Class 10 RD Sharma Solutions- Chapter 2 Polynomials - Exercise 2.1 | Set 1
Last Updated :
23 Jul, 2025
The Polynomials are fundamental algebraic expressions consisting of variables raised to the whole-number powers and combined using addition, subtraction, and multiplication. They form the basis of many concepts in algebra including solving equations, graphing functions, and mathematical modeling. In this chapter, we will explore polynomials in detail and solve various problems to understand their properties and applications better. Exercise 2.1 | Set 1 focuses on solving the fundamental problems to build a strong foundation in polynomials.
Polynomials
A polynomial is an algebraic expression that involves a sum of powers of one or more variables multiplied by the coefficients. For example: 3x2+2x−5 is a polynomial in the variable x. The degree of a polynomial is the highest power of the variable in the expression. Polynomials can be classified into different types based on their degree and number of terms such as monomials, binomials, and trinomials. They are crucial for solving algebraic equations and analyzing mathematical relationships.
Question 1. Find the zeros of each of the following quadratic polynomials and verify the relationship between the zeros and their coefficients:
(i) f(x) = x2 – 2x – 8
Solution:
Given that,
f(x) = x2 – 2x – 8
To find the zeros of the equation, 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
x = 4 and x = -2
Hence, the zeros of the quadratic equation are 4 and -2.
Now, Verification
As we know that,
Sum of zeros = – coefficient of x / coefficient of x^2
4 + (-2)= – (-2) / 1
2 = 2
Product of roots = constant / coefficient of x^2
4 x (-2) = (-8) / 1
-8 = -8
Hence the relationship between zeros and their coefficients are verified.
(ii) g(s) = 4s2 – 4s + 1
Solution:
Given that,
g(s) = 4s2 – 4s + 1
To find the zeros of the equation, put g(s) = 0
= 4s2 – 4s + 1 = 0
= 4s2 – 2s – 2s + 1= 0
= 2s(2s – 1) – (2s – 1) = 0
= (2s – 1)(2s – 1) = 0
s = 1/2 and s = 1/2
Hence, the zeros of the quadratic equation are 1/2 and 1/2.
Now, Verification
As we know that,
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
1/4 = 1/4
Hence the relationship between zeros and their coefficients are verified.
(iii) h(t)=t2 – 15
Solution:
Given that,
h(t) = t2 – 15 = t2 +(0)t – 15
To find the zeros of the equation, put h(t) = 0
= t2 – 15 = 0
= (t + √15)(t – √15)= 0
t = √15 and t = -√15
Hence, the zeros of the quadratic equation are √15 and -√15.
Now, Verification
As we know that,
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
Hence the relationship between zeros and their coefficients are verified.
(iv) f(x) = 6x2 – 3 – 7x
Solution:
Given that,
f(x) = 6x2 – 3 – 7x
To find the zeros of the equation, we put f(x) = 0
= 6x2 – 3 – 7x = 0
= 6x2 – 9x + 2x – 3 = 0
= 3x(2x – 3) + 1(2x – 3) = 0
= (2x – 3)(3x + 1) = 0
x = 3/2 and x = -1/3
Hence, the zeros of the quadratic equation are 3/2 and -1/3.
Now, Verification
As we know that,
Sum of zeros = – coefficient of x / coefficient of x2
3/2 + (-1/3) = – (-7) / 6
7/6 = 7/6
Product of roots = constant / coefficient of x2
3/2 x (-1/3) = (-3) / 6
-1/2 = -1/2
Hence the relationship between zeros and their coefficients are verified.
(v) p(x) = x2 + 2√2x – 6
Solution:
Given that,
p(x) = x2 + 2√2x – 6
To find the zeros of the equation, put p(x) = 0
= x2 + 2√2x – 6 = 0
= x2 + 3√2x – √2x – 6 = 0
= x(x + 3√2) – √2 (x + 3√2) = 0
= (x – √2)(x + 3√2) = 0
x = √2 and x = -3√2
Hence, the zeros of the quadratic equation are √2 and -3√2.
Now, Verification
As we know that,
Sum of zeros = – coefficient of x / coefficient of x2
√2 + (-3√2) = – (2√2) / 1
-2√2 = -2√2
Product of roots = constant / coefficient of x2
√2 x (-3√2) = (-6) / 2√2
-3 x 2 = -6/1
-6 = -6
Hence the relationship between zeros and their coefficients are verified.
(vi) q(x)=√3x2 + 10x + 7√3
Solution:
Given that,
q(x) = √3x2 + 10x + 7√3
To find the zeros of the equation, put q(x) = 0
= √3x2 + 10x + 7√3 = 0
= √3x2 + 3x +7x + 7√3x = 0
= √3x(x + √3) + 7 (x + √3) = 0
= (x + √3)(√3x + 7) = 0
x = -√3 and x = -7/√3
Hence, the zeros of the quadratic equation are -√3 and -7/√3.
Now, Verification
As we know that,
Sum of zeros = – coefficient of x / coefficient of x2
-√3 + (-7/√3) = – (10) /√3
(-3-7)/ √3 = -10/√3
-10/ √3 = -10/√3
Product of roots = constant / coefficient of x2
(-√3) x (-7/√3) = (7√3) / √3
7 = 7
Hence the relationship between zeros and their coefficients are verified.
(vii) f(x) = x2 – (√3 + 1)x + √3
Solution:
Given that,
f(x) = x2 – (√3 + 1)x + √3
To find the zeros of the equation, put f(x) = 0
= x2 – (√3 + 1)x + √3 = 0
= x2 – √3x – x + √3 = 0
= x(x – √3) – 1 (x – √3) = 0
= (x – √3)(x – 1) = 0
x = √3 and x = 1
Hence, the zeros of the quadratic equation are √3 and 1.
Now, Verification
Sum of zeros = – coefficient of x / coefficient of x2
√3 + 1 = – (-(√3 +1)) / 1
√3 + 1 = √3 +1
Product of roots = constant / coefficient of x2
1 x √3 = √3 / 1
√3 = √3
Hence the relationship between zeros and their coefficients are verified.
(viii) g(x) = a(x2+1)–x(a2+1)
Solution:
Given that,
g(x) = a(x2+1)–x(a2+1)
To find the zeros of the equation put g(x) = 0
= a(x2+1)–x(a2+1) = 0
= ax2 + a − a2x – x = 0
= ax2 − a2x – x + a = 0
= ax(x − a) − 1(x – a) = 0
= (x – a)(ax – 1) = 0
x = a and x = 1/a
Hence, the zeros of the quadratic equation are a and 1/a.
Now, Verification :
As we know that,
Sum of zeros = – coefficient of x / coefficient of x2
a + 1/a = – (-(a2 + 1)) / a
(a^2 + 1)/a = (a2 + 1)/a
Product of roots = constant / coefficient of x2
a x 1/a = a / a
1 = 1
Hence the relationship between zeros and their coefficients are verified.
(ix) h(s) = 2s2 – (1 + 2√2)s + √2
Solution:
Given that,
h(s) = 2s2 – (1 + 2√2)s + √2
To find the zeros of the equation 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
x = √2 and x = 1/2
Hence, the zeros of the quadratic equation are √3 and 1.
Now, Verification
As we know that,
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
Hence the relationship between zeros and their coefficients are verified.
(x) f(v) = v2 + 4√3v – 15
Solution:
Given that,
f(v) = v2 + 4√3v – 15
To find the zeros of the equation 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
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
Hence the relationship between zeros and their coefficients are verified.
(xi) p(y) = y2 + (3√5/2)y – 5
Solution:
Given that,
p(y) = y2 + (3√5/2)y – 5
To find the zeros of the equation put f(v) = 0
= y2 + (3√5/2)y – 5 = 0
= y2 – √5/2 y + 2√5y – 5 = 0
= y(y – √5/2) + 2√5 (y – √5/2) = 0
= (y + 2√5)(y – √5/2) = 0
This gives us 2 zeros,
y = √5/2 and y = -2√5
Hence, the zeros of the quadratic equation are √5/2 and -2√5.
Now, Verification
As we know that,
Sum of zeros = – coefficient of y / coefficient of y2
√5/2 + (-2√5) = – (3√5/2) / 1
-3√5/2 = -3√5/2
Product of roots = constant / coefficient of y2
√5/2 x (-2√5) = (-5) / 1
– (√5)2 = -5
-5 = -5
Hence the relationship between zeros and their coefficients are verified.
(xii) q(y) = 7y2 – (11/3)y – 2/3
Solution:
Given that,
q(y) = 7y2 – (11/3)y – 2/3
To find the zeros of the equation put q(y) = 0
= 7y2 – (11/3)y – 2/3 = 0
= (21y2 – 11y -2)/3 = 0
= 21y2 – 11y – 2 = 0
= 21y2 – 14y + 3y – 2 = 0
= 7y(3y – 2) – 1(3y + 2) = 0
= (3y – 2)(7y + 1) = 0
y = 2/3 and y = -1/7
Hence, the zeros of the quadratic equation are 2/3 and -1/7.
Now, Verification
As we know that,
Sum of zeros = – coefficient of y / coefficient of y2
2/3 + (-1/7) = – (-11/3) / 7
-11/21 = -11/21
Product of roots = constant / coefficient of y2
2/3 x (-1/7) = (-2/3) / 7
– 2/21 = -2/21
Hence the relationship between zeros and their coefficients are verified.
Question 2. For each of the following, find a quadratic polynomial whose sum and product respectively of the zeros are as given. Also, find the zeros of these polynomials by factorization.
(i) -8/3, 4/3
Solution:
As we know that the quadratic polynomial formed for the given sum and product of zeros is given by : f(x) = x2 + -(sum of zeros) x + (product of roots)
The sum of zeros = -8/3 and
Product of zero = 4/3
Therefore,
Required polynomial f(x) is,
= x2 – (-8/3)x + (4/3)
= x2 + 8/3x + (4/3)
To find the zeros we put f(x) = 0
= x2 + 8/3x + (4/3) = 0
= 3x2 + 8x + 4 = 0
= 3x2 + 6x + 2x + 4 = 0
= 3x(x + 2) + 2(x + 2) = 0
= (x + 2) (3x + 2) = 0
= (x + 2) = 0 and, or (3x + 2) = 0
Hence, the two zeros are -2 and -2/3.
(ii) 21/8, 5/16
Solution:
As we know that the quadratic polynomial formed for the given sum and product of zeros is given by : f(x) = x2 + -(sum of zeros) x + (product of roots)
The sum of zeros = 21/8 and
Product of zero = 5/16
Therefore,
The required polynomial f(x) is,
= x2 – (21/8)x + (5/16)
= x2 – 21/8x + 5/16
To find the zeros we put f(x) = 0
= x2 – 21/8x + 5/16 = 0
= 16x2 – 42x + 5 = 0
= 16x2 – 40x – 2x + 5 = 0
= 8x(2x – 5) – 1(2x – 5) = 0
= (2x – 5) (8x – 1) = 0
= (2x – 5) = 0 and, or (8x – 1) = 0
Hence, the two zeros are 5/2 and 1/8.
(iii) -2√3, -9
Solution:
As we know that the quadratic polynomial formed for the given sum and product of zeros is given by : f(x) = x2 + -(sum of zeros) x + (product of roots)
The sum of zeros = -2√3 and
Product of zero = -9
Therefore,
The required polynomial f(x) is,
= x2 – (-2√3)x + (-9)
= x2 + 2√3x – 9
To find the zeros we put f(x) = 0
= x2 + 2√3x – 9 = 0
= x2 + 3√3x – √3x – 9 = 0
= x(x + 3√3) – √3(x + 3√3) = 0
= (x + 3√3) (x – √3) = 0
= (x + 3√3) = 0 and, or (x – √3) = 0
Hence, the two zeros are -3√3and √3.
(iv) -3/2√5, -1/2
Solution:
As we know that the quadratic polynomial formed for the given sum and product of zeros is given by : f(x) = x2 + -(sum of zeros) x + (product of roots)
The sum of zeros = -3/2√5 and
Product of zero = -1/2
Therefore,
The required polynomial f(x) is,
= x2 – (-3/2√5)x + (-1/2)
= x2 + 3/2√5x – 1/2
To find the zeros we put f(x) = 0
= x2 + 3/2√5x – 1/2 = 0
= 2√5x2 + 3x – √5 = 0
= 2√5x2 + 5x – 2x – √5 = 0
= √5x(2x + √5) – 1(2x + √5) = 0
= (2x + √5) (√5x – 1) = 0
= (2x + √5) = 0 and, or (√5x – 1) = 0
Hence, the two zeros are -√5/2 and 1/√5.
Question 3. If α and β are the zeros of the quadratic polynomial f(x) = x2 – 5x + 4, find the value of 1/α + 1/β – 2αβ.
Solution:
Given that,
α and β are the roots of the quadratic polynomial f(x) where a = 1, b = -5 and c = 4
Using these values we can find,
Sum of the roots = α+β = -b/a = – (-5)/1 = 5,
Product of the roots = αβ = c/a = 4/1 = 4
We have to find 1/α +1/β – 2αβ
= [(α +β)/ αβ] – 2αβ
= (5)/ 4 – 2(4) = 5/4 – 8 = -27/ 4
Question 4. If α and β are the zeros of the quadratic polynomial p(y) = 5y2 – 7y + 1, find the value of 1/α+1/β.
Solution:
Given that,
α and β are the roots of the quadratic polynomial f(x) where a =5, b = -7 and c = 1,
Using these values we can find,
Sum of the roots = α+β = -b/a = – (-7)/5 = 7/5
Product of the roots = αβ = c/a = 1/5
We have to find 1/α +1/β
= (α +β)/ αβ
= (7/5)/ (1/5) = 7
Question 5. If α and β are the zeros of the quadratic polynomial f(x)=x2 – x – 4, find the value of 1/α+1/β–αβ.
Solution:
Given that,
α and β are the roots of the quadratic polynomial f(x) where a = 1, b = -1 and c = – 4
So, we can find,
Sum of the roots = α+β = -b/a = – (-1)/1 = 1
Product of the roots = αβ = c/a = -4 /1 = – 4
We have to find, 1/α +1/β – αβ
= [(α +β)/ αβ] – αβ
= [(1)/ (-4)] – (-4) = -1/4 + 4 = 15/ 4
Question 6. If α and β are the zeroes of the quadratic polynomial f(x) = x2 + x – 2, find the value of 1/α – 1/β.
Solution:
Given that:
α and β are the roots of the quadratic polynomial f(x) where a = 1, b = 1 and c = – 2
So, we can find
Sum of the roots = α+β = -b/a = – (1)/1 = -1,
Product of the roots = αβ = c/a = -2 /1 = – 2
We have to find, 1/α – 1/β
= [(β – α)/ αβ] = [β-α]/(αβ) x (α-β)/αβ = (√(α+β)2 -4αβ) / αβ = √(1+8) / 2 = 3/2
Question 7. If one of the zero of the quadratic polynomial f(x) = 4x2 – 8kx – 9 is negative of the other, then find the value of k.
Solution:
Given that,
The quadratic polynomial f(x) where a = 4, b = -8k and c = – 9
And, for roots to be negative of each other, let us assume that the roots α and – α.
Using these values we can find,
Sum of the roots = α – α = -b/a = – (-8k)/1 = 8k = 0 [∵ α – α = 0]
= k = 0
Question 8. If the sum of the zeroes of the quadratic polynomial f(t)=kt2 + 2t + 3k is equal to their product, then find the value of k.
Solution:
Given that,
The quadratic polynomial f(t)=kt2 + 2t + 3k, where a = k, b = 2 and c = 3k ,
Sum of the roots = Product of the roots
= (-b/a) = (c/a)
= (-2/k) = (3k/k)
= (-2/k) = 3
Hence k = -2/3
Question 9. If α and β are the zeros of the quadratic polynomial p(x) = 4x2 – 5x – 1, find the value of α2 β+α β2
Solution:
Given that,
α and β are the roots of the quadratic polynomial p(x) where a = 4, b = -5 and c = -1
Using these values we can find,
Sum of the roots = α+β = -b/a = – (-5)/4 = 5/4
Product of the roots = αβ = c/a = -1/4
We have to find, α^2 β+α β^2
= αβ(α +β)
= (-1/4)(5/4) = -5/16
Question 10. If α and β are the zeros of the quadratic polynomial f(t)=t2– 4t + 3, find the value of α4 β3+α3 β4.
Solution:
Given that,
α and β are the roots of the quadratic polynomial f(t) where a = 1, b = -4 and c = 3
Using these values we can find,
Sum of the roots = α+β = -b/a = – (-4)/1 = 4 ,
Product of the roots = αβ = c/a = 3/1 = 3
We have to find, α4 β3 + α3 β4
= α3 β3 (α +β)
= (αβ)3 (α +β)
= (3)3 (4) = 27 x 4 = 108
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Summary
Exercise 2.1 | Set 1 of RD Sharma's Class 10 Mathematics textbook introduces the basic concepts of polynomials. This section covers the definition of polynomials, their classification based on degree and number of terms, identification of polynomials, and basic operations with polynomials. Students learn to recognize polynomials in different forms, determine their degrees, identify coefficients and constants, and understand the concept of zeros of polynomials. The exercise also includes problems on evaluating polynomials for given values and finding the values of variables for which a polynomial equals zero.
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Calculus
Introduction to Differential CalculusDifferential calculus is a branch of calculus that deals with the study of rates of change of functions and the behaviour of these functions in response to infinitesimal changes in their independent variables.Some of the prerequisites for Differential Calculus include:Independent and Dependent Varia
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Limits in CalculusIn mathematics, a limit is a fundamental concept that describes the behaviour of a function or sequence as its input approaches a particular value. Limits are used in calculus to define derivatives, continuity, and integrals, and they are defined as the approaching value of the function with the inp
12 min read
Continuity of FunctionsContinuity of functions is an important unit of Calculus as it forms the base and it helps us further to prove whether a function is differentiable or not. A continuous function is a function which when drawn on a paper does not have a break. The continuity can also be proved using the concept of li
13 min read
DifferentiationDifferentiation in mathematics refers to the process of finding the derivative of a function, which involves determining the rate of change of a function with respect to its variables.In simple terms, it is a way of finding how things change. Imagine you're driving a car and looking at how your spee
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Differentiability of a Function | Class 12 MathsContinuity or continuous which means, "a function is continuous at its domain if its graph is a curve without breaks or jumps". A function is continuous at a point in its domain if its graph does not have breaks or jumps in the immediate neighborhood of the point. Continuity at a Point: A function f
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IntegrationIntegration, in simple terms, is a way to add up small pieces to find the total of something, especially when those pieces are changing or not uniform.Imagine you have a car driving along a road, and its speed changes over time. At some moments, it's going faster; at other moments, it's slower. If y
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Probability and Statistics
Basic Concepts of ProbabilityProbability is defined as the likelihood of the occurrence of any event. It is expressed as a number between 0 and 1, where 0 is the probability of an impossible event and 1 is the probability of a sure event.Concepts of Probability are used in various real life scenarios : Stock Market : Investors
7 min read
Bayes' TheoremBayes' Theorem is a mathematical formula used to determine the conditional probability of an event based on prior knowledge and new evidence. It adjusts probabilities when new information comes in and helps make better decisions in uncertain situations.Bayes' Theorem helps us update probabilities ba
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Probability Distribution - Function, Formula, TableA probability distribution is a mathematical function or rule that describes how the probabilities of different outcomes are assigned to the possible values of a random variable. It provides a way of modeling the likelihood of each outcome in a random experiment.While a Frequency Distribution shows
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Descriptive StatisticStatistics is the foundation of data science. Descriptive statistics are simple tools that help us understand and summarize data. They show the basic features of a dataset, like the average, highest and lowest values and how spread out the numbers are. It's the first step in making sense of informat
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What is Inferential Statistics?Inferential statistics is an important tool that allows us to make predictions and conclusions about a population based on sample data. Unlike descriptive statistics, which only summarize data, inferential statistics let us test hypotheses, make estimates, and measure the uncertainty about our predi
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Measures of Central Tendency in StatisticsCentral tendencies in statistics are numerical values that represent the middle or typical value of a dataset. Also known as averages, they provide a summary of the entire data, making it easier to understand the overall pattern or behavior. These values are useful because they capture the essence o
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Set TheorySet theory is a branch of mathematics that deals with collections of objects, called sets. A set is simply a collection of distinct elements, such as numbers, letters, or even everyday objects, that share a common property or rule.Example of SetsSome examples of sets include:A set of fruits: {apple,
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Practice