a² + 2b, because α + β = a and αβ = −b, so α² + β² = (α + β)² − 2αβ = a² + 2b.
Why? For a quadratic polynomial ax² + bx + c, the sum of the zeroes is −b/a and the product is c/a; the identity (α + β)² = α² + 2αβ + β² then gives α² + β².

The full 1 mark is for the correct option.

Answer:

(B) a² + 2b

✓ Answer matches the CBSE answer key, 2023 Main, Paper 30/2/1: option (B).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing a² − 2b because the constant term −b is read as b, making the product of the zeroes come out as b instead of −b.
Writing α² + β² = (α + β)² + 2αβ; the correct expansion has −2αβ.
Swapping the sum and product, taking α + β = −b and αβ = a, which leads to a wrong option.

Why this works

Compare x² − ax − b with the general form. The coefficient of x is −a and the constant term is −b. So the sum of the zeroes is α + β = −(−a)/1 = a, and the product is αβ = (−b)/1 = −b. Then α² + β² = (α + β)² − 2αβ = a² − 2(−b) = a² + 2b. Quick check: take p(x) = x² − 3x − 4, whose zeroes are 4 and −1. Then α² + β² = 16 + 1 = 17, and here a = 3, b = 4, so a² + 2b = 9 + 8 = 17.

Key idea to remember

Relationship between zeroes and coefficients of a quadratic polynomial: see Key ideas used on this page.

Explanation

NCERT states that if α and β are the zeroes of ax² + bx + c, then α + β = −b/a and αβ = c/a. Here the coefficient of x is −a and the constant term is −b, so α + β = a and αβ = −b. Using α² + β² = (α + β)² − 2αβ gives a² + 2b.

Important formulas

α + β = −b/a · For p(x) = ax² + bx + c; here it gives α + β = a.
αβ = c/a · Here c = −b, so αβ = −b.
α² + β² = (α + β)² − 2αβ · This identity gives the required expression a² + 2b.

Q7. If one zero of the polynomial \(6x^2 + 37x - (k - 2)\) is reciprocal of the other, then what is the value of \(k\) ?

Mark practised

(A) \(-4\)
(B) \(-6\)
(C) \(6\)
(D) \(4\)

Marks: 1
Level: Standard
Asked in: 2 paper sets
Years: 2023
Topic: 2.3 Relationship between Zeroes and Coefficients of a Polynomial
Source: (CBSE 2023 Mathematics Standard, Set 30/2/1, Q9, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) −4, because if the zeroes are α and 1/α, their product is 1, so c/a = −(k−2)/6 = 1, giving k = −4.
Why? When one zero is the reciprocal of the other, their product is 1. By the relationship between zeroes and coefficients of a quadratic polynomial, the product of zeroes equals c/a, so equating c/a to 1 determines k.

The full 1 mark is for the correct option.

Answer:

(A) −4

✓ Answer matches the CBSE answer key, 2023 Main, Paper 30/2/1: option (A).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Writing the constant term as k − 2 instead of −(k − 2); this changes the sign of c in 6x² + 37x − (k − 2).
Using the sum of zeroes, α + 1/α = −37/6, and trying to find α; the reciprocal condition fixes the product, not the sum.
Thinking the product of reciprocal zeroes is −1 because the constant term is negative; a number and its reciprocal always have product 1.

Why this works

The condition 'one zero is the reciprocal of the other' is a condition on the product of the zeroes, not on their sum. If the zeroes are α and 1/α, then α × 1/α = 1. For p(x) = 6x² + 37x − (k−2), comparing with ax² + bx + c gives a = 6 and c = −(k−2). The relationship between zeroes and coefficients says the product of zeroes is c/a, so −(k−2)/6 = 1. Then −(k−2) = 6, so k − 2 = −6, and k = −4. Quick check: with k = −4, p(x) = 6x² + 37x + 6 = (6x + 1)(x + 6), whose zeroes are −1/6 and −6, and −6 is the reciprocal of −1/6.

Key idea to remember

Relationship between zeroes and coefficients of a quadratic polynomial: see Key ideas used on this page.

Explanation

For a quadratic polynomial ax² + bx + c with zeroes α and β, the product of zeroes is αβ = c/a. If one zero is the reciprocal of the other, the zeroes are α and 1/α, so αβ = 1. Therefore c/a = 1. Here a = 6 and c = −(k−2), so −(k−2)/6 = 1, which gives k = −4.

Important formulas

αβ = c/a · Product of the zeroes of ax² + bx + c, where a ≠ 0.

Q8. If \(\alpha\) and \(\beta\) are the zeroes of the polynomial \(x^2 - 1\), then the value of \((\alpha + \beta)\) is

Mark practised

(A) 2
(B) 1
(C) -1
(D) 0

Marks: 1
Level: Standard
Asked in: 2 paper sets
Years: 2023
Topic: 2.3 Relationship between Zeroes and Coefficients of a Polynomial
Source: (CBSE 2023 Mathematics Standard, Set 30/6/1, Q6, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(D) 0, because for x² − 1, b = 0, so α + β = −b/a = 0.
Why? In ax² + bx + c, b is the coefficient of x and the sum of the zeroes is −b/a. When the x-term is absent, the numerator −b is zero, so the sum is zero.

The full 1 mark is for the correct option.

Answer:

(D) 0

✓ Answer matches the CBSE answer key, 2023 Main, Paper 30/6/1: option (D).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 2, because x² − 1 is wrongly read as (x − 1)², which would give two equal zeroes.
Choosing 1, because the constant term −1 is taken with the opposite sign as the sum.
Choosing −1, because the product of the zeroes is used instead of the sum.

Q9. The number of quadratic polynomials having zeroes \(-5\) and \(-3\) is

Mark practised

(A) 1
(B) 2
(C) 3
(D) more than 3

Marks: 1
Level: Standard
Asked in: 2 paper sets
Years: 2023
Topic: 2.3 Relationship between Zeroes and Coefficients of a Polynomial
Source: (CBSE 2023 Mathematics Standard, Set 30/6/1, Q3, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(D) more than 3, because every quadratic polynomial k(x² + 8x + 15), with k ≠ 0, has zeroes −5 and −3, so there are infinitely many.
Why? By the relationship between zeroes and coefficients, a quadratic with given zeroes is k[x² − (α+β)x + αβ] with k ≠ 0, so the zeroes do not fix a unique polynomial.

The full 1 mark is for the correct option.

Answer:

(D) more than 3

✓ Answer matches the CBSE answer key, 2023 Main, Paper 30/6/1: option (D).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing option A because only x² + 8x + 15 is written, forgetting that multiplying it by any non-zero real k gives another valid quadratic polynomial.
Choosing option B by considering only two choices of k, such as the polynomials x² + 8x + 15 and −x² − 8x − 15, without noticing that k can be any real number except 0.
Thinking the two given zeroes determine exactly one polynomial, whereas the relationship gives a whole family k[x² − (α+β)x + αβ].

Why this works

For zeroes −5 and −3, sum of zeroes = −8 and product of zeroes = 15. By the relationship in NCERT Example 4, every quadratic polynomial with these zeroes is k[x² − (sum)x + product] = k(x² + 8x + 15), where k is any non-zero real number. Since there are infinitely many choices of k, the number of such polynomials is more than 3. Quick check: x² + 8x + 15 = (x + 5)(x + 3), so its value at x = −5 and x = −3 is 0.

Key idea to remember

Relationship between Zeroes and Coefficients of a Polynomial: see Key ideas used on this page.

Explanation

NCERT shows that if α and β are zeroes of a quadratic polynomial ax² + bx + c, then the polynomial can be written as k(x − α)(x − β), where k is a non-zero constant. Here the zeroes are −5 and −3, so every quadratic polynomial with these zeroes is k(x + 5)(x + 3), which simplifies to k(x² + 8x + 15). Since k can be any non-zero real number, the number of such polynomials is more than 3.

Important formulas

\(k[x^2 - (\alpha+\beta)x + \alpha\beta]\) · General quadratic polynomial with zeroes α and β; k is any non-zero real number.

Q10. If one zero of a quadratic polynomial \(kx^2 + 4x + k\) is 1, then the value of \(k\) is :

Mark practised

(A) 2
(B) -2
(C) 4
(D) -4

Marks: 1
Level: Basic
Asked in: 2 paper sets
Years: 2024
Topic: 2.3 Relationship between Zeroes and Coefficients of a Polynomial
Source: (CBSE 2024 Mathematics Basic, Set 430/4/1, Q6, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) -2, because p(1) = k(1)² + 4(1) + k = 2k + 4 = 0, so k = -2.
Why? By the definition of a zero of a polynomial, if the given zero is substituted into p(x), the result is 0; this gives a linear equation in k.

The full 1 mark is for the correct option.

Answer:

(B) -2

✓ Answer matches the CBSE answer key, 2024 Main, Paper 430/4/1: option (B).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Forgetting that the constant term is also k, so the student leaves out the last k and gets -4 as the answer.
Solving 2k + 4 = 0 but writing the answer as 2 instead of -2.
Putting x = 0 instead of the given zero.

Q11. The zeroes of the quadratic polynomial \(2x^2 - 3x - 9\) are :

Mark practised

(A) \(3, \frac{-3}{2}\)
(B) \(-3, \frac{-3}{2}\)
(C) \(-3, \frac{3}{2}\)
(D) \(3, \frac{3}{2}\)

Marks: 1
Level: Standard
Asked in: 2 paper sets
Years: 2024
Topic: 2.3 Relationship between Zeroes and Coefficients of a Polynomial
Source: (CBSE 2024 Mathematics Standard, Set 30/3/1, Q6, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) 3, -3/2, because 2x² - 3x - 9 = (2x + 3)(x - 3), so x = 3 or x = -3/2.
Why? A real number k is a zero of p(x) when p(k) is zero; setting each factor to zero gives the zeroes.

The full 1 mark is for the correct option.

Answer:

(A) 3, -3/2

✓ Answer matches the CBSE answer key, 2024 Main, Paper 30/3/1: option (A).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Writing the zero of 2x + 3 as 3/2 instead of -3/2.
Writing the zero of x - 3 as -3 instead of 3.
Splitting the middle term with numbers whose product is 18 instead of -18, so the factors have the wrong signs.

Why this works

To factorise 2x² - 3x - 9 by splitting the middle term, look for two numbers whose product is 2 × (-9) = -18 and whose sum is -3. Those numbers are -6 and 3, because -6 + 3 = -3 and -6 × 3 = -18. So 2x² - 3x - 9 = 2x² - 6x + 3x - 9 = 2x(x - 3) + 3(x - 3) = (2x + 3)(x - 3). Setting 2x + 3 to zero gives x = -3/2, and setting x - 3 to zero gives x = 3. So the zeroes are 3 and -3/2, which is option (A).

Key idea to remember

Zeroes of a Polynomial: see Key ideas used on this page.

Explanation

The zeroes of a polynomial p(x) are the values of x at which p(x) equals zero. Factorising the quadratic gives p(x) = (2x + 3)(x - 3). A product equals zero only when one of its factors equals zero, so set 2x + 3 and x - 3 each to zero.

Important formulas

p(k) = 0 · A real number k is a zero of a polynomial p(x) when this holds.
AB = 0 ⇒ A = 0 or B = 0 · Used to read the zeroes from the factorised form.

Q12. If \(\alpha\) and \(\beta\) are the zeroes of the polynomial \(p(x) = 2x^2 + 6x - 6\) , then the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\) is equal to :

Mark practised

(A) -1
(B) 1
(C) -3
(D) 3

Marks: 1
Level: Basic
Asked in: 2 paper sets
Compartment: 2025
Topic: 2.3 Relationship between Zeroes and Coefficients of a Polynomial
Source: (CBSE 2025 Compartment Mathematics Basic, Set 430/S/1, Q11, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 1, because 1/α + 1/β = (α + β)/(αβ) = (−6/2) ÷ (−6/2) = 1.
Why? For zeroes α and β of ax² + bx + c, the sum is −b/a and the product is c/a; dividing the sum by the product gives the sum of the reciprocals.

The full 1 mark is for the correct option.

Answer:

(B) 1

✓ Answer matches the CBSE answer key, 2025 Compartment, Paper 430/S/1: option (B).

Open CBSE’s official marking scheme for the 2025 compartment exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing option C, −3, because α + β = −3, without then dividing by αβ to find the reciprocal sum.
Using the constant term −6 as the product of the zeroes instead of c/a = −3; that gives (−3)/(−6) = 0.5, which is not an option.
Using −6, the negative of b, as the sum of the zeroes instead of −b/a = −3; that gives (−6)/(−3) = 2, which is not an option.

Why this works

For the quadratic polynomial ax² + bx + c, NCERT gives α + β = −b/a and αβ = c/a. Here the coefficients are 2, 6 and −6, so α + β = −6/2 = −3 and αβ = −6/2 = −3. Then 1/α + 1/β = (α + β)/(αβ) = (−3)/(−3) = 1. The a cancels, giving −b/c = −6/(−6) = 1. Another correct method also earns full marks.

Key idea to remember

Relationship between Zeroes and Coefficients of a Polynomial: see Key ideas used on this page.

Explanation

For a quadratic polynomial ax² + bx + c with zeroes α and β, NCERT gives α + β = −b/a and αβ = c/a. The sum of the reciprocals is (α + β)/(αβ), so the question is solved by substituting the two relations.

Important formulas

α + β = −b/a · Sum of the zeroes of ax² + bx + c.
αβ = c/a · Product of the zeroes of ax² + bx + c.
1/α + 1/β = (α + β)/(αβ) = −b/c · Sum of reciprocals of the zeroes.

Q13. If \(-4\) is a zero of the polynomial \(p(x) = x^2 - x - (2 + 2k)\) , then the value of \(k\) is :

Mark practised

(A) 3
(B) 9
(C) 6
(D) \(-9\)

Marks: 1
Level: Standard
Asked in: 2 paper sets
Years: 2025
Topic: 2.3 Relationship between Zeroes and Coefficients of a Polynomial
Source: (CBSE 2025 Mathematics Standard, Set 30/2/1, Q16, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 9, because p(−4) = 16 + 4 − (2 + 2k) = 0 gives 18 − 2k = 0, so k = 9.
Why? By the definition of a zero of a polynomial, substituting the given zero makes the polynomial equal to 0, which gives a linear equation in k.

The full 1 mark is for the correct option.

Answer:

(B) 9

✓ Answer matches the CBSE answer key, 2025 Main, Paper 30/2/1: option (B).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Writing (−4)² as −16; the square of −4 is 16, so p(−4) starts with 16, not −16.
Expanding −(2 + 2k) as −2 + 2k; it is −2 − 2k, and this sign error leads to option D, −9.
Solving 18 − 2k = 0 by moving 18 to the other side without changing its sign; the correct step is 2k = 18, so the answer is 9, not −9.

Q14. \(\alpha, \beta\) are zeroes of the polynomial \(2x^2+5x+1\). The value of \(\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)\) is :

Mark practised

(A) \(-\frac{5}{4}\)
(B) 5
(C) \(\frac{5}{4}\)
(D) -5

Marks: 1
Level: Basic
Asked in: 2 paper sets
Years: 2025
Topic: 2.3 Relationship between Zeroes and Coefficients of a Polynomial
Source: (CBSE 2025 Mathematics Basic, Set 430/4/1, Q9, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(D) -5, because 1/α + 1/β = (α+β)/(αβ) = (-5/2)/(1/2) = -5.
Why? For a quadratic polynomial ax² + bx + c, the sum of zeroes is -b/a and the product is c/a, so the reciprocal sum is the quotient of these two.

The full 1 mark is for the correct option.

Answer:

(D) -5

✓ Answer matches the CBSE answer key, 2025 Main, Paper 430/4/1: option (D).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Forgetting the minus sign in the sum, so writing the sum as 5/2 instead of -5/2; this gives 5, option B.
Using 1/(α+β) for the sum of reciprocals instead of (α+β)/(αβ); this is not the correct combination.
Taking the product αβ as -1/2 instead of 1/2; the final sign then becomes wrong.

Why this works

Here a is 2, b is 5 and c is 1. From the relationship between zeroes and coefficients, α + β = -5/2 and αβ = 1/2. Then 1/α + 1/β = (α+β)/(αβ) = (-5/2)/(1/2) = -5. The coefficient b is positive, so the sum of the zeroes is negative, and the product is positive; therefore the reciprocal sum is negative.

Key idea to remember

Relationship between Zeroes and Coefficients of a Polynomial: see Key ideas used on this page.

Explanation

For a quadratic polynomial ax² + bx + c, the zeroes α and β satisfy α + β = -b/a and αβ = c/a. This solution rewrites the sum of reciprocals as (α+β)/(αβ), so the coefficients give the value directly without finding the zeroes.

Important formulas

α + β = -b/a · Sum of zeroes of a quadratic polynomial ax² + bx + c.
αβ = c/a · Product of zeroes of a quadratic polynomial ax² + bx + c.
1/α + 1/β = (α+β)/(αβ) · Taking LCM of the two denominators.

Q15. The sum and the product of zeroes of the polynomial \(p(x) = x^2 + 5x + 6\) are respectively

Mark practised

(A) 5, -6
(B) -5, 6
(C) 2, 3
(D) -2, -3 1

Marks: 1
Level: Basic
Years: 2023
Topic: 2.3 Relationship between Zeroes and Coefficients of a Polynomial
Source: (CBSE 2023 Mathematics Basic, Set 430/1/1, Q10, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) -5, 6, because for ax² + bx + c, sum of zeroes = -b/a = -5 and product = c/a = 6.
Why? For a quadratic polynomial ax² + bx + c, the sum of the zeroes equals -b/a and the product equals c/a, so comparing coefficients gives the two required values.

The full 1 mark is for the correct option.

Answer:

(B) -5, 6

✓ Answer matches the CBSE answer key, 2023 Main, Paper 430/1/1: option (B).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing C or D because x² + 5x + 6 factorises as (x + 2)(x + 3), which gives the zeroes -2 and -3, while the question asks for their sum and product.
Writing the sum as 5 because the negative sign in -b/a is dropped.

Why this works

For a quadratic polynomial p(x) = ax² + bx + c, if α and β are its zeroes, then p(x) = a(x - α)(x - β). Expanding gives ax² - a(α + β)x + aαβ. Comparing this with ax² + bx + c gives α + β = -b/a and αβ = c/a. For x² + 5x + 6, -b/a = -5/1 = -5 and c/a = 6/1 = 6. As a quick check, x² + 5x + 6 = (x + 2)(x + 3), so the zeroes are -2 and -3; their sum is -5 and their product is 6, which matches the printed choice (B).

Key idea to remember

Relationship between Zeroes and Coefficients of a Polynomial: see Key ideas used on this page.

Explanation

For a quadratic polynomial ax² + bx + c, the sum of its zeroes is -b/a and the product of its zeroes is c/a. Compare x² + 5x + 6 with ax² + bx + c, substitute the coefficient of x and the constant term into these relations, and divide by the coefficient of x². This gives the sum as -5 and the product as 6.

Important formulas

α + β = -b/a · Sum of the zeroes of the quadratic polynomial ax² + bx + c
αβ = c/a · Product of the zeroes of the quadratic polynomial ax² + bx + c

Q16. Which of the following is a quadratic polynomial with zeroes \(\frac{5}{3}\) and 0 ?

Mark practised

(A) \(3x(3x - 5)\)
(B) \(3x(x - 5)\)
(C) \(x^2 - \frac{5}{3}\)
(D) \(\frac{5}{3}x^2\)

Marks: 1
Level: Standard
Years: 2023
Topic: 2.3 Relationship between Zeroes and Coefficients of a Polynomial
Source: (CBSE 2023 Mathematics Standard, Set 30/1/2, Q8, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) 3x(3x − 5), because putting x = 0 and x = 5/3 in 3x(3x − 5) gives 0.
Why? By the definition of a zero of a polynomial, a number k is a zero if p(k) = 0; the factorised form becomes 0 at both given numbers.

The full 1 mark is for the correct option.

Answer:

(A) 3x(3x − 5)

✓ Answer matches the CBSE answer key, 2023 Main, Paper 30/1/2: option (A).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing option B because it also contains the factor x, without checking that its second zero is 5, not 5/3.
Choosing option C because 5/3 appears in the expression, although x² − 5/3 has no factor x and therefore does not become zero at x = 0.
Choosing option D because 5/3 is the coefficient, although (5/3)x² is zero only at x = 0 and misses the zero 5/3.

Why this works

Substitution checks the answer. For option A: at x = 0, 3(0)(3(0) − 5) = 0; at x = 5/3, 3(5/3)(3(5/3) − 5) = 5(5 − 5) = 0. So both required zeroes are zeroes of option A. Option B fails at x = 5/3 because its second factor is 5/3 − 5, which is not 0. Option C has no factor x, so it is not 0 at x = 0. Option D is zero only at x = 0, so it misses 5/3.

Key idea to remember

Zero of a polynomial: see Key ideas used on this page.

Explanation

A real number k is a zero of a polynomial p(x) if p(k) = 0. When a quadratic is written as a product of factors, it is zero exactly when one factor is zero; the factor 3x is zero at 0 and the factor 3x − 5 is zero at 5/3.

Important formulas

p(k) = 0 · Definition: k is a zero of polynomial p(x) when the value of the polynomial at k is 0.
α + β = −b/a, αβ = c/a · Relationship between zeroes and coefficients of ax² + bx + c; it can verify a factorised quadratic.

Q17. If α, β are zeroes of a polynomial p(x) = 2x² - x - 1 then α² + β² is equal to

Mark practised

(A) -3/4
(B) 5/4
(C) 1/4
(D) 3/4

Marks: 1
Level: Standard
Years: 2023
Topic: 2.3 Relationship between Zeroes and Coefficients of a Polynomial
Source: (CBSE 2023 Mathematics Standard, Set 30/1/3, Q5, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 5/4, because α + β = 1/2, αβ = −1/2, so α² + β² = (α+β)² − 2αβ = (1/2)² − 2(−1/2) = 5/4.
Why? For ax² + bx + c, the sum of zeroes is −b/a and the product is c/a; the identity (α+β)² = α² + β² + 2αβ converts these into the required value.

The full 1 mark is for the correct option.

Answer:

(B) 5/4

✓ Answer matches the CBSE answer key, 2023 Main, Paper 30/1/3: option (B).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Taking b as 1 instead of the actual coefficient −1 of x, so the sum of the zeroes comes out wrong.
Using the constant term −1 as the product of the zeroes without dividing by 2.
Adding 2αβ instead of subtracting it when expanding (α+β)².

Why this works

The formulas give the sum and product of the zeroes without solving the quadratic. For p(x) = 2x² − x − 1, the coefficient of x² is 2, the coefficient of x is −1, and the constant term is −1. So α + β = 1/2 and αβ = −1/2. Then α² + β² = (α+β)² − 2αβ = (1/2)² − 2(−1/2) = 5/4, which is option B.

Key idea to remember

Relationship between Zeroes and Coefficients of a Polynomial: see Key ideas used on this page.

Explanation

For a quadratic polynomial ax² + bx + c, the sum of its zeroes is −b/a and the product is c/a. The question asks for α² + β², which is not one of these direct results, so the identity (α+β)² = α² + β² + 2αβ connects the sum and product to the required expression.

Important formulas

α + β = −b/a · Sum of zeroes of ax² + bx + c.
αβ = c/a · Product of zeroes of ax² + bx + c.
α² + β² = (α + β)² − 2αβ · Identity used to find the sum of squares of the zeroes.

Q18. If one zero of the polynomial \(x^2 + 3x + k\) is 2, then the value of \(k\).

Mark practised

(A) \(-10\)
(B) \(10\)
(C) \(5\)
(D) \(-5\)

Marks: 1
Level: Standard
Years: 2023
Topic: 2.3 Relationship between Zeroes and Coefficients of a Polynomial
Source: (CBSE 2023 Mathematics Standard, Set 30/6/2, Q8, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) -10, because 2 is a zero, so 2² + 3(2) + k = 0, giving k = -10.
Why? By the definition of a zero of a polynomial, a zero makes the polynomial's value 0, so substituting 2 gives the equation to solve for k.

The full 1 mark is for the correct option.

Answer:

(A) -10

✓ Answer matches the CBSE answer key, 2023 Main, Paper 30/6/2: option (A).

Open CBSE’s official marking scheme for the 2023 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Forgetting the x² term, so only 3(2) is used and the wrong answer -6 is obtained.
Finding 4 + 6 = 10 and choosing 10 as the answer, without solving 10 + k = 0 for k.
Putting 0 in place of k in the polynomial and finding the zeroes of x² + 3x, instead of using the given zero 2 to find k.

Q19. The zeroes of the polynomial \(x^2 + 5x + 6\) are :

Mark practised

(A) 2, 3
(B) 3, -2
(C) -3, -2
(D) -3, 2

Marks: 1
Level: Basic
Compartment: 2024
Topic: 2.3 Relationship between Zeroes and Coefficients of a Polynomial
Source: (CBSE 2024 Compartment Mathematics Basic, Set 430(B), Q2, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) -3, -2, because x² + 5x + 6 = (x + 2)(x + 3), giving x = -2 or x = -3.
Why? A zero of a polynomial is a number at which the polynomial's value becomes zero; factorising the quadratic into linear factors gives these numbers.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2024 Compartment, Paper 430(B): option (C).

Open CBSE’s official marking scheme for the 2024 compartment exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing 2 and 3 because the factorisation gives (x + 2)(x + 3), without solving the factors to get -2 and -3.
Choosing -3 and 2, forgetting that the factor (x + 2) gives -2, not 2.

Why this works

For x² + 5x + 6, look for two numbers whose sum is 5 and product is 6. Those numbers are 2 and 3, so split 5x into 2x + 3x: x² + 5x + 6 = x² + 2x + 3x + 6 = x(x + 2) + 3(x + 2) = (x + 2)(x + 3). The polynomial vanishes at x = -2 and x = -3. Quick check: (-2)² + 5(-2) + 6 = 0, and (-3)² + 5(-3) + 6 = 0. The wrong options contain at least one positive zero, but both zeroes here are negative.

Key idea to remember

Zero of a polynomial: see Key ideas used on this page.

Explanation

A real number k is a zero of a polynomial p(x) if p(k) becomes zero. To find the zeroes of a quadratic polynomial, factorise it into linear factors; each factor gives one zero.

Important formulas

If α and β are the zeroes of ax² + bx + c, then α + β = -b/a and αβ = c/a. · For x² + 5x + 6, this gives sum -5 and product 6.

Q20. A quadratic polynomial whose zeroes are -8 and 3, is

Mark practised

(A) \((x + 8)(x + 3)\)
(B) \(x^2 + 5x + 24\)
(C) \((x - 8)(x - 3)\)
(D) \(x^2 + 5x - 24\)

Marks: 1
Level: Basic
Years: 2024
Topic: 2.3 Relationship between Zeroes and Coefficients of a Polynomial
Source: (CBSE 2024 Mathematics Basic, Set 430/3/2, Q8, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(D) x² + 5x − 24, because (x + 8)(x − 3) = x² + 5x − 24, which has zeroes −8 and 3.
Why? A quadratic polynomial with zeroes α and β has factors x − α and x − β; expanding the factors built from the two zeroes gives the required polynomial.

The full 1 mark is for the correct option.

Answer:

(D) x² + 5x − 24

✓ Answer matches the CBSE answer key, 2024 Main, Paper 430/3/2: option (D).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing option A, because it has the correct factor for the first zero but the second factor gives the wrong second zero.
Choosing option C, because it has the correct factor for the second zero but the first factor gives the wrong first zero.
Writing the x-coefficient as the sum of the zeroes without the minus sign in x² − (sum)x + product, so the sign of the x-term is wrong.

Why this works

A quadratic polynomial with zeroes α and β has the form k(x − α)(x − β). Here α = −8 and β = 3, so the factors are (x + 8) and (x − 3). Expanding (x + 8)(x − 3) gives x² + 5x − 24. Quick check: at x = −8, 64 − 40 − 24 = 0, and at x = 3, 9 + 15 − 24 = 0. The other options fail on at least one of the two required zeroes. Another correct method also earns full marks.

Key idea to remember

Relationship between Zeroes and Coefficients of a Polynomial: see Key ideas used on this page.

Explanation

For a quadratic polynomial ax² + bx + c with zeroes α and β, the NCERT text derives ax² + bx + c = k(x − α)(x − β), which gives α + β = −b/a and αβ = c/a. Here α = −8 and β = 3, so α + β = −5 and αβ = −24. Therefore x² − (α + β)x + αβ becomes x² + 5x − 24.

Important formulas

k(x − α)(x − β) · Any quadratic polynomial with zeroes α and β, where k is a non-zero constant.
x² − (α + β)x + αβ · The expanded form of the factor form above.
α + β = −b/a, αβ = c/a · For ax² + bx + c with zeroes α and β.

Q21. If 1 is a zero of the polynomial \(x^2 + ax + 2\), then 'a' is :

Mark practised

(A) 5
(B) 3
(C) -3
(D) -5

Marks: 1
Level: Basic
Years: 2024
Topic: 2.3 Relationship between Zeroes and Coefficients of a Polynomial
Source: (CBSE 2024 Mathematics Basic, Set 430(B), Q2, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) -3, because p(1) = 1 + a + 2 = a + 3, and since 1 is a zero, a + 3 = 0, so a = -3.
Why? A real number k is a zero of p(x) exactly when p(k) = 0; substituting the given zero turns the condition into a linear equation for a.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2024 Main, Paper 430(B): option (C).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Leaving out the constant term 2 while substituting the zero, so the equation used has only 1 + a.
Choosing 3 instead of -3 after moving the constant to the other side.
Trying to use the sum or product of zeroes with an unknown second zero, instead of substituting the given zero directly.

Q22. If one zero of the polynomial \((k - 2)x^2 - 10x + 3\) is reciprocal of the other, then the value of 'k' is :

Mark practised

(A) 3
(B) 5
(C) -5
(D) -3

Marks: 1
Level: Basic
Years: 2024
Topic: 2.3 Relationship between Zeroes and Coefficients of a Polynomial
Source: (CBSE 2024 Mathematics Basic, Set 430(B), Q3, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 5, because the product of the zeroes is 3/(k−2), reciprocal zeroes have product 1, so 3/(k−2) = 1, giving k = 5.
Why? For a quadratic polynomial ax² + bx + c, the product of the zeroes is c/a; if one zero is the reciprocal of the other, their product is 1.

The full 1 mark is for the correct option.

Answer:

(B) 5

✓ Answer matches the CBSE answer key, 2024 Main, Paper 430(B): option (B).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Taking the coefficient of x² as k instead of k − 2, so the product is taken as 3/k and the wrong option A (3) is reached.
Using the sum of the zeroes, −b/a, instead of the product, so the reciprocal condition is not used correctly.
Taking the constant term as −3, so the product becomes −3/(k−2) instead of 3/(k−2).
Setting the product equal to 0 instead of 1.

Why this works

Let the zeroes be α and β. 'One zero is the reciprocal of the other' means β = 1/α, so αβ = 1. For (k−2)x² − 10x + 3, comparing with ax² + bx + c gives a = k−2, b = −10, c = 3. The product of the zeroes is c/a = 3/(k−2). Equating the two products gives 3/(k−2) = 1, so k−2 = 3, hence k = 5. Quick check: with k = 5, the polynomial is 3x² − 10x + 3 = (3x − 1)(x − 3), whose zeroes are 1/3 and 3; 3 is the reciprocal of 1/3, and their product is 1.

Key idea to remember

Relationship between Zeroes and Coefficients of a Polynomial: see Key ideas used on this page.

Explanation

For a quadratic polynomial ax² + bx + c, if α and β are its zeroes, then α + β = −b/a and αβ = c/a. The phrase 'one zero is the reciprocal of the other' means αβ = 1. So we equate c/a to 1 and solve for k.

Important formulas

αβ = c/a · Product of the zeroes of ax² + bx + c; used here because reciprocal zeroes have product 1.
α + β = −b/a · Sum of the zeroes; not needed in this question but part of the same relationship.

Q23. If \(\alpha, \beta\) are the zeroes of the polynomial \(6x^2 - 5x - 4\) , then \(\frac{1}{\alpha} + \frac{1}{\beta}\) is equal to :

Mark practised

(A) \(\frac{5}{4}\)
(B) \(-\frac{5}{4}\)
(C) \(\frac{4}{5}\)
(D) \(\frac{5}{24}\)

Marks: 1
Level: Standard
Years: 2024
Topic: 2.3 Relationship between Zeroes and Coefficients of a Polynomial
Source: (CBSE 2024 Mathematics Standard, Set 30/1/3, Q16, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) −5/4, because 1/α + 1/β = (α+β)/(αβ) = (−b/a)/(c/a) = (5/6)/(−2/3) = −5/4.
Why? For a quadratic polynomial ax² + bx + c, the sum of the zeroes is α + β = −b/a and the product is αβ = c/a, so the sum of reciprocals becomes a single fraction.

The full 1 mark is for the correct option.

Answer:

(B) −5/4

✓ Answer matches the CBSE answer key, 2024 Main, Paper 30/1/3: option (B).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Taking the product as a positive number, 4/6, instead of −4/6, which turns the answer into 5/4. The constant term is −4, so the product is negative.
Writing 1/α + 1/β as 1/(α+β), which would be 6/5 here; these two expressions are not equal.
Confusing the two relations and putting the product over the sum, (−2/3)/(5/6) = −4/5, instead of the sum over the product.

Why this works

Compare 6x² − 5x − 4 with ax² + bx + c: here a is 6, b is −5 and c is −4. So α + β = −b/a = 5/6 and αβ = c/a = −4/6 = −2/3. The zeroes are not zero since their product is −2/3, so the reciprocals are defined. Then 1/α + 1/β = (α+β)/(αβ) = (5/6)/(−2/3) = −5/4. Quick check: 6x² − 5x − 4 = (2x + 1)(3x − 4), so the zeroes are −1/2 and 4/3; their reciprocals add to −2 + 3/4 = −5/4. Another correct method also earns full marks.

Key idea to remember

Relationship between Zeroes and Coefficients of a Polynomial: see Key ideas used on this page.

Explanation

A zero of a polynomial is a value of x for which the polynomial evaluates to zero. If α and β are the zeroes of ax² + bx + c, then α + β = −b/a and αβ = c/a. The expression 1/α + 1/β is rewritten as (α+β)/(αβ), so only the sum and product of the zeroes are needed, not the zeroes themselves.

Important formulas

α + β = −b/a · Sum of the zeroes of ax² + bx + c.
αβ = c/a · Product of the zeroes.
1/α + 1/β = (α+β)/(αβ) · Combining the reciprocals over a common denominator.

Q24. If α and β are zeroes of the polynomial 5x² + 3x - 7, the value of (1/α) + 1/β is

Mark practised

(A) -3/7
(B) 3/5
(C) 3/7
(D) -5/7

Marks: 1
Level: Standard
Years: 2024
Topic: 2.3 Relationship between Zeroes and Coefficients of a Polynomial
Source: (CBSE 2024 Mathematics Standard, Set 30/2/1, Q9, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) 3/7, because (1/α)+(1/β) = (α+β)/(αβ) = (−3/5)/(−7/5) = 3/7.
Why? For a quadratic polynomial ax² + bx + c, the sum of its zeroes is −b/a and the product is c/a, so the sum of the reciprocals of the zeroes is (α+β)/(αβ) = −b/c.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2024 Main, Paper 30/2/1: option (C).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Writing the sum of zeroes as 3/5 instead of −3/5; this leads to the wrong value −3/7.
Taking the product of zeroes as 7/5 instead of −7/5; this also leads to the wrong value −3/7.
Using the constant term −7 directly as the product instead of dividing by a to get −7/5; the arithmetic (−3/5)/(−7) gives 3/35, which is not among the options.

Why this works

Here α and β are the zeroes of 5x² + 3x − 7, meaning the values of x for which the polynomial becomes 0. 1/α is the reciprocal of α, and 1/β is the reciprocal of β. To add them without finding the actual zeroes, put them over one denominator: (1/α)+(1/β) = β/(αβ) + α/(αβ) = (α+β)/(αβ). Compare 5x² + 3x − 7 with ax² + bx + c; here b is 3 and c is −7, the constant term. By the relationship between zeroes and coefficients, α+β = −b/a = −3/5 and αβ = c/a = −7/5. Therefore (α+β)/(αβ) = (−3/5)/(−7/5) = 3/7.

Key idea to remember

Relationship between Zeroes and Coefficients of a Polynomial: see Key ideas used on this page.

Explanation

The relationship says that if α and β are zeroes of ax² + bx + c, then α+β = −b/a and αβ = c/a. This solution uses those two facts together: (1/α)+(1/β) = (α+β)/(αβ) = (−b/a)/(c/a) = −b/c. For 5x² + 3x − 7, b is 3 and c is −7, so −b/c = −3/(−7) = 3/7.

Important formulas

α + β = −b/a · Sum of the zeroes of ax² + bx + c.
αβ = c/a · Product of the zeroes of ax² + bx + c.
(1/α) + (1/β) = (α+β)/(αβ) · Sum of the reciprocals of the zeroes.

Q25. If \(\alpha\) and \(\beta\) are zeroes of the polynomial \(2x^2 - 9x + 5\), then value of \(\alpha^2 + \beta^2\) is

Mark practised

(A) \(\frac{1}{4}\)
(B) \(\frac{61}{4}\)
(C) 1
(D) \(\frac{71}{4}\)

Marks: 1
Level: Standard
Years: 2024
Topic: 2.3 Relationship between Zeroes and Coefficients of a Polynomial
Source: (CBSE 2024 Mathematics Standard, Set 30/2/2, Q6, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 61/4, because α² + β² = (α+β)² − 2αβ = (9/2)² − 2(5/2) = 61/4.
Why? For zeroes α and β of ax² + bx + c, α + β = −b/a and αβ = c/a; then α² + β² = (α+β)² − 2αβ.

The full 1 mark is for the correct option.

Answer:

(B) 61/4

✓ Answer matches the CBSE answer key, 2024 Main, Paper 30/2/2: option (B).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Writing α² + β² as (α+β)² and dropping −2αβ, which gives 81/4 instead of 61/4.
Swapping the two relationships for 2x² − 9x + 5, using 5/2 as the sum and 9/2 as the product.
Subtracting only αβ = 5/2 instead of 2αβ, which leads to 71/4.

Why this works

For the quadratic 2x² − 9x + 5, a = 2, b = −9, c = 5. The relationship between zeroes and coefficients gives α + β = −b/a = 9/2 and αβ = c/a = 5/2. The identity α² + β² = (α+β)² − 2αβ changes the sum of squares into these known values: (9/2)² − 2(5/2) = 81/4 − 5 = 61/4. As a quick check, 2(5/2) = 5 = 20/4, so 81/4 − 20/4 = 61/4. The correct choice is B.

Key idea to remember

Relationship between Zeroes and Coefficients of a Polynomial: see Key ideas used on this page.

Explanation

For a quadratic polynomial ax² + bx + c, the sum of the zeroes is −b/a and the product is c/a. This question asks for α² + β², which is not one of these two quantities, so the identity α² + β² = (α+β)² − 2αβ is used to express it through the sum and product.

Important formulas

α + β = −b/a · For zeroes α, β of ax² + bx + c.
αβ = c/a · Product of the zeroes of ax² + bx + c.
α² + β² = (α+β)² − 2αβ · Rewrites the required expression using sum and product.

Q26. If \(\alpha\) and \(\beta\) (\(\alpha \gt \beta\)) are the zeroes of the polynomial \(-x^2 + 8x + 9\), then \((\alpha - \beta)\) is equal to

Mark practised

(A) -10
(B) 10
(C) \(\pm 10\)
(D) 8

Marks: 1
Level: Standard
Years: 2024
Topic: 2.3 Relationship between Zeroes and Coefficients of a Polynomial
Source: (CBSE 2024 Mathematics Standard, Set 30/2/3, Q12, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) 10, because sum of zeroes = 8, product = −9, so (α−β)² = 8² − 4(−9) = 100, giving α−β = 10.
Why? For a quadratic polynomial ax² + bx + c with zeroes α and β, the sum of zeroes is −b/a and the product is c/a. The identity (α − β)² = (α + β)² − 4αβ connects the difference to the sum and product, and the condition α > β selects the positive root.

The full 1 mark is for the correct option.

Answer:

(B) 10

✓ Answer matches the CBSE answer key, 2024 Main, Paper 30/2/3: option (B).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing option C, ±10, because the ± sign was kept; the condition α > β makes α − β positive, so only 10 is correct.
Choosing option D, 8, because 8 is the coefficient of x; the question asks for α − β, not the sum of the zeroes.
Forgetting that the coefficient of x² is −1, so the sum of the zeroes is −8/(−1) = 8, not −8.

Why this works

The polynomial is −x² + 8x + 9, so the coefficient of x² is −1, the coefficient of x is 8, and the constant term is 9. Using the relationship between zeroes and coefficients, the sum is −b/a = 8 and the product is c/a = −9. Then (α − β)² = (α + β)² − 4αβ = 8² − 4(−9) = 100. Since α > β, α − β is the positive square root, so α − β = 10. Another correct method, factorising the polynomial, also earns full marks.

Key idea to remember

Relationship between Zeroes and Coefficients of a Polynomial: see Key ideas used on this page.

Explanation

For a quadratic polynomial ax² + bx + c, if α and β are its zeroes, then α + β = −b/a and αβ = c/a. This solution uses these relations to find the sum and product without factorising, then uses the identity (α − β)² = (α + β)² − 4αβ.

Important formulas

α + β = −b/a · Sum of the zeroes of the quadratic polynomial ax² + bx + c.
αβ = c/a · Product of the zeroes of the quadratic polynomial ax² + bx + c.
(α − β)² = (α + β)² − 4αβ · Links the difference of the zeroes to their sum and product.

Q27. A quadratic polynomial, one of whose zeroes is \(2 + \sqrt{5}\) and the sum of whose zeroes is 4, is :

Mark practised

(A) \(x^2 + 4x - 1\)
(B) \(x^2 - 4x - 1\)
(C) \(x^2 - 4x + 1\)
(D) \(x^2 + 4x + 1\)

Marks: 1
Level: Standard
Years: 2024
Topic: 2.3 Relationship between Zeroes and Coefficients of a Polynomial
Source: (CBSE 2024 Mathematics Standard, Set 30/4/2, Q17, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(B) x² - 4x - 1, because the other zero is 2 - √5 and x² - (2 + √5 + 2 - √5)x + (2 + √5)(2 - √5) = x² - 4x - 1.
Why? For a quadratic polynomial ax² + bx + c with zeroes α and β, α + β = -b/a and αβ = c/a; with leading coefficient 1, the polynomial is x² - (α + β)x + αβ.

The full 1 mark is for the correct option.

Answer:

(B) x² - 4x - 1

✓ Answer matches the CBSE answer key, 2024 Main, Paper 30/4/2: option (B).

Open CBSE’s official marking scheme for the 2024 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Wrongly treating 4 - (2 + √5) as 4 - 2 + √5, so the second zero becomes 2 + √5 and the sum becomes 4 + 2√5 instead of 4.
Wrongly expanding (2 + √5)(2 - √5) as 4 + 5, giving 9; the identity (a + b)(a - b) = a² - b² gives 4 - 5 = -1.
Choosing option A because it has constant term -1, without checking that its zeroes add to -4, not 4.

Why this works

Let the zeroes be α and β. One zero is α = 2 + √5 and the sum is 4, so β = 4 - (2 + √5) = 2 - √5. Their product is (2 + √5)(2 - √5) = 4 - 5 = -1. A quadratic polynomial with leading coefficient 1 is x² - (sum of zeroes)x + (product of zeroes), which is x² - 4x - 1. Quick check: put x = 2 + √5 in x² - 4x - 1; (2 + √5)² - 4(2 + √5) - 1 = (9 + 4√5) - (8 + 4√5) - 1 = 0. Also the sum is (2 + √5) + (2 - √5) = 4, matching the given condition. Another correct method also earns full marks.

Key idea to remember

Relationship between zeroes and coefficients of a quadratic polynomial: see Key ideas used on this page.

Explanation

The relationship between zeroes and coefficients says that if α and β are the zeroes of a quadratic polynomial ax² + bx + c, then α + β = -b/a and αβ = c/a. So, when the coefficient of x² is 1, the polynomial is x² - (α + β)x + αβ. In this question the sum of zeroes is 4 and the product is -1, so the polynomial is x² - 4x - 1.

Important formulas

α + β = -b/a · Sum of zeroes of ax² + bx + c.
αβ = c/a · Product of zeroes of ax² + bx + c.
x² - (α + β)x + αβ · Quadratic polynomial when the coefficient of x² is 1.

Q28. If \(\alpha\) and \(\beta\) are the zeroes of polynomial \(3x^2 + 6x + k\) such that \(\alpha + \beta + \alpha\beta = -\frac{2}{3}\), then the value of \(k\) is :

Mark practised

(A) -8
(B) 8
(C) -4
(D) 4

Marks: 1
Level: Standard
Years: 2025
Topic: 2.3 Relationship between Zeroes and Coefficients of a Polynomial
Source: (CBSE 2025 Mathematics Standard, Set 30/1/1, Q1, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(D) 4, because α + β = -6/3 = -2 and αβ = k/3, so -2 + k/3 = -2/3 gives k = 4.
Why? For a quadratic polynomial ax² + bx + c, the sum of the zeroes is -b/a and the product of the zeroes is c/a; substituting these into the given relation gives an equation in k.

The full 1 mark is for the correct option.

Answer:

(D) 4

✓ Answer matches the CBSE answer key, 2025 Main, Paper 30/1/1: option (D).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Forgetting the minus sign in the sum of zeroes and taking the sum as 2; this leads to -8.
Taking the product of the zeroes as k instead of k/3; this leads to 4/3, which is not among the choices.
While solving -2 + k/3 = -2/3, moving -2 to the right side as -2 instead of +2; this also leads to -8.

Why this works

Compare 3x² + 6x + k with ax² + bx + c. Here a = 3, b = 6 and c = k. So α + β = -b/a = -6/3 = -2 and αβ = c/a = k/3. Put these in the given relation: -2 + k/3 = -2/3. Add 2 to both sides: k/3 = -2/3 + 2 = 4/3, so k = 4. Check with k = 4: the left side α + β + αβ becomes -2 + 4/3, which is -2/3, matching the given relation. The other choices fail this check: for -8, the left side is -2 + (-8/3) = -14/3; for 8, it is -2 + 8/3 = 2/3; for -4, it is -2 + (-4/3) = -10/3.

Key idea to remember

Relationship between Zeroes and Coefficients of a Polynomial: see Key ideas used on this page.

Explanation

If α and β are the zeroes of the quadratic polynomial ax² + bx + c, then α + β = -b/a and αβ = c/a. In this question the polynomial is 3x² + 6x + k, so a = 3, b = 6 and c = k. Substituting α + β and αβ into the given relation gives an equation that is solved for k.

Important formulas

α + β = -b/a · Sum of the zeroes of ax² + bx + c
αβ = c/a · Product of the zeroes of ax² + bx + c

Q29. The sum of the zeroes of the polynomial \(p(x) = 5x - 7x^2 + 3\) is :

Mark practised

(A) \(\frac{-7}{5}\)
(B) \(\frac{7}{5}\)
(C) \(\frac{5}{7}\)
(D) \(\frac{-5}{7}\)

Marks: 1
Level: Standard
Years: 2025
Topic: 2.3 Relationship between Zeroes and Coefficients of a Polynomial
Source: (CBSE 2025 Mathematics Standard, Set 30/1/3, Q11, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(C) 5/7, because for p(x) = -7x² + 5x + 3, the sum of the zeroes is -b/a = -(5)/(-7) = 5/7.
Why? For a quadratic polynomial ax² + bx + c, the sum of the zeroes is -b/a, so only the coefficient of x and the coefficient of x² are needed.

The full 1 mark is for the correct option.

Answer:

✓ Answer matches the CBSE answer key, 2025 Main, Paper 30/1/3: option (C).

Open CBSE’s official marking scheme for the 2025 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Taking the coefficient of x as a and the coefficient of x² as b without writing the polynomial in standard form, which gives 7/5 (option B) instead of 5/7.
Using the ratio coefficient of x divided by coefficient of x², which gives -5/7 (option D), instead of the negative of that ratio.

Why this works

Write p(x) in the standard form ax² + bx + c. Here the polynomial is -7x² + 5x + 3, so the coefficient of x² is -7 and the coefficient of x is 5. By the relationship between zeroes and coefficients, the sum of the zeroes is -b/a = -5/(-7) = 5/7. No factorisation is needed; the sum is read directly from the coefficients.

Key idea to remember

Relationship between Zeroes and Coefficients of a Polynomial: see Key ideas used on this page.

Explanation

For a quadratic polynomial ax² + bx + c with zeroes α and β, the factors are (x - α) and (x - β). Comparing coefficients gives α + β = -b/a and αβ = c/a. This question uses only the sum formula: identify a and b, then substitute directly.

Important formulas

α + β = -b/a · Sum of the zeroes of the quadratic polynomial ax² + bx + c, a ≠ 0.

Q30. At which of the following points, the quadratic polynomial \(p(x) = -3x + 18x^2 - 1\) intersects the positive \(x\)-axis ?

Mark practised

(A) \(\left(\frac{1}{6}, 0\right)\)
(B) \(\left(-\frac{1}{3}, 0\right)\)
(C) \(\left(-\frac{1}{6}, 0\right)\)
(D) \(\left(\frac{1}{3}, 0\right)\)

Marks: 1
Level: Standard
Years: 2026
Topic: 2.2 Geometrical Meaning of the Zeroes of a Polynomial
Source: (CBSE 2026 Mathematics Standard, Set 30(B), Q1, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(D) (1/3, 0), because p(x) = (6x + 1)(3x - 1), so the zeroes are x = -1/6 and x = 1/3, and only x = 1/3 is positive.
Why? A point (k, 0) lies on the graph of y = p(x) exactly when p(k) = 0, so the zeroes of p(x) give the x-coordinates of the x-axis intersection points.

The full 1 mark is for the correct option.

Answer:

(D) (1/3, 0)

✓ Answer matches the CBSE answer key, 2026 Main, Paper 30(B): option (D).

Open CBSE’s official marking scheme for the 2026 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing (1/6, 0) just because it is positive; p(1/6) = -1, not 0.
Factorising correctly but then selecting (-1/6, 0), forgetting that the question asks for the positive x-axis.

Why this works

The x-axis is the line y = 0, so the graph of y = p(x) meets the x-axis exactly where p(x) = 0. Write the polynomial in standard form: p(x) = 18x² - 3x - 1. Factorising: 18x² - 3x - 1 = (6x + 1)(3x - 1). So p(x) = 0 when x = -1/6 or x = 1/3. Check: p(1/3) = 18×(1/3)² - 3×(1/3) - 1 = 0, and p(-1/6) = 18×(-1/6)² - 3×(-1/6) - 1 = 0. Of these two zeroes, only x = 1/3 is positive, so the intersection with the positive x-axis is (1/3, 0). Check the tempting option: p(1/6) = 18×(1/6)² - 3×(1/6) - 1 = -1, so (1/6, 0) is not on the graph.

Key idea to remember

Zeroes of a polynomial: see Key ideas used on this page.

Explanation

A real number k is a zero of a polynomial p(x) if p(k) = 0. Geometrically, the zeroes of a polynomial are the x-coordinates of the points where the graph of y = p(x) intersects the x-axis. This solution sets p(x) = 0 and finds the zeroes, then picks the one that is positive.

Important formulas

p(k) = 0 · Definition of a zero of a polynomial: k is a zero if the value of the polynomial at k is 0.
y = ax² + bx + c · A quadratic polynomial; its x-axis intersection points have x-coordinates equal to its zeroes.

Q31. The graph of a quadratic polynomial \(f(x)\) passes through \((5, 0)\), \((0, -1)\) and \((-2, 0)\). The two factors of the polynomial are

Mark practised

(A) \((x + 2), (x - 5)\)
(B) \((x + 5), (x - 2)\)
(C) \((x + 1), (x - 5)\)
(D) \((x - 1), (x + 2)\)

Marks: 1
Level: Standard
Years: 2026
Topic: 2.2 Geometrical Meaning of the Zeroes of a Polynomial
Source: (CBSE 2026 Mathematics Standard, Set 30(B), Q2, 1 mark)
Status: Verified solution

Step-by-step solution

Step 1: Choose the option
1 mark
(A) \((x + 2), (x - 5)\), because the graph cuts the \(x\)-axis at the two given points, so those \(x\)-coordinates are the zeroes and each zero \(k\) gives the factor \(x - k\).
Why? A zero of a polynomial is the \(x\)-coordinate of a point where its graph meets the \(x\)-axis, and each zero \(k\) gives the factor \(x - k\).

The full 1 mark is for the correct option.

Answer:

(A) (x + 2), (x - 5)

✓ Answer matches the CBSE answer key, 2026 Main, Paper 30(B): option (A).

Open CBSE’s official marking scheme for the 2026 main exam to see how the board gives marks for this question.

Common mistakes to avoid

Choosing (B) because you read the \(x\)-intercepts \(5\) and \(-2\) and write them as \(x + 5\) and \(x - 2\); the zero \(5\) gives the factor \(x - 5\), not \(x + 5\).
Choosing (C) or (D) because you try to use the \(y\)-intercept as a factor; a point on the \(y\)-axis is not a zero, so it gives no factor.
Checking only that \(x = 5\) makes one factor zero and forgetting to check that \(x = -2\) makes the other factor zero; both zeroes must appear in the factor pair.

Why this works

The points \((5, 0)\) and \((-2, 0)\) lie on the \(x\)-axis. The zeroes of a polynomial are exactly the \(x\)-coordinates of the points where its graph meets the \(x\)-axis, so \(5\) and \(-2\) are the zeroes. For a quadratic polynomial, if \(k\) is a zero, then \(x - k\) is a factor. Hence the factors are \(x - 5\) and \(x + 2\), which is option (A). The third point checks this: if \(f(x) = a(x - 5)(x + 2)\), then the point \((0, -1)\) gives \(f(0) = a(-5)(2) = -10a = -1\), so \(a = \frac{1}{10}\). Thus \(f(x) = \frac{1}{10}(x - 5)(x + 2)\), and \(f(0) = -1\) is satisfied.

Key idea to remember

Geometrical Meaning of the Zeroes of a Polynomial: see Key ideas used on this page.

Explanation

The zeroes of a polynomial \(p(x)\) are precisely the \(x\)-coordinates of the points where the graph of \(y = p(x)\) intersects the \(x\)-axis. Here the graph meets the \(x\)-axis at \(x = 5\) and \(x = -2\), so these are the zeroes. For a quadratic polynomial, a zero \(k\) corresponds to a factor \(x - k\), giving the factors \(x - 5\) and \(x + 2\).

Important formulas

If \(k\) is a zero of \(p(x)\), then \(x - k\) is a factor of \(p(x)\). · Turns each \(x\)-intercept into a factor.
The zeroes of \(p(x)\) are the \(x\)-coordinates where \(y = p(x)\) intersects the \(x\)-axis. · Links the given graph points to the zeroes.

1 / 31

Key ideas used on this page

Relationship between Zeroes and Coefficients of a Polynomial
Used in questions 1 2 4 5 8 9 12 14 15 17 20 22 23 24 25 26 28 29
Zero of a polynomial
Used in questions 10 13 16 18 19 21
Relationship between zeroes and coefficients of a quadratic polynomial
Used in questions 6 7 27
Zeroes of a Polynomial
Used in questions 11 30
A polynomial of degree n has at most n zeroes
Used in question 3
Geometrical Meaning of the Zeroes of a Polynomial
Used in question 31

How to use these questions

Try each question on paper first. Then read the steps: the mark beside each step shows what the examiner looks for. If a step is new to you, revise Section 2.2 and Section 2.3 of the NCERT chapter.

To revise the lessons behind these questions, read the NCERT chapter, exactly as NCERT publishes it on its own website.

Official papers behind this page

Every question on this page comes from an official CBSE Class 10 Mathematics board paper. The board publishes the question papers on cbse.gov.in and the marking schemes on cbse.gov.in, where you can see how marks are given.

Year	Exam	Course	Question paper	Marking scheme
2026	Main	Standard (041)	Open CBSE's official question paper for the 2026 main exam (one ZIP file of about 61.9 MB, Class 10 Maths Standard and Basic together) to read the full paper.	Open CBSE's official marking scheme for the 2026 main exam (one ZIP file of about 22.8 MB, Class 10 Maths Standard (041)) to see how marks are given.
2025	Main	Standard (041)	Open CBSE's official question paper for the 2025 main exam (one ZIP file of about 54.5 MB, Class 10 Maths Standard (041)) to read the full paper.	Open CBSE's official marking scheme for the 2025 main exam (one ZIP file of about 56.9 MB, Class 10 Maths Standard and Basic together) to see how marks are given.
2025	Main	Basic (241)	Open CBSE's official question paper for the 2025 main exam (one ZIP file of about 49.9 MB, Class 10 Maths Basic (241)) to read the full paper.	Open CBSE's official marking scheme for the 2025 main exam (one ZIP file of about 56.9 MB, Class 10 Maths Standard and Basic together) to see how marks are given.
2025	Compartment	Basic (241)	Open CBSE's official question paper for the 2025 compartment exam (one ZIP file of about 2.4 MB, Class 10 Maths Basic (241)) to read the full paper.	Open CBSE's official marking scheme for the 2025 compartment exam (one ZIP file of about 18.7 MB, Class 10 Maths Basic (241)) to see how marks are given.
2024	Main	Standard (041)	Open CBSE's official question paper for the 2024 main exam (one ZIP file of about 35.3 MB, Class 10 Maths Standard (041)) to read the full paper.	Open CBSE's official marking scheme for the 2024 main exam (one ZIP file of about 23.4 MB, Class 10 Maths Standard (041)) to see how marks are given.
2024	Main	Basic (241)	Open CBSE's official question paper for the 2024 main exam (one ZIP file of about 43.4 MB, Class 10 Maths Basic (241)) to read the full paper.	Open CBSE's official marking scheme for the 2024 main exam (one ZIP file of about 23.8 MB, Class 10 Maths Basic (241)) to see how marks are given.
2024	Compartment	Basic (241)	Open CBSE's official question paper for the 2024 compartment exam (one ZIP file of about 2.2 MB, Class 10 Maths Basic (241)) to read the full paper.	Open CBSE's official marking scheme for the 2024 compartment exam (one ZIP file of about 5.6 MB, Class 10 Maths Basic (241)) to see how marks are given.
2023	Main	Standard (041)	Open CBSE's official question paper for the 2023 main exam (one ZIP file of about 44 MB, Class 10 Maths Standard (041)) to read the full paper.	Open CBSE's official marking scheme for the 2023 main exam (one ZIP file of about 17.4 MB, Class 10 Maths Standard (041)) to see how marks are given.
2023	Main	Basic (241)	Open CBSE's official question paper for the 2023 main exam (one ZIP file of about 49.4 MB, Class 10 Maths Basic (241)) to read the full paper.	Open CBSE's official marking scheme for the 2023 main exam (one ZIP file of about 26 MB, Class 10 Maths Basic (241)) to see how marks are given.

Sources and session

Each question is reproduced as printed in its CBSE paper, with its year, paper code and question number; only the layout is tidied. The same question printed in several paper sets counts once, and the card says how often it was printed in 2023–2026. These counts describe past papers; they are not predictions.

Every solution is original and follows NCERT Mathematics Class X (reprint 2026-27). The mark beside each step follows the official CBSE marking scheme where the scheme shows it; a mark labelled “(suggested)” follows the board’s usual pattern for that type of question.

Reference: CBSE Class 10 Mathematics board question papers via cbse.gov.in; all solutions are original. learncbse.net is an independent website and is not affiliated with CBSE or NCERT.

1-mark questions on Polynomials

All 31 Polynomials 1-mark questions

Q1. If the zeroes of the quadratic polynomial \(x^2 + (a + 1)x + b\) are 2 and -3, then

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q2. The number of quadratic polynomials having zeroes -1 and 3 is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q3. Which of the following statements is true for a polynomial \(p(x)\) of degree 3?

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q4. If \(\alpha\) and \(\beta\) are two zeroes of a polynomial \(f(x) = px^2 - 2x + 3p\) and \(\alpha + \beta = \alpha\beta\), then value of \(p\) is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q5. If α, β are the zeroes of a polynomial p(x) = x² + x - 1, then 1/α + 1/β equals to

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q6. If \(\alpha\) and \(\beta\) are the zeroes of the quadratic polynomial \(p(x) = x^2 - ax - b\) , then the value of \(\alpha^2 + \beta^2\) is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q7. If one zero of the polynomial \(6x^2 + 37x - (k - 2)\) is reciprocal of the other, then what is the value of \(k\) ?

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q8. If \(\alpha\) and \(\beta\) are the zeroes of the polynomial \(x^2 - 1\), then the value of \((\alpha + \beta)\) is

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q9. The number of quadratic polynomials having zeroes \(-5\) and \(-3\) is

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q10. If one zero of a quadratic polynomial \(kx^2 + 4x + k\) is 1, then the value of \(k\) is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q11. The zeroes of the quadratic polynomial \(2x^2 - 3x - 9\) are :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q12. If \(\alpha\) and \(\beta\) are the zeroes of the polynomial \(p(x) = 2x^2 + 6x - 6\) , then the value of \(\frac{1}{\alpha} + \frac{1}{\beta}\) is equal to :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q13. If \(-4\) is a zero of the polynomial \(p(x) = x^2 - x - (2 + 2k)\) , then the value of \(k\) is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q14. \(\alpha, \beta\) are zeroes of the polynomial \(2x^2+5x+1\). The value of \(\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)\) is :

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q15. The sum and the product of zeroes of the polynomial \(p(x) = x^2 + 5x + 6\) are respectively

Step-by-step solution

Step 1: Choose the option

Answer:

Common mistakes to avoid

Q16. Which of the following is a quadratic polynomial with zeroes \(\frac{5}{3}\) and 0 ?

Step-by-step solution

Step 1: Choose the option