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NCERT Class 10 Maths Chapter 2 Exercise 2.1 Solutions:How to Find Zeroes from Graphs (Key Points)

1.NCERT Class 10 Maths Chapter 2 Exercise 2.1 Solutions:How to Find Zeroes from Graphs (Key Points):

In this article “NCERT Class 10 Maths Chapter 2 Exercise 2.1 Solutions”,we will find zeroes of Polynomials (Representation in Graphs).

Also Read This Article:- Coordinate Plane in Class 9

2.NCERT Class 10 Maths Chapter 2 Exercise 2.1 Key Points

Key Concepts & Summary:Chapter 2.1 Polynomials
1.Basic Definitions & Degree of a Polynomial
(1.)Polynomial:An algebraic expression of the form P(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0 ,where powers of x are non-negative integers.
(2.)Degree of a Polynomial:The highest power of the variable x in a polynomial p(x).
(3.)Non-Polynomial Expressions:Expressions with negative or fractional exponents (e.g. \frac{1}{x-1}, \sqrt{x} + 2,) are not polynomials.
(4.)Types of Polynomials (Based on Degree)
\begin{array}{|l|c|c|c|}\hline\text{Types of} & \text{Degree} & \text{General standard} & \text{Example} \\\text{Polynomial} & & \text{Form} & \\\hline(1.) \text{ Linear polynomial} & 1 & ax + b \ (a \neq 0) & 2x + 3 \\(2.) \text{ Quadratic polynomial} & 2 & ax^2 + bx + c \ (a \neq 0) & 2x^2 + 3x - 5 \\(3.) \text{ Cubic Polynomial} & 3 & ax^3 + bx^2 + cx + d \ (a \neq 0) & x^3 - 4x \\\hline\end{array}
(5.)Value and Zeroes of a Polynomial
Value of p(x) at x=k:Obtained by substituting x=k into p(x),denoted as p(k).
(i)Zero of a Polynomial:A real number k is called a zero of the polynomial p(x) if p(k)=0.
(ii) Zero of a Linear Polynomial:For p(x)=ax+b,setting P(k) = 0 \Rightarrow k = -\frac{b}{a} = -\frac{\text{constant term}}{\text{coefficient of x}} .
(6.)Geometrical Meaning of the Zeroes of a Polynomial
(ii)The zeroes of a polynomial p(x) are precisely the x-coordinates of the points where the graph of y=p(x) intersects the x-axis.
(ii)Linear Polynomial (y=ax+b):
(iii)Graph shape:Straight Line Intersects the x-axis at exactly 1 point: \left(-\frac{b}{a}, 0\right)
(iv) Has exactly 1 zero.
(v)Quadratic Polynomial (y = ax^2 + bx + c):
(vi)Graph shape:Parabola
(a)Opens upwards  (\cup) if a > 0
(b)Opens downwards (\cap) if a < 0 Possible Cases for Zeroes:
Case 1:Intersects x-axis at 2 distinct points \Rightarrow 2 distinct zeroes.
Case 2:Touches x-axis at 1 point (coincident points)  \Rightarrow 1 zero or 2 equal zeroes.
Case 3:Does not touch or intersect the x-axis \Rightarrow No real zeroes.
(vii)Cubic Polynomial (y = ax^3 + bx^2 + cx + d):
(a)Can intersect the x-axis at a maximum of 3 points.
(b)Has at most 3 zeroes.
(7.)Important Rule to Remember A polynomial p(x) of degree n can intersect the x-axis at most n points,and therefore has at most n zeroes (8.)Value of zeroes of a Quadratic polynomial
P(x) = ax^2 + bx + c \ (a \neq 0) \\ \text{Sum of zeroes} = -\left(\frac{\text{coefficient of } x}{\text{coefficient of } x^2}\right) = -\frac{b}{a} \\ \text{Product of zeroes} = \frac{\text{constant term}}{\text{coefficient of } x^2} = \frac{c}{a}
If \alpha \text{ and } \beta are zeroes of polynomial then quadratic polynomial
K\left[x^2 - (\text{sum of zeroes})x + \text{product of polynomial}\right] \\ \Rightarrow K \left[x^2 - (\alpha + \beta)x + \alpha\beta\right] \\ \text{Hence } \alpha + \beta = -\frac{b}{a} \text{ and } \alpha\beta = \frac{b}{a}

3.How to Find Zeroes from Graphs

Example:1.The graphs of y=p(x) are given in Figure below for some polynomials p(x).Find the number of zeroes of p(x),in each case.

Example:1(i).

Solution:Polynomial not touches the x-axis. So,number of zeroes=0
Example:1(ii)

Solution:Polynomial intersects the x-axis at one point. So,number of zeroes=1
Example:1(iii)

 Solution:Polynomial intersects the x-axis at three point. So,number of zeroes=3
Example:1(iv)

Solution:Polynomial intersects the x-axis at two point. So,number of zeroes=2
Example:1(v)

Solution:Polynomial intersects the x-axis at four point. So,number of zeroes=4
Example:1(vi)

Solution:Polynomial intersects the x-axis at three point. So,number of zeroes=3
By solving the above examples,one can understand the NCERT Class 10 Maths Chapter 2 Exercise 2.1 Solutions.

Also Read This Article:- Distance Between Two Points Class 9

4.Practice Question for Students

The graphs of y=p(x) are given in Figure below for some polynomials p(x).Find the number of zeroes of p(x),in each case.
graph (1.)

(2.)

Answer:(1.)2 (2.)1
By solving the above questions, one can understand the NCERT Class 10 Maths Chapter 2 Exercise 2.1 Solutions well because the concept is well understood when you solve the questions practically.

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Also Read This Article:- Zero of a function

5.Frequently Asked Questions Related to NCERT Class 10 Maths Chapter 2 Exercise 2.1

Q:1.What is a linear polynomial?

Ans:A polynomial of degree 1 is called a linear polynomial.
For example 2x-3

Q:2.Define a Quadratic Polynomial

Ans:A polynomial of degree 2 is called a quadratic polynomial.
Example: 2x^2 + 3x - \frac{3}{5}

Q:3.What is meant by a three dimensional polynomial?

Ans:A polynomial of degree 3 is called a cubic polynomial.
Example: -x^2 + x^3
By answering the above questions,you can know about the primary terms of NCERT Class 10 Maths Chapter 2 Exercise 2.1 Solutions.

6.छात्र-छात्राओं से आज का प्रश्न (Today’s Questions to Students):

🎯 विनर्स कॉर्नर:क्या आप इस सवाल का सही जवाब दे सकते हैं?अपने नाम के साथ नीचे कमेंट करें!सही जवाब देने वाले Top Students के नाम हमारी आगे post/is article ke update में Photo ya Special Mention के साथ publish की जाएगी।अपना जवाब अभी दर्ज करें!👇
निम्नलिखित भिन्न को आरोही क्रम में लिखो। (Write the following fractions in ascending order.)
\frac{5}{9}, \frac{3}{5}, \frac{9}{12}
*पिछली प्रश्नोत्तरी का उत्तर*
पुल की लम्बाई=समय×चाल
=15 मिनट×5 किमी प्रति घंटा
=15 मिनट× \frac{5 \times 1000}{60}  मीटर/मिनट
=1250 मीटर
*Previous Quiz Solution*
Bridge length=time×speed
=15 min×5 kmph
=15 minutes× \frac{5 \times 1000}{60} m/min
=1250 m
*”This article has been prepared by **Satyam Coaching Centre** on the **Satyam Mathematics** blog.”*

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