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Important Questions for Class 10 Maths Chapter 2 Polynomials

Important Questions For Class 10 Maths Chapter 2 Polynomials - Explore detailed answers and step by step solutions which will help you ace your Class 10 Math examination.
authorImageNeha Tanna24 Mar, 2025
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Important Questions for Class 10 Maths Chapter 2

 Important Questions for Class 10 Maths Chapter 2: Here are some Important Questions for Class 10 Maths Chapter 2 Polynomials based on the NCERT book. The answers to these questions will aid students in getting ready for and performing well on the CBSE Class 10 Maths exam.

The Important Questions for Class 10 Maths Chapter 2 below cover the most recent syllabus that the board has mandated. They are also predicated on the most recent exam format.

Students can thoroughly prepare for the board exam by practicing the Important Questions for Class 10 Maths Chapter 2. Students will discover thorough responses in addition to the pertinent questions. They can consult the solutions if they run into trouble answering the questions.

Important Questions for Class 10 Maths Chapter 2 Overview

In order to help students quickly review the topic by solving these problems, we have also included a few more significant exam-oriented questions for topic 2 Polynomials of Class 10 Math. Expressions with varying degrees of unknown variables are covered in this chapter. The zeros of the polynomial are the solutions to these formulas. The linear polynomial has degree 1, the quadratic polynomial has degree 2, and the cubic polynomial has degree 3.

Important Questions for Class 10 Maths Chapter 2 Polynomials

Below we have provided Important Questions for Class 10 Maths Chapter 2 to help students prepare better for their Class 10 maths exams. Students can prepare these Important Questions for Class 10 Maths Chapter 2 Polynomials before their exams to understand the concepts better.

CBSE Class 10 Sample Paper

Q.1: Find the value of “p” from the polynomial x 2 + 3x + p, if one of the zeroes of the polynomial is 2.

Solution:

As 2 is the zero of the polynomial.
We know that if α is a zero of the polynomial p(x), then p(α) = 0 Substituting x = 2 in x 2 + 3x + p, ⇒ 2 2 + 3(2) + p = 0 ⇒ 4 + 6 + p = 0 ⇒ 10 + p = 0 ⇒ p = -10

Q.2: Does the polynomial a 4 + 4a 2 + 5 have real zeroes?

Solution:

In the given polynomial, let a 2 = x.
Now, the polynomial becomes, x  2 + 4x + 5 Comparing with ax 2 + bx + c, Here, b 2 – 4ac = 4 2 – 4(1)(5) = 16 – 20 = -4
So, D = b  2 – 4ac < 0 As the discriminant (D) is negative,
the given polynomial does not have real roots or zeroes.

Read More - How to Score Well in Class 10: 5 Study Tips You Need

Q.3: Compute the zeroes of the polynomial 4x 2 – 4x – 8. Also, establish a relationship between the zeroes and coefficients.

Solution:

Let the given polynomial be
p(x) = 4x  2 – 4x – 8
To find the zeroes, take p(x) = 0 Now,
factorise the equation 4x  2 – 4x – 8 = 0 4x 2 – 4x – 8 = 0 4(x 2 – x – 2) = 0 x 2 – x – 2 = 0 x 2 – 2x + x – 2 = 0 x(x – 2) + 1(x – 2) = 0 (x – 2)(x + 1) = 0 x = 2, x = -1
So, the roots of 4x  2 – 4x – 8 are -1 and 2.
Relation between the sum of zeroes and coefficients: -1 + 2 = 1 = -(-4)/4
i.e. (- coefficient of x/ coefficient of x  2 ) Relation between the product of zeroes and coefficients: (-1) × 2 = -2 =  -8/4 i.e (constant/coefficient of x 2 )

Q.4: Find the quadratic polynomial if its zeroes are 0, √5.

Solution:

A quadratic polynomial can be written using the sum and product of its zeroes as:
 x 2 – (α + β)x + αβ Where α and β are the roots of the polynomial. Here, α = 0 and β = √5 So, the polynomial will be: x 2 – (0 + √5)x + 0(√5) = x 2 – √5x

Q.5: Find the value of “x” in the polynomial 2a 2 + 2xa + 5a + 10 if (a + x) is one of its factors.

Solution:

Let f(a) = 2a 2 + 2xa + 5a + 10 Since, (a + x) is a factor of 2a 2 + 2xa + 5a + 10, f(-x) = 0 So, f(-x) = 2x 2 – 2x 2 – 5x + 10 = 0 -5x + 10 = 0 5x = 10 x = 10/5
Therefore, x = 2

Q.6: How many zeros does the polynomial (x – 3) 2 – 4 have? Also, find its zeroes.

Solution:

Given polynomial is (x – 3) 2 – 4
Now, expand this expression. => x  2 + 9 – 6x – 4 = x 2 – 6x + 5
 As the polynomial has a degree of 2, the number of zeroes will be 2. Now, solve x 2 – 6x + 5 = 0 to get the roots. So, x 2 – x – 5x + 5 = 0 => x(x – 1) -5(x – 1) = 0 => (x – 1)(x – 5) = 0 x = 1, x = 5 So, the roots are 1 and 5.

Q.7: α and β are zeroes of the quadratic polynomial x 2 – 6x + y. Find the value of ‘y’ if 3α + 2β = 20.

Solution:

Let, f(x) = x² – 6x + y From the given, 3α + 2β = 20———————(i) From f(x), α + β = 6———————(ii) And, αβ = y———————(iii) Multiply equation
(ii) by 2. Then, subtract the whole equation from equation (i), => α = 20 – 12 = 8 Now, substitute this value in equation (ii), => β = 6 – 8 = -2 Substitute the values of α and β in equation (iii) to get the value of y, such as; y = αβ = (8)(-2) = -16

Q.8: If the zeroes of the polynomial x 3 – 3x 2 + x + 1 are a – b, a, a + b, then find the value of a and b.

Solution:

Let the given polynomial be: p(x) = x 3 – 3x 2 + x + 1 Given, The zeroes of the p(x) are a – b, a, and a + b.
Now, compare the given polynomial equation with general expression. px  3 + qx 2 + rx + s = x 3 – 3x 2 + x + 1 Here, p = 1, q = -3, r = 1 and s = 1 For sum of zeroes: Sum of zeroes will be = a – b + a + a + b -q/p = 3a Substitute the values q and p. -(-3)/1 = 3a a = 1
So, the zeroes are 1 – b, 1, 1 + b. For the product of zeroes: Product of zeroes = 1(1 – b)(1 + b) -s/p = 1 – 𝑏  2 => -1/1 = 1 – 𝑏 2 Or, 𝑏 2 = 1 + 1 =2 So, b = √2 Thus, 1 – √2, 1, 1 + √2 are the zeroes of equation 𝑥 3 − 3𝑥 2 + 𝑥 + 1.

Q.9: Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes, respectively.

(i) 1/4, -1

(ii) 1, 1

(iii) 4, 1

Solution:

(i) From the formulas of sum and product of zeroes, we know, Sum of zeroes = α + β Product of zeroes = αβ Given, Sum of zeroes = 1/4 Product of zeroes = -1
Therefore, if α and β are zeroes of any quadratic polynomial, then the polynomial can be written as:- x  2 – (α + β)x + αβ = x 2 – (1/4)x + (-1) = 4x 2 – x – 4 Thus, 4x 2 – x – 4 is the required quadratic polynomial.
(ii) Given, Sum of zeroes = 1 = α + β Product of zeroes = 1 = αβ Therefore, if α and β are zeroes of any quadratic polynomial, then the polynomial can be written as:- x 2 – (α + β)x + αβ = x 2 – x + 1
Thus, x  2 – x + 1 is the quadratic polynomial.
(iii) Given, Sum of zeroes, α + β = 4 Product of zeroes, αβ = 1 Therefore, if α and β are zeroes of any quadratic polynomial, then the polynomial can be written as:- x 2 – (α + β)x + αβ = x 2 – 4x + 1
Thus, x  2 – 4x +1 is the quadratic polynomial.

Q.10: Obtain all other zeroes of 3x 4 + 6x 3 – 2x 2 – 10x – 5, if two of its zeroes are √(5/3) and-√(5/3).

Solution: Since this is a polynomial of degree 4, hence there will be a total of 4 roots.

√(5/3) and-√(5/3) are zeroes of polynomial f(x). ∴ [x-√(5/3)] [x+√(5/3)] = x 2 -(5/3) Therefore, 3x 2 + 6x + 3 = 3x(x + 1) +3 (x + 1) = (3x + 3)(x + 1) = 3(x + 1)(x + 1) = 3(x + 1)(x + 1) Hence, x + 1 = 0 i.e. x = – 1 , – 1 is a zero of p(x).
So, its zeroes are given by: x = −1 and x = −1.
Therefore, all four zeroes of the given polynomial are:
√(5/3) and-√(5/3), −1 and −1.
 
Q.11: Find a quadratic polynomial whose zeroes are reciprocals of the zeroes of the polynomial f(x) = ax 2 + bx + c, a ≠ 0, c ≠ 0. Solution: Let α and β be the zeroes of the polynomial
 f(x) = ax 2 + bx + c. So, α + β = -b/a αβ = c/a
According to the given, 1/α and 1/β are the zeroes of the required quadratic polynomial.
Now, the sum of zeroes = (1/α) + (1/β) = (α + β)/αβ = (-b/a)/ (c/a) = -b/c Product of two zeroes = (1/α) (1/β) = 1/αβ = 1/(c/a) = a/c
The required quadratic polynomial = k[x  2 – (sum of zeroes)x + (product of zeroes)] = k[x 2 – (-b/c)x + (a/c)] = k[x 2 + (b/c) + (a/c)]
 
Q.12: Divide the polynomial f(x) = 3x  2 – x 3 – 3x + 5 by the polynomial g(x) = x – 1 – x 2 and verify the division algorithm. Solution:
Given, f(x) = 3x 2 – x 3 – 3x + 5 g(x) = x – 1 – x 2 Dividing f(x) = 3x 2 – x 3 – 3x + 5 by g(x) = x – 1 – x 2
Here, Quotient = q(x) = x – 2 Remainder = r(x) = 3 By division algorithm of polynomials, Dividend = (Quotient × Divisor) + Remainder
So, [q(x) × g(x)] + r(x) = (x – 2)(x – 1 – x  2 ) + 3 = x 2 – x – x 3 – 2x + 2 + 2x 2 + 3 = 3x 2 – x 3 – 3x + 5 = f(x)
Hence, the division algorithm is verified.
Q.13: For what value of k, is the polynomial f(x) = 3x 4 – 9x 3 + x 2 + 15x + k completely divisible by 3x 2 – 5?
Solution: Given, f(x) = 3x 4 – 9x 3 + x 2 + 15x + k g(x) = 3x 2 – 5 Dividing f(x) by g(x), Given that f(x) is completely divisible by 3x 2 – 5.
So, the remainder = 0 k + 10 = 0 k = -10 
Q.14: If 4 is a zero of the cubic polynomial x  3 – 3x 2 – 10x + 24, find its other two zeroes.
Solution: Given cubic polynomial  is p(x) = x 3 – 3x 2 – 10x + 24 4 is a zero of p(x). So, (x – 4) is the factor of p(x). Let us divide the given polynomial by (x – 4). Here, the quotient = x 2 + x – 6 = x 2 + 3x – 2x – 6 = x(x + 3) – 2(x + 3) = (x – 2)(x + 3) Therefore, the other two zeroes of the given cubic polynomial are 2 and -3.
Important Questions for Class 10 Maths
Chapter 1 Real Numbers
Chapter 2 Polynomials
Chapter 3 Linear Equations In Two Variables
Chapter 4 Quadratic Equations
Chapter 5 Arithmetic Progression
Chapter 6 Triangles
Chapter 7 Coordinate Geometry
Chapter 8 Introduction to Trigonometry
Chapter 9 Applications of Trigonometry
Chapter 10 Circles
Chapter 11 Constructions
Chapter 12 Areas Related to Circles
Chapter 13 Surface Areas and Volumes
Chapter 14 Statistics
Chapter 15 Probability
 

Important Questions For Class 10 Maths Chapter 2 FAQs

What is the most important chapter in class 10th maths?

Chapters on Real Numbers, Polynomials, and Quadratic Equations are critical, as they form the backbone of higher mathematical concepts.

Is 10th maths easy or hard?

No, Class 10th maths is not difficult. Average students or below-average students can pass the exam with flying colors if they use the proper strategy and implement it.

How to score 95% in class 10?

By consistent practice you can easily score 95% in class 10
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