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By Karan Singh Bisht
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Updated on 11 Jun 2025, 18:07 IST
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:
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.
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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.
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
The RD Sharma Class 10 Polynomials chapter covers several essential topics that help students understand the concept of polynomials in depth. Some key topics include:
You can easily access the RD Sharma Class 10 Chapter 2 solutions for Polynomials on Infinity Learn. These solutions offer detailed explanations of every problem in the chapter and serve as an excellent study resource for students. By referring to the RD Sharma Class 10 Solutions Polynomials, you can understand the step-by-step methods used to solve problems and improve your grasp on the topics.
Yes, you can Download RD Sharma Class 10 Chapter 2 PDF with Solutions from Infinity Learn offering free resources. The RD Sharma Class 10 PDF for this chapter is designed to help students study at their own pace, with easy access to all the solutions and concepts covered in the chapter. This PDF is a great tool for revision and practice.
The Polynomials RD Sharma Class 10 solutions are invaluable for exam preparation. They provide clear, step-by-step explanations of complex problems, which can help you understand and solve polynomial problems with confidence.
These solutions are also beneficial for practicing and reinforcing key concepts from Class 10 Maths Chapter 2. By using the RD Sharma Chapter 2 Class 10 solutions, students can achieve better results and a thorough understanding of polynomials in Class 10.