Class 10 Math Chapter 2 Polynomials Solutions

Here, Class 10 Math Chapter 2 Polynomials Solutions are provided. Subject matter experts have created these solutions to help students in their board exam. They have created solutions in such a way that students can revise the entire solutions in very little time and can score good marks in the board exam. Every solution is provided in detail and in very simple language.

Class 10 Math Chapter 2 Polynomials Solutions are very important for board exam. Solving these Polynomials solutions will boost the confidence and students can score better in exam. Solutions are provided according to latest NCERT syllabus. Experts created these step-by-step solutions, which will definitely be helpful for students.

Polynomials Solutions

Exercise 2.1

class 10th maths Polynomials chapter 2 solution1
class 10th maths Polynomials chapter 2 solution2
class 10th maths Polynomials chapter 2 solution3

Exercise 2.2

class 10th maths Polynomials chapter 2 solution4
class 10th maths Polynomials chapter 2 solution5
class 10th maths Polynomials chapter 2 solution6
class 10th maths Polynomials chapter 2 solution7
class 10th maths Polynomials chapter 2 solution8
class 10th maths Polynomials chapter 2 solution9

Chapter 1: Real Numbers Solutions
Chapter 3: Pair of Linear Equations in Two Variables Solutions

We have given solutions of all exercise of class 10 math chapter 2. These solutions will be helpful for students understanding core concept of Polynomials chapters.

To understand the chapter thoroughly, it is essential to practice all the questions provided within it, as well as any additional questions. Solving previous years’ question papers also proves to be very helpful in exam preparation. You can also find other study material resources on our website.

Important Questions for NCERT Chapter 2: Polynomials

Question :Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(𝐢)x2−2x−8(𝐢𝐢)4s2−4s+1(𝐢𝐢𝐢)6x2−3−7x\textbf{(i)} x^2 – 2x – 8 \quad \textbf{(ii)} 4s^2 – 4s + 1 \quad \textbf{(iii)} 6x^2 – 3 – 7x

Answer :

x2−2x−8=(x−4)(x+2)x^2 – 2x – 8 = (x – 4)(x + 2)
The value of x2−2x−8 is zero when x−4=0 or x+2=0, i.e., when x=4 or x=−2\text{The value of } x^2 – 2x – 8 \text{ is zero when } x – 4 = 0 \text{ or } x + 2 = 0\text{, i.e., when } x = 4 \text{ or } x = -2
Therefore, the zeroes of x2−2x−8 are 4 and −2.\text{Therefore, the zeroes of } x^2 – 2x – 8 \text{ are } 4 \text{ and } -2 .
Sum of zeroes=4−2=2=−(−2)1=−(Coefficient of x)Coefficient of x2\text{Sum of zeroes} = 4 – 2 = 2 = \frac{-(-2)}{1} = \frac{-(\text{Coefficient of } x)}{\text{Coefficient of } x^2}
Product of zeroes=4×(−2)=−8=(−8)1=Constant termCoefficient of x2\text{Product of zeroes} = 4 \times (-2) = -8 = \frac{(-8)}{1} = \frac{\text{Constant term}}{\text{Coefficient of } x^2}
4s2−4s+1=(2s−1)24s^2 – 4s + 1 = (2s – 1)^2
The value of 4s2−4s+1 is zero when 2s−1=0, i.e.,\text{The value of } 4s^2 – 4s + 1 \text{ is zero when } 2s – 1 = 0\text{, i.e.,}
s=12Therefore,thezeroesof4s2−4s+1are1/2and1/2.s = \frac{1}{2} Therefore, the zeroes of 4s^2 – 4s + 1 are 1/2 and 1/2.
Sum of zeroes=12+12=1=−(−4)4=−( Coefficient of s)( Coefficient of s2)\text{Sum of zeroes} = \frac{1}{2} + \frac{1}{2} = 1 = \frac{-(-4)}{4} = \frac{-(\text{ Coefficient of } s)}{(\text{ Coefficient of } s^2)}
Product of zeroes=12×12=14=Constant termCoefficient of s2\text{Product of zeroes} = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = \frac{\text{Constant term}}{\text{Coefficient of } s^2}
(iii)6x2−3−7x=6x2−7x−3=(3x+1)(2x−3)(iii) 6x^2 – 3 – 7x = 6x^2 – 7x – 3 = (3x + 1)(2x – 3)
Thevalueof6x2−3−7xiszerowhen3x+1=0orThe value of 6x^2 – 3 – 7x is zero when 3x + 1 = 0 or
2x−3=0, i.e.,2x – 3 = 0\text{, i.e.,}
x=−13 or x=32x = \frac{-1}{3} \text{ or } x = \frac{3}{2}
Therefore,thezeroesof6x2−3−7xare−13and32.Therefore, the zeroes of 6x^2 – 3 – 7x are \frac{-1}{3} and \frac{3}{2}.
Sum of zeroes=−13+32=76=−(−7)6=−( Coefficient of x)Coefficient of x2\text{Sum of zeroes} = \frac{-1}{3} + \frac{3}{2} = \frac{7}{6} = \frac{-(-7)}{6} = \frac{-(\text{ Coefficient of } x)}{\text{Coefficient of } x^2}
Product of zeroes=−13×32=−12=−36=Constant termCoefficient of x2\text{Product of zeroes} = \frac{-1}{3} \times \frac{3}{2} = \frac{-1}{2} = \frac{-3}{6} = \frac{\text{Constant term}}{\text{Coefficient of } x^2}

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

14,−1\frac{1}{4}, -1

Answer :

14,−1\frac{1}{4}, -1
Let the polynomial be ax2+bx+c and its zeroes be α and β.\text{Let the polynomial be } ax^2 + bx + c \text{ and its zeroes be } \alpha \text{ and } \beta.
α+β=14=−ba\alpha + \beta = \frac{1}{4} = \frac{-b}{a}
αβ=−1=−44=ca\alpha\beta = -1 = \frac{-4}{4} = \frac{c}{a}
If a=4, then b=−1,c=−4\text{If } a = 4\text{, then } b = -1, c = -4
Therefore, the quadratic polynomial is 4x2−x−4.\text{Therefore, the quadratic polynomial is } 4x^2 – x – 4.

Leave a Comment

Your email address will not be published. Required fields are marked *