Polynomials: Important Questions and Step-by-Step Solutions for Class 10

Polynomials are one of the foundational concepts in Class 10 Mathematics. This chapter introduces students to important ideas such as the degree of a polynomial, zeros of a polynomial, relationships between zeros and coefficients, and methods of factorisation. These concepts are not only essential for examinations but also form the basis for advanced topics in algebra, calculus, and higher mathematics. In this guide, you will find important questions along with detailed step-by-step solutions designed according to the Class 10 syllabus and examination pattern.

Our subject experts have provided detailed solutions for these problems based on the latest CBSE syllabus and the NCERT textbook. This material helps students revise the chapter easily and perform well in the final examination.

Table of Contents

• Key Concepts in Polynomials for Class 10
• Important Formulas in Polynomials Class 10
• Section A: 1-Mark Questions (MCQ & Very Short Answer)
• Section B: 2-Mark Questions (Short Answer)
• Section C: 3-Mark Questions
• Section D: 4-Mark Questions (Long Answer)
• Section E: Previous Year Questions

Key Concepts in Polynomials for Class 10

A polynomial is an algebraic expression consisting of variables, coefficients, and non-negative integer exponents combined using operations such as addition, subtraction, and multiplication. For example, 2x³ + 3x² − 4x + 5 is a polynomial.

• Zeroes of a Polynomial:

A zero (or root) is a value of x where p(x) = 0. 

p(α) = 0 ⇒ α is a zero

• Types of Polynomials:

Classified by degree: Linear (degree 1), Quadratic (degree 2), and Cubic (degree 3). The degree determines the maximum number of zeroes.

A polynomial of degree 1 has a maximum of 1 zero. A polynomial of degree 2 has a maximum of 2 zeroes and a polynomial of degree 3 has a maximum of 3 zeroes.

• Sum & Product of Zeroes:

For a quadratic ax² + bx + c, if α and β are zeroes, there is a direct relationship with the coefficients.

α + β = −b/a

αβ = c/a

• Forming a Polynomial:

If you know the zeroes, you can always work backward and construct the polynomial using the standard formula.

x² − (α+β)x + αβ

• Cubic Polynomial Relations:

For ax³ + bx² + cx + d with zeroes α, β, γ, the relationships extend naturally from the quadratic case.

α + β + γ = −b/a

αβ + βγ + γα = c/a

αβγ = −d/a

• Division Algorithm:

For any two polynomials p(x) and g(x), where g(x) ≠ 0, we can always write p(x) = g(x)·q(x) + r(x), where degree of r(x) < degree of g(x).

p(x) = g(x)·q(x) + r(x)

Important Formulas in Polynomials Class 10

Section A: 1-Mark Questions (MCQ & Very Short Answer)

Q1: The graph of a polynomial p(x) cuts the x-axis at 3 points and touches it at 2 other points. The number of zeroes of p(x) is:

(a) 2 (b) 3

(c) 5 (d) 1

Solution: Every point where the graph intersects or touches the x-axis corresponds to a zero. The graph cuts the x-axis at 3 points and touches it at 2 points. Total = 5.

Therefore, the number of zeroes of p(x) is 5

Answer: (c) 5  

Q2: If the zeroes of the quadratic polynomial x² + (a+1)x + b are 2 and −3, then:

(a) a = −7, b = −1

(b) a = 0, b = −6 

(c) a = 2, b = −6

(d) a = 0, b = 6

Solution: Sum of zeroes = 2 + (−3) = −1 = −(a+1)/1

⇒ a = 0. 

Product of zeroes = 2×(−3) = −6 = b/1 

 ⇒ b = −6.

Answer: (b) a = 0, b = −6

Q3: A polynomial of degree 5 in x has how many zeroes at most?

(a) 4 (b) 5 

(c) 3 (d) 6

Solution: A polynomial of degree n can have at most n real zeroes. So a degree-5 polynomial has at most 5 zeroes.

Answer: (b) 5

Q4: If one zero of the polynomial p(x) = x² + 3x + k is 2, the value of k is:

(a) 1 (b) 5

(c) −10 (d) 10

Solution: Since 2 is a zero, p(2) = 0

 ⇒ 4 + 6 + k = 0 

 ⇒ k = −10.

Answer: (c) −10

Q5: Find the number of zeroes of the polynomial p(x) = x² − 4. 

Solution: Set p(x) = 0: x² = 4 

 ⇒ x = ±2. There are 2 zeroes: x = 2 and x = −2.

Answer: 2 zeroes (x = 2 and x = −2)

Q6: Write a quadratic polynomial whose zeroes are −4 and 2.

Solution: Sum of zeroes: α + β = −4 + 2 = −2

Product of zeroes: αβ = (−4)(2) = −8

Polynomial: x² − (−2)x + (−8) = x² + 2x − 8

Answer: x² + 2x − 8

Section B: 2-Mark Questions (Short Answer)

Q7: Find the zeroes of the polynomial p(x) = 4x² − 4x − 8 and verify the relationship between the zeroes and coefficients.

Solution: 4x² − 4x − 8 = 0

⇒  4(x² − x − 2) = 0 

⇒  x² − x − 2 = 0

Factorise: (x − 2)(x + 1) = 0 

⇒  x = 2 and x = −1

Verification: Sum: 2 + (−1) = 1 = −(−4)/4 = 1 

Product: 2 × (−1) = −2 = (−8)/4 = −2 

Zeroes of p(x) = 4x² − 4x − 8 are 2 and −1. Relationships verified.

Q8: Find the quadratic polynomial whose zeroes are √3 and −√3.

Solution: Sum of zeroes: √3 + (−√3) = 0

Product of zeroes: (√3)(−√3) = −3

Polynomial: x² − (0)x + (−3) = x² − 3

Answer: x² − 3

Q9: Does the polynomial a⁴ + 4a² + 5 have real zeroes? Justify your answer.

Solution: Let a² = x. Then the polynomial becomes x² + 4x + 5.

Discriminant: D = b² − 4ac = 16 − 20 = −4

Since D < 0, this polynomial has no real roots.

The polynomial a⁴ + 4a² + 5 has no real zeroes. 

Q10: Find the quadratic polynomial, the sum and product of whose zeroes are −3 and 2 respectively.

Solution: Using k[x² − (α+β)x + αβ], substitute α+β = −3 and αβ = 2.

Polynomial = x² − (−3)x + 2 = x² + 3x + 2

Answer: x² + 3x + 2

Q11: If α and β are zeroes of p(x) = x² − ax − b, find the value of (α + β + αβ).

Solution: Sum of zeroes: α + β = a (since −(−a)/1 = a)

Product of zeroes: αβ = −b/1 = −b

Therefore: α + β + αβ = a + (−b) = a − b

Answer: a − b

Section C: 3-Mark Questions

Q12: α and β are zeroes of x² − 6x + y. Find the value of y if 3α + 2β = 20.

Solution: From the polynomial: α + β = 6 and αβ = y

Given: 3α + 2β = 20 

⇒ multiply α + β = 6 by 2 to get 2α + 2β = 12

Subtracting: (3α + 2β) − (2α + 2β) = 20 − 12

⇒ α = 8

Then β = 6 − 8 = −2

Therefore: y = αβ = 8 × (−2) = −16

Answer: y = −16

Q13: If the zeroes of the polynomial x³ − 3x² + x + 1 are a−b, a, a+b, find the values of a and b.

Solution: Sum of zeroes: (a−b) + a + (a+b) = 3a = 3 (since −(−3)/1 = 3)

⇒ a = 1

Product of zeroes: (a−b)(a)(a+b) = −1 (since −1/1 = −1)

Substituting a = 1: 1(1−b)(1+b) = −1 

⇒ 1 − b² = −1 

⇒ b² = 2 

⇒ b = ±√2

a = 1, b = ±√2. The zeroes are 1−√2, 1, 1+√2.

Q14: Divide p(x) = x³ − 3x² + 5x − 3 by g(x) = x² − 2, and verify the division algorithm.

Solution: Perform long division of x³ − 3x² + 5x − 3 by x² − 2.

Quotient q(x) = x − 3, Remainder r(x) = 7x − 9

Verification: g(x)·q(x) + r(x) = (x² − 2)(x − 3) + 7x − 9 = x³ − 3x² + 5x − 3 = p(x) q(x) = x − 3, r(x) = 7x − 9. Division algorithm verified.

Q15: Find a quadratic polynomial whose zeroes are reciprocals of the zeroes of f(x) = ax² + bx + c (a ≠ 0, c ≠ 0).

Solution: Let α, β be zeroes of f(x). Then α + β = −b/a and αβ = c/a.

New zeroes are 1/α and 1/β.

New sum = 1/α + 1/β = (α+β)/αβ = (−b/a)/(c/a) = −b/c

New product = (1/α)(1/β) = 1/αβ = a/c

Required polynomial = k[x² − (−b/c)x + (a/c)] = k[x² + (b/c)x + (a/c)]

Answer: cx² + bx + a (taking k = c)

Section D: 4-Mark Questions (Long Answer)

Q16: Obtain all other zeroes of 3x⁴ + 6x³ − 2x² − 10x − 5 if two of its zeroes are √(5/3) and −√(5/3).

Solution: Two zeroes are √(5/3) and −√(5/3)

 So (x − √(5/3))(x + √(5/3)) = x² − 5/3 is a factor.

Multiply through by 3: (3x² − 5) is a factor of the polynomial.

Divide 3x⁴ + 6x³ − 2x² − 10x − 5 by 3x² − 5:

After long division: Quotient = x² + 2x + 1 = (x + 1)²

So 3x⁴ + 6x³ − 2x² − 10x − 5 = (3x² − 5)(x + 1)²

Remaining zeroes from (x + 1)² = 0 

⇒  x = −1 (repeated)

All zeroes: √(5/3), −√(5/3), −1, −1

Q17: For what value of k is the polynomial f(x) = 3x⁴ − 9x³ + x² + 15x + k completely divisible by 3x² − 5?

Solution: For complete divisibility, the remainder must be zero when f(x) is divided by 3x² − 5.

Perform long division of 3x⁴ − 9x³ + x² + 15x + k by 3x² − 5.

The quotient obtained is x²−3x+2, and the remainder is k+10.

For complete divisibility: k + 10 = 0 

⇒ k = -10

Answer: k = −10

Q18: Verify that 3, −1, and −1/3 are the zeroes of the cubic polynomial p(x) = 3x³ − 5x² − 11x − 3. Also, verify the relationships between the zeroes and the coefficients.

Solution: Verification:

p(3) = 81 − 45 − 33 − 3 = 0 

p(−1) = −3 − 5 + 11 − 3 = 0 

p(−1/3) = 3(−1/27) − 5(1/9) − 11(−1/3) − 3 = −1/9 − 5/9 + 11/3 − 3 = 0 

Sum of zeroes: 3 + (−1) + (−1/3) = 5/3 = −(−5)/3 = 5/3 = -b/a

Sum of products (pairwise): 3(−1) + (−1)(−1/3) + (−1/3)(3) = −3 + 1/3 − 1 = −11/3 = c/a 

Product of zeroes: 3 × (−1) × (−1/3) = 1 = −(−3)/3 = 1 = −d/a

All three are verified as zeroes, and all coefficient relationships hold.

Section E: Previous Year Questions

Q: Zeroes of the polynomial p(x) = x² − 3√2x + 4 are: (1 Mark) (2025)

Solution: Setting up the factorisation:

 x² − 3√2x + 4 

= x² − 2√2x − √2x + 4 

= (x − 2√2)(x − √2) = 0

Answer: x = 2√2 and x = √2

Q: If α and β are zeroes of p(x) = kx² − 30x + 45k and α + β = αβ, find the value of k. (1 Mark) 2025

Solution: α + β = 30/k and αβ = 45k/k = 45. Setting equal: 30/k = 45 → k = 30/45 = 2/3.

Answer: k = 2/3

Q: If the sum of zeroes of p(x) = (p+1)x² + (2p+3)x + (3p+4) is −1, find the value of p. (2 Marks – 2025)

Solution: Sum of zeroes = −(2p+3)/(p+1) = −1 

⇒  2p+3 = p+1 

⇒  p = −2.

Answer: p = −2

Q: If α and β are the zeroes of the polynomial 2x² + 5x + k, and α² + β² + αβ = 21/4, find k. (3 Marks-2024)

Solution: α + β = −5/2 and αβ = k/2. 

Now α² + β² + αβ = (α+β)² − αβ = 25/4 − k/2 = 21/4. 

Solving: k/2 = 1 

⇒ k = 2.

Answer: k = 2

Q: The zeroes of the polynomial x² − √2x − 12 are p and q. Find a polynomial whose zeroes are 2p and 2q. (2 Marks- 2023)

Solution: p + q = √2, pq = −12. 

New sum = 2p + 2q = 2√2.

New product = 4pq = −48. 

Polynomial = x² − 2√2x − 48.

Answer: x² − 2√2x − 48

Q: If one zero of the quadratic polynomial f(x) = 4x² − 8kx − 9 is the negative of the other, find the value of k. (2 Marks – 2022)

Solution: If one zero is α and the other is −α, their sum = 0. 

Sum of zeroes = 8k/4 = 2k = 0 

⇒ k = 0.

Answer: k = 0

Q: Find the zeroes of the quadratic polynomial 6x² − 3 − 7x and verify the relationship between the zeroes and the coefficients. (3 Marks -2021)

Solution: 6x² − 7x − 3 = (3x + 1)(2x − 3) = 0. 

Zeroes: x = −1/3 and x = 3/2.

Sum = −1/3 + 3/2 = 7/6 = 7/6 

Product = (−1/3)(3/2) = −1/2 = −3/6 

The zeroes of the quadratic polynomial 6x² − 3 − 7x are −1/3 and 3/2. Relationships verified.

Q: Find all the zeroes of 2x⁴ − 3x³ − 3x² + 6x − 2, if you know that two of its zeroes are √2 and −√2. (4 Marks - 2020)

Solution: Since √2 and −√2 are zeroes, (x² − 2) is a factor. 

Dividing 2x⁴ − 3x³ − 3x² + 6x − 2 by (x² − 2) gives quotient 2x² − 3x + 1 = (2x − 1)(x − 1). 

Remaining zeroes: x = 1/2 and x = 1.

All zeroes: √2, −√2, 1/2, and 1

Q: If α and β are the zeroes of the polynomial x² − 4√3x + 3, find the value of α + β − αβ. (1 Mark -2019)

Solution: α + β = 4√3 (coefficient rule) and αβ = 3. So α + β − αβ = 4√3 − 3.

Answer: 4√3 − 3