RD Sharma Solutions for Class 10 Maths Chapter 2 Polynomials

RD Sharma Solutions for Class 10 Maths Chapter 2 – Polynomials are designed to help students master this crucial topic and excel in their board exams. Mathematics is a subject that offers great scoring potential in Class 10, and to support students in achieving high marks, we at Infinity Learn have developed comprehensive RD Sharma Class 10 Polynomials solutions. These RD Sharma solutions are crafted by our expert faculty to provide clear, detailed explanations of important concepts and solve problems step-by-step.

These solutions are an invaluable resource for any student aiming for top marks in their Mathematics exams. The RD Sharma Class 10 Chapter 2 solutions are specifically tailored to help students strengthen their exam preparation, improve their understanding, and develop problem-solving skills. 

By practicing these solutions, students will enhance their conceptual knowledge and become familiar with different methods of solving problems. This practice also builds confidence, which is crucial for performing well in exams.

Here’s an overview of the key topics covered in Class 10 Maths Chapter 2:

• Understanding polynomials and their types
• Geometrical representation of linear and quadratic polynomials
• The geometric meaning of the zeros of a polynomial
• Exploring the relationship between the zeros and coefficients of a polynomial

For easy access, the RD Sharma Class 10 PDF of this chapter is available for download, allowing students to study and practice at their own pace. By working through these RD Sharma Class 10 Solutions Polynomials, students will gain a deeper understanding of the chapter and be well-equipped to tackle related problems in their exams. The Polynomials RD Sharma Class 10 solutions also serve as a valuable reference to improve problem-solving skills and boost exam performance.

If you are looking for a structured guide, you can access the RD Sharma Chapter 2 Class 10 solutions and start practicing today to ensure that you achieve excellent results in your Mathematics exam.

Download RD Sharma Class 10 chapter 2 PDF with Solutions

RD Sharma Class 10 Chapter 2 PDF includes detailed solutions, examples, and extra questions to help you master polynomials and other topics. Click here to download the RD Sharma Class 10 Chapter 2 PDF.

Access Answers to RD Sharma Solutions for Class 10 Maths Chapter 2- Polynomials

1. Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients:

Q. g(s) = 4s2 – 4s + 1

Solution: Given,

g(s) = 4s2 – 4s + 1

To find the zeros, we put g(s) = 0

⇒ 4s² – 4s + 1 = 0

⇒ 4s² – 2s – 2s + 1= 0

⇒ 2s(2s – 1) – (2s – 1) = 0

⇒ (2s – 1)(2s – 1) = 0

This gives us 2 zeros, for

s = 1/2 and s = 1/2

Hence, the zeros of the quadratic equation are 1/2 and 1/2.

Now, for verification,

Sum of zeros = – coefficient of s / coefficient of s²

1/2 + 1/2 = – (-4) / 4

1 = 1

Product of roots = constant / coefficient of s²

1/2 x 1/2 = 1/4

Q. f(x) = x² – 2x – 8

Solution: Given,

f(x) = x2 – 2x – 8

To find the zeros, we put f(x) = 0

 x² – 2x – 8 = 0

  x² – 4x + 2x – 8 = 0

 x(x – 4) + 2(x – 4) = 0

 (x – 4)(x + 2) = 0

This gives us 2 zeros, for

x = 4 and x = -2

Hence, the zeros of the quadratic equation are 4 and -2.

Now, for verification,

Sum of zeros = – coefficient of x / coefficient of x²

4 + (-2)= – (-2) / 1

2 = 2

Product of roots = constant / coefficient of x²

4 x (-2) = (-8) / 1

-8 = -8

1/4 = 1/4

Q. h(t)=t² – 15

Solution: Given,

h(t) = t² – 15 = t2 +(0)t – 15

To find the zeros, we put h(t) = 0

⇒ t2 – 15 = 0

⇒ (t + √15)(t – √15)= 0

This gives us 2 zeros, for

t = √15 and t = -√15

Hence, the zeros of the quadratic equation are √15 and -√15.

Now, for verification,

Sum of zeros = – coefficient of t / coefficient of t2

√15 + (-√15) = – (0) / 1

0 = 0

Product of roots = constant / coefficient of t2

√15 x (-√15) = -15/1

-15 = -15

Q. h(s) = 2s² – (1 + 2√2)s + √2

Solution: Given,

h(s) = 2s2 – (1 + 2√2)s + √2

To find the zeros, we put h(s) = 0

⇒ 2s² – (1 + 2√2)s + √2 = 0

⇒ 2s² – 2√2s – s + √2 = 0

⇒ 2s(s – √2) -1(s – √2) = 0

⇒ (2s – 1)(s – √2) = 0

This gives us 2 zeros, for

x = √2 and x = 1/2

Hence, the zeros of the quadratic equation are √3 and 1.

Now, for verification,

Sum of zeros = – coefficient of s / coefficient of s²

√2 + 1/2 = – (-(1 + 2√2)) / 2

(2√2 + 1)/2 = (2√2 +1)/2

Product of roots = constant / coefficient of s²

1/2 x √2 = √2 / 2

√2 / 2 = √2 / 2

Q. f(v) = v² + 4√3v – 15

Solution: Given,

f(v) = v² + 4√3v – 15

To find the zeros, we put f(v) = 0

⇒ v² + 4√3v – 15 = 0

⇒ v² + 5√3v – √3v – 15 = 0

⇒ v(v + 5√3) – √3 (v + 5√3) = 0

⇒ (v – √3)(v + 5√3) = 0

This gives us 2 zeros, for

v = √3 and v = -5√3

Hence, the zeros of the quadratic equation are √3 and -5√3.

Now, for verification,

Sum of zeros = – coefficient of v / coefficient of v²

√3 + (-5√3) = – (4√3) / 1

-4√3 = -4√3

Product of roots = constant / coefficient of v²

√3 x (-5√3) = (-15) / 1

-5 x 3 = -15

-15 = -15