CBSE Class 10 Maths Polynomial NCERT Solutions
This resource provides comprehensive NCERT Solutions for Chapter 2: Polynomials, designed for CBSE Class 10 Mathematics. It covers multiple-choice questions (MCQs) that test fundamental concepts of polynomials, including finding the value of coefficients when zeroes are given and constructing quadratic polynomials from their zeroes. The solutions offer step-by-step explanations, making complex problems accessible. This guide is ideal for students preparing for their board examinations, offering clear methods to solve problems related to zeroes, coefficients, and polynomial formation, thereby reinforcing their understanding and boosting confidence for exam revision.
Quick info
| Board | CBSE |
|---|---|
| Class | Class 10 |
| Subject | Maths (Exemplar) |
| Session | 2026 |
| Language | English |
| Type | NCERT Solutions |
| Chapter | 2. Polynomial |
Chapter summary
This chapter focuses on polynomials, covering key concepts like zeroes of a polynomial and their relationship with coefficients. The NCERT Solutions for Chapter 2 provide detailed explanations for solving problems, including finding unknown coefficients given the zeroes and forming quadratic polynomials when zeroes are provided. The exercises are designed to reinforce understanding of these core principles, preparing students for various question types they might encounter.
Learning outcomes
- Understand the definition and types of polynomials.
- Identify the zeroes of a polynomial.
- Apply the relationship between zeroes and coefficients of a quadratic polynomial.
- Construct a quadratic polynomial given its zeroes.
- Solve problems involving unknown coefficients in polynomial expressions.
Topics covered
Paper topics
- Polynomials
- Zeroes of a polynomial
- Quadratic polynomials
- Relationship between zeroes and coefficients
- Sum of zeroes
- Product of zeroes
- Forming quadratic polynomials
- Solving for coefficients
Important topics
- Relationship between zeroes and coefficients of a quadratic polynomial
- Constructing a quadratic polynomial from its zeroes
- Finding unknown coefficients using given zeroes
- Understanding the definition of zeroes
PDF preview
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Questions and Solutions
Question 1
(A) <math>\frac{4}{3}</math>
(B) <math>\frac{-4}{3}</math>
(C) <math>\frac{2}{3}</math>
(D) 3
We are given that -3 is one of the zeroes of the quadratic polynomial <math>(k-1)x^2 + kx + 1</math>. This means that when we substitute x = -3 into the polynomial, the result must be 0.
Let P(x) = <math>(k-1)x^2 + kx + 1</math>.
Since -3 is a zero, P(-3) = 0.
Substituting x = -3 into the polynomial:
<math display="block">(k-1)(-3)^2 + k(-3) + 1 = 0</math>
Now, we simplify the equation:
<math display="block">(k-1)(9) - 3k + 1 = 0</math>
<math display="block">9k - 9 - 3k + 1 = 0</math>
Combine like terms:
<math display="block">6k - 8 = 0</math>
Now, solve for k:
<math display="block">6k = 8</math>
<math display="block">k = \frac{8}{6}</math>
Simplify the fraction:
<math display="block">k = \frac{4}{3}</math>
Therefore, the value of k is <math>\frac{4}{3}</math>. The correct option is (A).
Question 2
(A) <math>x^2 - x + 12</math>
(B) <math>x^2 + x + 12</math>
(C) <math>\frac{x^2}{2} - \frac{x}{2} - 6</math>
(D) <math>2x^2 + 2x - 24</math>
Let the zeroes of the quadratic polynomial be <math>\alpha = -3</math> and <math>\beta = 4</math>.
We know that for a quadratic polynomial, the sum of the zeroes is given by <math>\alpha + \beta</math> and the product of the zeroes is given by <math>\alpha\beta</math>.
Calculate the sum of the zeroes:
<math display="block">\alpha + \beta = -3 + 4 = 1</math>
Calculate the product of the zeroes:
<math display="block">\alpha\beta = (-3) \times 4 = -12</math>
A quadratic polynomial can be expressed in the form <math>x^2 - (\text{sum of zeroes})x + (\text{product of zeroes})</math>.
Substituting the calculated values:
<math display="block">x^2 - (1)x + (-12) = x^2 - x - 12</math>
Now, let's check the given options. Option (C) is <math>\frac{x^2}{2} - \frac{x}{2} - 6</math>. If we multiply this by 2, we get <math>x^2 - x - 12</math>, which matches our derived polynomial.
Alternatively, we can check if the zeroes -3 and 4 satisfy the polynomial in option (C).
For <math>x = -3</math>: <math>\frac{(-3)^2}{2} - \frac{(-3)}{2} - 6 = \frac{9}{2} + \frac{3}{2} - 6 = \frac{12}{2} - 6 = 6 - 6 = 0</math>.
For <math>x = 4</math>: <math>\frac{(4)^2}{2} - \frac{4}{2} - 6 = \frac{16}{2} - 2 - 6 = 8 - 2 - 6 = 0</math>.
Since both zeroes satisfy the polynomial, option (C) is correct.
Question 3
(A) <math>a = -7, b = -1</math>
(B) <math>a = 5, b = -1</math>
(C) <math>a = 2, b = -6</math>
(D) <math>a = 0, b = -6</math>
We are given that the zeroes of the quadratic polynomial <math>P(x) = x^2 + (a+1)x + b</math> are 2 and -3.
According to the properties of polynomial zeroes, if 2 and -3 are the zeroes, then substituting these values for x in P(x) should result in 0.
Substitute x = 2:
<math display="block">P(2) = (2)^2 + (a+1)(2) + b = 0</math>
<math display="block">4 + 2a + 2 + b = 0</math>
<math display="block">6 + 2a + b = 0</math>
This gives us the first equation:
<math display="block">2a + b = -6 \quad \quad (1)</math>
Substitute x = -3:
<math display="block">P(-3) = (-3)^2 + (a+1)(-3) + b = 0</math>
<math display="block">9 - 3a - 3 + b = 0</math>
<math display="block">6 - 3a + b = 0</math>
This gives us the second equation:
<math display="block">-3a + b = -6 \quad \quad (2)</math>
Now we have a system of two linear equations with two variables, a and b:
- <math>2a + b = -6</math>
- <math>-3a + b = -6</math>
Subtract equation (2) from equation (1) to eliminate b:
<math display="block">(2a + b) - (-3a + b) = -6 - (-6)</math>
<math display="block">2a + b + 3a - b = -6 + 6</math>
<math display="block">5a = 0</math>
<math display="block">a = 0</math>
Now substitute the value of a = 0 into equation (1):
<math display="block">2(0) + b = -6</math>
<math display="block">0 + b = -6</math>
<math display="block">b = -6</math>
So, the values are <math>a = 0</math> and <math>b = -6</math>. The correct option is (D).
Common mistakes
- Incorrectly applying the sum and product of zeroes formula.
- Errors in algebraic manipulation when solving for coefficients.
- Confusing the signs in the relationship between zeroes and coefficients.
- Mistakes in forming the quadratic polynomial from given zeroes.
Revision tips
- Review the relationship between zeroes and coefficients for quadratic polynomials.
- Practice constructing polynomials from given zeroes.
- Work through each MCQ to solidify understanding of concepts.
- Pay close attention to algebraic steps in solving for unknown variables.
Practice MCQs
Q1. If one of the zeroes of the quadratic polynomial (k-1) + kx + 1 is -3, what is the value of k?
Explanation: Since -3 is a zero, substituting (k-1)(-3)^2 + k(-3) + 1 = 0 leads to 9(k-1) - 3k + 1 = 0, which simplifies to 6k - 8 = 0, giving /6 = 4/3.
Q2. Which quadratic polynomial has zeroes -3 and 4?
Explanation: For a quadratic polynomial with zeroes \(\) and \(\), the polynomial is of the form - (\(\) + \(\))x + \(\)\(\). Here, sum of zeroe= 1, and product of zeroe* 4 = -12. So, the polynomial is - (1)x + (-12) = - x - 12. Multiplying by 1/2 gives /2 - x/2 - 6.
Q3. If the zeroes of the quadratic polynomial + (a+1)x + b are 2 and -3, what are the values of a and b?
Explanation: Given zeroes are 2 and -3. Substituting x=2 into the polynomial gives 4 + 2(a+1) + b = 0, simplifying to 2a + b = -6. Substituting x=-3 gives 9 - 3(a+1) + b = 0, simplifying to -3a + b = -12. Solving the system of equations 2a + b = -6 and -3a + b = -12 yields a = 0 and b = -6.
Frequently asked questions
What is the main focus of the NCERT Solutions for Class 10 Maths Chapter 2: Polynomial?
These solutions focus on understanding the concept of zeroes of a polynomial and the relationship between the zeroes and coefficients of a quadratic polynomial. They also cover how to form a quadratic polynomial given its zeroes.
How do these solutions help in exam preparation?
The solutions provide clear, step-by-step explanations for various types of problems, including MCQs. This helps students grasp the methods required to solve questions accurately and efficiently, boosting their confidence for exams.
What is the relationship between zeroes and coefficients in a quadratic polynomial?
For a quadratic polynomial ax^2 + bx + c, if \(\alpha\) and \(\beta\) are its zeroes, then the sum of zeroes \(\alpha + \beta = -b/a\) and the product of zeroes \(\alpha\)\(\beta = c/a\).
How can I form a quadratic polynomial if I know its zeroes?
A quadratic polynomial can be formed using the formula: x^2 - (sum of zeroes)x + (product of zeroes). If the zeroes are \(\alpha\) and \(\beta\), the polynomial is x^2 - (\(\alpha + \beta\))x + \(\alpha\)\(\beta\).
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