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# NCERT Solutions for Class 10 Maths Chapter 2: Polynomials - Exercise 2.2

Last updated date: 07th Sep 2024
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## NCERT Solutions for Class 10 Maths Chapter 2 Polynomials

NCERT Solution for Class 10 Maths PDF includes the solved Exercise 2.2 from Chapter 2 Polynomials. These solutions were developed by math professionals who have updated based on the current academic year 2024-25. The Answers for Class 10 Math Ex 2.2, Chapter 2 are provided here in an accessible and step-by-step manner. Also, This article has covered every crucial topics, such as Zeroes and Coefficients of Polynomials. After going over these NCERT Solutions for Class 10 Maths, students would undoubtedly be able to tackle these problems with ease.

Table of Content
1. NCERT Solutions for Class 10 Maths Chapter 2 Polynomials
2. Glance on Class 10 Maths Exercise 2.2 Polynomials Chapter 2
3. Topics Covered in Class 10 Maths Chapter 2 Exercise 2.2
4. Access Class 10 Maths Chapter 2 Exercise 2.2 Solutions – Polynomials
4.1Exercise 2.2
5. NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.2
5.1Definition
5.2Degree of a Polynomial
5.3Types of Polynomial
5.4Value of a Polynomial
5.5Zero of a Polynomial
5.6Graph of a Polynomial
5.7Meaning of the Zeroes of a Quadratic Polynomial
5.8The Following Observations Can Be Made On Class 10 Maths ex 2.2
5.9Note in Exercise 2.2 Class 10 Math:
6. NCERT Solutions Class 10 Maths Chapter 2 Exercises
7. Other Related Links for Chapter 2 Polynomials Of Class 10
8. Chapter-Specific NCERT Solutions for Class 10 Maths
FAQs

## Glance on Class 10 Maths Exercise 2.2 Polynomials Chapter 2

• Class 10 Maths Exercise 2.2 dives into polynomials, which are algebraic expressions formed with variables and their non-negative integer powers.

• This exercise concentrates on quadratic polynomials, which are of degree 2.

• The focus is on finding the zeroes of these polynomials. The zeroes are the values of the variable (x in this case) that make the polynomial equal to zero.

• The exercise also explores the relationship between the zeroes and the coefficients of a polynomial.

• There's a surprising connection –  the sum of the zeroes is equal to the negative of the coefficient of the first-degree term (linear term), and the product of the zeroes is equal to the constant term.

• Exercise 2.2 helps you with identifying polynomials

• Understanding the concept of degree of a polynomial

• Finding zeroes of quadratic polynomials

• Discovering the relationship between zeroes and coefficients

## Topics Covered in Class 10 Maths Chapter 2 Exercise 2.2

1. Understanding Polynomials

2. Types of Polynomials

3. Degree of a Polynomial

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## Access Class 10 Maths Chapter 2 Exercise 2.2 Solutions – Polynomials

### Exercise 2.2

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

(i) ${{\text{x}}^{\text{2}}}-\text{2x}-\text{8}$

Ans:

Given: ${{\text{x}}^{\text{2}}}-\text{2x}-\text{8}$.

Now factorize the given polynomial to get the roots.

$\Rightarrow \text{(x}-\text{4)(x+2)}$

The value of ${{\text{x}}^{\text{2}}}-\text{2x}-\text{8}$ is zero.

when $\text{x}-\text{4=0}$ or $\text{x+2=0}$. i.e., $\text{x = 4}$ or $\text{x = }-\text{2}$

Therefore, the zeroes of ${{\text{x}}^{\text{2}}}-\text{2x}-\text{8}$ are $\text{4}$ and $-2$.

Now, Sum of zeroes$\text{=4}-\text{2= 2 =}-\dfrac{\text{2}}{\text{1}}\text{=}-\dfrac{\text{Coefficient of x}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}$

$\therefore Sum\,of\,zeroes=-\dfrac{Coefficient\,of\,x}{Coefficient\,of\,{{x}^{2}}}$

Product of zeroes $\text{=4 }\!\!\times\!\!\text{ (}-\text{2)=}-\text{8=}\dfrac{\text{(}-\text{8)}}{\text{1}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}$

$\therefore \Pr oduct\,of\,zeroes=\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}$ .

(ii) $4{{s}^{2}}-4s+1$

Ans:

Given: $\text{4}{{\text{s}}^{\text{2}}}-\text{4s+1}$

Now factorize the given polynomial to get the roots.

$\Rightarrow {{\text{(2s}-\text{1)}}^{\text{2}}}$

The value of $\text{4}{{\text{s}}^{\text{2}}}-\text{4s+1}$ is zero.

when $\text{2s}-\text{1=0}$, $\text{2s}-\text{1=0}$. i.e., $\text{s =}\dfrac{\text{1}}{\text{2}}$ and $\text{s =}\dfrac{\text{1}}{\text{2}}$

Therefore, the zeroes of $\text{4}{{\text{s}}^{\text{2}}}-\text{4s+1}$ are $\dfrac{\text{1}}{\text{2}}$ and $\dfrac{\text{1}}{\text{2}}$.

Now, Sum of zeroes$\text{=}\dfrac{\text{1}}{\text{2}}\text{+}\dfrac{\text{1}}{\text{2}}\text{=1=}\dfrac{\text{(}-\text{4)}}{\text{4}}\text{=}\dfrac{-\text{(Coefficient of s)}}{\text{Coefficient of }{{\text{s}}^{\text{2}}}}$

$\therefore \text{Sum of zeroes =}\dfrac{-\text{(Coefficient of s)}}{\text{Coefficient of }{{\text{s}}^{\text{2}}}}$

Product of zeroes$=\dfrac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{2}}=\dfrac{\text{1}}{\text{4}}=\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{s}}^{\text{2}}}}$

$\therefore Product of zeroes=\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{s}}^{\text{2}}}}$ .

(iii) $6{{x}^{2}}-3-7x$

Ans:

Given: $\text{6}{{\text{x}}^{\text{2}}}-\text{3}-\text{7x }$

$\Rightarrow \text{6}{{\text{x}}^{\text{2}}}-\text{7x}-\text{3}$

Now factorize the given polynomial to get the roots.

$\Rightarrow \text{(3x+1)(2x}-\text{3)}$

The value of  $\text{6}{{\text{x}}^{\text{2}}}-\text{3}-\text{7x }$ is zero.

when $\text{3x+1=0}$ or $\text{2x}-\text{3=0}$. i.e., $\text{x =}\dfrac{-\text{1}}{\text{3}}$ or $\text{x =}\dfrac{\text{3}}{\text{2}}$.

Therefore, the zeroes of  $\text{6}{{\text{x}}^{\text{2}}}-\text{3}-\text{7x }$ are $\dfrac{\text{-1}}{\text{3}}$ and $\dfrac{\text{3}}{\text{2}}$.

Now, Sum of zeroes$\text{=}\dfrac{-\text{1}}{\text{3}}\text{+}\dfrac{\text{3}}{\text{2}}\text{=}\dfrac{\text{7}}{\text{6}}\text{=}\dfrac{-\text{(}-\text{7)}}{\text{6}}\text{=}\dfrac{-\text{(Coefficient of x)}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}$

$\therefore Sum of zeroes\text{=}\dfrac{-\text{(Coefficient of x)}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}$

Product of zeroes$\text{=}\dfrac{-\text{1}}{\text{3}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{3}}{\text{2}}\text{=}\dfrac{-\text{3}}{\text{6}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}$

$\therefore \text{Product of zeroes=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}$

(iv) $4{{u}^{2}}+8u$

Ans:

Given: $\text{4}{{\text{u}}^{\text{2}}}\text{+8u}$

$\Rightarrow \text{4}{{\text{u}}^{\text{2}}}\text{+8u+0}$

$\Rightarrow \text{4u(u+2)}$

The value of $\text{4}{{\text{u}}^{\text{2}}}\text{+8u}$ is zero.

when $\text{4u=0}$ or $\text{u+2=0}$. i.e., $\text{u = 0}$ or $\text{u =}-\text{2}$

Therefore, the zeroes of $\text{4}{{\text{u}}^{\text{2}}}\text{+8u}$ are $\text{0}$ and $\text{-2}$.

Now, Sum of zeroes$\text{=0+(}-\text{2)=}-\text{2=}\dfrac{-\text{8}}{\text{4}}\text{=}\dfrac{-\text{(Coefficient of u)}}{\text{Coefficient of }{{\text{u}}^{\text{2}}}}$

$\therefore \text{Sum of zeroes=}\dfrac{-\text{(Coefficient of u)}}{\text{Coefficient of }{{\text{u}}^{\text{2}}}}$

Product of zeroes$\text{=0 }\!\!\times\!\!\text{ (}-\text{2)= 0 =}\dfrac{\text{0}}{\text{4}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{u}}^{\text{2}}}}$

$\therefore \text{Product of zeroes=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{u}}^{\text{2}}}}$

(v) ${{t}^{2}}-15$

Ans:

Given: ${{\text{t}}^{\text{2}}}-\text{15}$

$\Rightarrow {{\text{t}}^{\text{2}}}-\text{0t}-\text{15}$

Now factorize the given polynomial to get the roots.

$\Rightarrow \text{(t}-\sqrt{\text{15}}\text{)(t +}\sqrt{\text{15}}\text{)}$

The value of ${{\text{t}}^{\text{2}}}-\text{15}$ is zero.

when $\text{t}-\sqrt{\text{15}}\text{=0}$ or $\text{t+}\sqrt{\text{15}}\text{=0}$, i.e., $\text{t=}\sqrt{\text{15}}$ or $\text{t=}-\sqrt{\text{15}}$

Therefore, the zeroes of ${{\text{t}}^{\text{2}}}-\text{15}$ are $\sqrt{\text{15}}$ and $-\sqrt{\text{15}}$.

Now, Sum of zeroes$\text{=}\sqrt{\text{15}}\text{+(}-\sqrt{\text{15}}\text{)=0=}\dfrac{-\text{0}}{\text{1}}\text{=}\dfrac{-\text{(Coefficient of t)}}{\text{Coefficient of }{{\text{t}}^{\text{2}}}}$

$\therefore \text{Sum of zeroes =}\dfrac{-\text{(Coefficient of t)}}{\text{Coefficient of }{{\text{t}}^{\text{2}}}}$

Product of zeroes$\text{=}\left( \sqrt{\text{15}} \right)\times \left( -\sqrt{\text{15}} \right)\text{=}-\text{15=}\dfrac{-\text{15}}{\text{1}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{t}}^{\text{2}}}}$

$\therefore \text{Product of zeroes =}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{t}}^{\text{2}}}}$ .

(vi) $3{{x}^{2}}-x-4$

Ans

Given: $\text{3}{{\text{x}}^{\text{2}}}-\text{x}-\text{4}$

Now factorize the given polynomial to get the roots.

$\Rightarrow \left( 3x-4 \right)(x+1)$

The value of $\text{3}{{\text{x}}^{\text{2}}}-\text{x}-\text{4}$ is zero.

when $\text{3x}-\text{4=0}$ or $\text{x+1=0}$, i.e., $\text{x=}\dfrac{\text{4}}{\text{3}}$ or $\text{x=}-\text{1}$

Therefore, the zeroes of $\text{3}{{\text{x}}^{\text{2}}}-\text{x}-\text{4}$ are $\dfrac{\text{4}}{\text{3}}$ and $\text{-1}$.

Now, Sum of zeroes$\text{=}\dfrac{\text{4}}{\text{3}}\text{+(}-\text{1)=}\dfrac{\text{1}}{\text{3}}\text{=}\dfrac{-\text{(}-\text{1)}}{\text{3}}\text{=}\dfrac{-\text{(Coefficient of x)}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}$

$\therefore \text{Sum of zeroes=}\dfrac{-\text{(Coefficient of x)}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}$

Product of zeroes$\text{=}\dfrac{\text{4}}{\text{3}}\times \text{(}-\text{1)=}\dfrac{-\text{4}}{\text{3}}\text{=}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}$

$\therefore \text{Product of zeroes =}\dfrac{\text{Constant term}}{\text{Coefficient of }{{\text{x}}^{\text{2}}}}$ .

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

(i) $\dfrac{1}{4},-1$

Ans:

Given: $\dfrac{\text{1}}{\text{4}}\text{,-1}$

Let the zeroes of polynomial be $\text{ }\!\!\alpha\!\!\text{ }$ and $\text{ }\!\!\beta\!\!\text{ }$.

Then,

$\text{ }\!\!\alpha\!\!\text{ + }\!\!\beta\!\!\text{ =}\dfrac{\text{1}}{\text{4}}$

$\text{ }\!\!\alpha\!\!\text{ }\!\!\beta\!\!\text{ =}-\text{1}$

Hence, the required polynomial is ${{\text{x}}^{\text{2}}}-\left( \alpha +\beta \right)\text{x+}\alpha \beta$.

$\Rightarrow {{x}^{2}}-\dfrac{1}{4}x-1$

$\Rightarrow 4{{x}^{2}}-x-4$

Therefore, the quadratic polynomial is $\text{4}{{\text{x}}^{\text{2}}}-\text{x}-\text{4}$.

(ii) $\sqrt{2},\dfrac{1}{3}$

Ans:

Given: $\sqrt{\text{2}}\text{,}\dfrac{\text{1}}{\text{3}}$

Let the zeroes of polynomial be $\text{ }\!\!\alpha\!\!\text{ }$ and $\text{ }\!\!\beta\!\!\text{ }$.

Then,

$\text{ }\!\!\alpha\!\!\text{ + }\!\!\beta\!\!\text{ =}\sqrt{\text{2}}$

$\text{ }\!\!\alpha\!\!\text{ }\!\!\beta\!\!\text{ =}\dfrac{\text{1}}{\text{3}}$

Hence, the required polynomial is ${{\text{x}}^{\text{2}}}-\left( \alpha +\beta \right)\text{x+}\alpha \beta$.

$\Rightarrow {{\text{x}}^{\text{2}}}-\sqrt{2}\text{x+}\dfrac{1}{3}$

$\Rightarrow 3{{\text{x}}^{\text{2}}}-3\sqrt{2}\text{x+1}$

Therefore, the quadratic polynomial is $3{{\text{x}}^{\text{2}}}-3\sqrt{2}\text{x+1}$.

(iii) $0,\sqrt{5}$

Ans:

Given: $\text{0,}\sqrt{\text{5}}$

Let the zeroes of polynomial be $\text{ }\!\!\alpha\!\!\text{ }$ and $\text{ }\!\!\beta\!\!\text{ }$.

Then,

$\text{ }\!\!\alpha\!\!\text{ + }\!\!\beta\!\!\text{ = 0}$

$\text{ }\!\!\alpha\!\!\text{ }\!\!\beta\!\!\text{ =}\sqrt{\text{5}}$

Hence, the required polynomial is ${{\text{x}}^{\text{2}}}-\left( \alpha +\beta \right)\text{x+}\alpha \beta$.

$\Rightarrow {{\text{x}}^{\text{2}}}-0\text{x+}\sqrt{5}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+}\sqrt{5}$

Therefore, the quadratic polynomial is ${{\text{x}}^{\text{2}}}\text{+}\sqrt{\text{5}}$.

(iv) $1,1$

Ans:

Given: $\text{1,1}$

Let the zeroes of polynomial be $\text{ }\!\!\alpha\!\!\text{ }$ and $\text{ }\!\!\beta\!\!\text{ }$.

Then,

$\text{ }\!\!\alpha\!\!\text{ + }\!\!\beta\!\!\text{ =1}$

$\text{ }\!\!\alpha\!\!\text{ }\!\!\beta\!\!\text{ =1}$

Hence, the required polynomial is ${{\text{x}}^{\text{2}}}-\left( \alpha +\beta \right)\text{x+}\alpha \beta$.

$\Rightarrow {{\text{x}}^{\text{2}}}-\text{1x+1}$

Therefore, the quadratic polynomial is ${{\text{x}}^{\text{2}}}-\text{x+1}$.

(v) $-\dfrac{1}{4},\dfrac{1}{4}$

Ans:

Given: $-\dfrac{\text{1}}{\text{4}}\text{,}\dfrac{\text{1}}{\text{4}}$

Let the zeroes of polynomial be $\text{ }\!\!\alpha\!\!\text{ }$ and $\text{ }\!\!\beta\!\!\text{ }$.

Then,

$\text{ }\!\!\alpha\!\!\text{ + }\!\!\beta\!\!\text{ =}-\dfrac{\text{1}}{\text{4}}$

$\text{ }\!\!\alpha\!\!\text{ }\!\!\beta\!\!\text{ =}\dfrac{\text{1}}{\text{4}}$

Hence, the required polynomial is ${{\text{x}}^{\text{2}}}-\left( \alpha +\beta \right)\text{x+}\alpha \beta$.

$\Rightarrow {{\text{x}}^{\text{2}}}-\left( -\dfrac{1}{4} \right)\text{x+}\dfrac{1}{4}$

$\Rightarrow \text{4}{{\text{x}}^{\text{2}}}\text{+ x +1}$

Therefore, the quadratic polynomial is  $\text{4}{{\text{x}}^{\text{2}}}\text{+ x +1}$ .

(vi) $4,1$

Ans

Given: $\text{4,1}$

Let the zeroes of polynomial be $\text{ }\!\!\alpha\!\!\text{ }$ and $\text{ }\!\!\beta\!\!\text{ }$.

Then,

$\text{ }\!\!\alpha\!\!\text{ + }\!\!\beta\!\!\text{ = 4}$

$\text{ }\!\!\alpha\!\!\text{ }\!\!\beta\!\!\text{ =1}$

Hence, the required polynomial is ${{\text{x}}^{\text{2}}}-\left( \alpha +\beta \right)\text{x+}\alpha \beta$.

$\Rightarrow {{\text{x}}^{\text{2}}}-\text{4x+1}$

Therefore, the quadratic polynomial is  ${{\text{x}}^{\text{2}}}-\text{4x+1}$.

## NCERT Solutions for Class 10 Maths Chapter 2 Exercise 2.2

### Definition

If x be a variable and x be a positive integer with a1, a2,  a3, ……., an be constants then, f(x) = anxn + an-1x-1 + ….. + a1x + a0   is called a polynomial in x.

In this polynomial anxn + an-1x-1, ………, a1x and a0 are called the terms of the polynomial, and an, an-1,…., a1, a0 are called their coefficients.

Example: p(x) = 4x -1  is a polynomial in variable x.

q(y) = 2y2 – 3y + 5 is a polynomial in variable y.

Note: Expressions like 3x2 - 2√x + 7, 1/(3x-5), etc. are not polynomials as some powers of x are not integers.

### Degree of a Polynomial

The exponent of the highest degree term in a polynomial is called its degree.

Example:

1. f(x) = 4x -1 is a polynomial in x of degree 1.

2. q(y) = 2y2 - 3y + 5 is a polynomial in y of degree 2.

### Types of Polynomial

Let us see the different types of polynomials.

1. Constant Polynomial: It is a polynomial of degree zero.

Ex: f(x) = 5, g(x) = 2, etc.

The constant polynomial f(x) = 0 is called the zero polynomial. The degree of the zero polynomial is not defined as f(x) = 0, g(x) =0x, h(x) = 0x2, p(x) = 0x3, etc. are all equal to the zero polynomial.

2. Linear Polynomial: It is a polynomial of degrees.

Ex: f(x) = 4x -1.

A polynomial q(y) = 3y2 + 4 is not linear as it has degree 2.

In general, any linear polynomial in x with real coefficients is of the form f(x) = ax + b where a & b are real numbers and a 0.

Note: A linear polynomial may be a monomial (having only one term. Ex: q(x) = 5x) or a binomial (having 2 terms. Ex: f(x) = 4x-1)

3. Quadratic Polynomial: It is a polynomial of degree 2.

Ex: f(x) = 5x2 - 3x + 4.

In general, a quadratic polynomial is of the form f(x) = ax2 + bx + c, where a, b & c are real numbers and a 0.

4. Cubic Polynomial: A polynomial of degree 3 is called a cubic polynomial.

Ex: f(x) = 2x3 – 5x2 + 7x – 9.

The general form of a cubic polynomial is f(x) = ax3 + bx2 + cx + d, where a, b, c, d are real numbers and a 0.

### Value of a Polynomial

If f(x) is a polynomial and ‘a’ is a real number then the real number obtained by replacing x by ‘a’ , is called the value f(x) at x = a and is denoted by f(a). For example: if f(x) = 2x2 - 3x -2, then its value

1. at x = 1 is given by f(1) = 2(1)2 - 3 1 -2 = 2-3-2 = -3

2. at x = -2, is given by f(-2) = 2(-2)2 -3(-2)-2 = 8 + 6 -2 = 12

### Zero of a Polynomial

Consider the cubic polynomial f(x) = x3 – 6x2 + 11x - 6. The value of this polynomial at x =1, 2, 3 are respectively

f(1) =13 – 6 * 12 + 11 * 1 – 6 = 1-6 + 11 -6 =0

f(2) = 23 – 6 * 22 + 11 * 2 – 6 = 8 – 24 + 22 – 6 = 0

f(3) = 33 – 6* 32 + 11 * 3 -6 = 27  -54 + 33 -6 = 0

1, 2, 3 are called the zeroes of the cubic polynomial f(x) = x3 -6x2 + 11x - 6.

A real number is 0 of a polynomial f(x), if f() = 0.

To find this we solve the equation f(x) = 0.

For Example: if the polynomial is f(x) = x -3 then putting x -3 = 0, we get x = 3

Then the real number 3 is the zero of the polynomial f(x) = x -3.

There can be more than one zero of a polynomial as can be seen in the example f(x) = x3-6x2 + 11x - 6.

### Graph of a Polynomial

The graph of a polynomial f(x) is the set of all points (x, y), where y = f(x). This graph is a smooth free hand curve, passing through the points (x1, y1), (x2, y2), (x3,y3), ……, etc. y1, y2, y3, …… are the estimated values of the polynomial f(x) at x1, x2, x3,..... respectively.

Geometrical meaning of the zeroes of a linear polynomial

We take an example f(x) = 4x - 2

Let y  4x - 2

We make a table with values of y corresponding to different values of x.

 A B C x 0 2 ½ y -2 6 0

The points A(0, 1) and B (2, 6) are plotted on the graph paper on a suitable scale. A line drawn runs through these points to derive the graph of the given polynomial. It can be seen that the graph intersects the x-axis at C (½, 0), where y = f(x) = 0 so that the zero of the given polynomial is ½.

### Meaning of the Zeroes of a Quadratic Polynomial

Graph of a quadratic polynomial f(x) = x2 - 2x - 8

Let y = x2 - 2x - 8

We construct a table containing the values of y corresponding to various values of x:

 x -4 -3 -2 -1 0 1 2 3 4 5 6 y = x2 - 2x - 8 16 7 0 -5 -8 -9 -8 -5 0 7 16

We now plot the points (-4, 16), ( -3, 7), (-2, 0) , (-1, -5), (0 -8), (1, 9), (2, 8), (3, -5), (4, 0) , (5, 7) and (6, 16) on a graph paper and draw a smooth free hand curve passing through these points. Hence, the curve derived demonstrates the graph of the polynomial f(x) = x2 -2x -8. Such a curve is called a parabola. The lowest point of the graph (1, 9) is called the vertex of the parabola.

### The Following Observations Can Be Made On Class 10 Maths ex 2.2

1. The graph is symmetrical about the line parallel to the y-axis through the vertex (1, -9). This line is called the axis of the parabola.

2. The concavity of the parabola is towards the positive direction of the y-axis (this happens when the coefficient of x2 is positive.

3. As  x2 -2x -8 = (x-4) (x + 2), the polynomial f(x) has two factors (x-4) & (x+2), so that the parabola cuts the x-axis at two distinct points (4,0) & (-2, 0). The x-coordinates of these two points: 4 & -2 are the two zeroes of f(x).

### Note in Exercise 2.2 Class 10 Math:

1. The number of zeroes of a polynomial depends on its degree. For a polynomial of degree n, the number of zeroes can be utmost n or less. For example, for a quadratic, polynomial the no. of zeroes can be 2 or less.

2. For a quadratic polynomial, its graph (ex: a parabola)  may cut the x-axis at two points in fig (i), or at one point in fig (ii) or at no point at all in fig (iii).

## Conclusion

Your one-stop resource for learning how to solve these expressions might be Vedanta's Ex 2.2 Class 10 NCERT Solutions. These solutions simplify difficult subjects like factorization and finding zeroes into simple steps with illustrations. Pay particular attention to comprehending how a polynomial's zeroes and coefficients relate to one another.You'll come across a number of polynomial problems throughout the chapter, and this is the key to solving them. By using these solutions for practice, you'll get more confident and ace the test!

## NCERT Solutions Class 10 Maths Chapter 2 Exercises

 Chapter 2 - Polynomials All Exercises in PDF Format Exercise 2.1 1 Question & Solutions

## Chapter-Specific NCERT Solutions for Class 10 Maths

Given below are the chapter-wise NCERT Solutions for Class 10 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 10 Maths Chapter 2: Polynomials - Exercise 2.2

1. State the condition in which the graph of quadratic polynomial p(x) = ax2 + bx + c is an upward parabola or downward parabola.

For a > 0, the parabola is upward and for a < 0, the parabola is downward.

2. Where will I get the best reference for NCERT Solutions for Class 10 Chapter 2 -- Polynomial Exercise 2.2?

You will get the best reference for NCERT Solutions for Class 10 Chapter 2 - Polynomial Exercise 2.2 on the official website of Vedantu, the leading education portal in India. The solutions and the reference notes for NCERT Solutions Chapter 2 -  Polynomials are created by the in-house subject experts stepwise as per the latest guidelines of NCERT (CBSE). These notes and solutions will give you a better understanding of the topic and will help you with revision for your exams.

3. Can I download the NCERT Solutions for Class 10 Chapter 2 -- Polynomials Exercise 2.2 from the Vedantu app?

You can definitely download the free pdf of NCERT for Class 10 Chapter 2 -- Polynomials Exercise 2.2 from the Vedantu app. You can also register for Maths online tuitions on www. Vedantu.com.

4. What are the main topics in Exercise 2.2 of Chapter 2 of Class 10 Maths to learn?

Exercise 2.2 of Chapter 2 of Class 10 Maths textbook focuses on polynomials and their types. Values, degrees, and how to plot them on the graph are also covered. Polynomial equations are also solved and answers are explained. The solutions PDF has answers to all the NCERT questions and also extra questions from practice developed by experts. The notes and answers in the solutions PDF will help you understand the concepts better and easily.

5. How many questions are there in Exercise 2.2 of Chapter 2 of Class 10 Maths?

Exercise 2.2 from the Chapter 2 of Class 10 Maths book has a few questions. The answers to all these questions and more for practice can be found in the solution PDF by Vedantu. Before the exercise, the examples are also explained in depth in the PDF. This helps the students understand the concept better. The PDF explains the concepts required to answer each question. All the questions and answers are developed by experts.

6. How many examples are there in the NCERT textbook before Exercise 2.2 in Chapter 2 of Class 10 Maths?

There are five examples before Exercise 2.2 of Chapter 2 of Class 10 Maths in the NCERT textbook. These examples discuss the relationship between coefficients and zeros. This is the topic on which Exercise 2.2 of Chapter 2 is based. All examples are explained step by step and give an overview of Exercise 2.2 of Chapter 2 through this. The NCERT Solutions PDF also has extra questions in addition to the exercise questions to help students understand the concept.

7. What is the sum and the product of the zeros of the polynomial, according to Exercise 2.2 of Chapter 2 of Class 10 Maths?

The zeroes of a polynomial are the values of the equation. They are the factors of p(x). In the concept covered before Exercise 2.2 of Chapter 2 of Class 10 Maths, we understand that the sum of the zeros of an equation is equal to -(coefficient of x) divided by the coefficient of x square. The product of the zero is the constant term divided by the coefficient of x square. The concept is explained with the help of examples in the textbook before the Exercise and the NCERT Solution PDF.

8. How can I practice more questions like in Exercise 2.2 of Chapter 2 of Class 10 Maths NCERT textbook?

Students can refer to the NCERT Solutions PDF for Exercise 2.2 of Chapter 2 of Class 10 Maths free of cost on the Vedantu website and the Vedantu app, for NCERT Solutions and extra questions. The solutions PDF provides solutions for all the questions asked in the exercise. It has enough questions for the students to practice. The PDF also explains concepts in an easy-to-understand language by experts for the students to grasp them better. For extra practice, students should refer to the pdf and practice as many questions as possible.