NCERT Exemplar Class 9 Maths Solutions Chapter 2 Polynomials 2021

NCERT Exemplar Class 9 Maths Solutions Chapter 2 Polynomials

Edited By Ravindra Pindel | Updated on Aug 31, 2022 11:39 AM IST

NCERT exemplar Class 9 Maths solutions chapter 2 discusses polynomials, their properties, and solving polynomials for their zeroes. These NCERT exemplar Class 9 Maths chapter 2 solutions are designed by our subject-matter experts of mathematics. These solutions deliver accurate and elaborate answers to the questions of NCERT Class 9 Maths Book. These NCERT exemplar Class 9 Maths chapter 2 solutions build the concepts of polynomials by providing the students with a step-by-step approach. CBSE 9 maths Syllabus is followed for curating these NCERT exemplar Class 9 Maths solutions chapter 2.

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This Story also Contains

NCERT Exemplar Class 9 Maths Solutions Chapter 2-Exercise 2.1
NCERT Exemplar Class 9 Maths Solutions Chapter 2-Exercise 2.2
NCERT Exemplar Class 9 Maths Solutions Chapter 2-Exercise 2.3
NCERT Exemplar Class 9 Maths Solutions Chapter 2-Exercise 2.4
NCERT Exemplar Class 9 Maths Solutions Chapter 2 Contains the Solutions of the Questions Based on the Mentioned Topics:
NCERT Class 9 Maths Exemplar Solutions for Other Chapters:
Features of NCERT Exemplar Class 9 Maths Solutions Chapter 2:

NCERT Exemplar Class 9 Maths Solutions Chapter 2-Exercise 2.1

Question:1

Which one of the following is a polynomial?
$(A) \frac{x^{2}}{2} - \frac{2}{x^{2}} (B) \sqrt{2 x} - 1 (C) x^{2} + \frac{3 x^{\frac{3}{2}}}{\sqrt{x}} (D) \frac{x - 1}{x + 2}$

Answer:

x^{2} + \frac{3 x^{\frac{3}{2}}}{\sqrt{x}}

Solution
Polynomial: A polynomial is an expression consisting of variables and coefficients and also non negative powers. It contains different powers of the same variable,
Here

\frac{x^{2}}{2} - \frac{2}{x^{2}}

is not a polynomial because power of x must be a non-negative integer.

\sqrt{2 x} - 1

is also not a polynomial because degree of variable should always be a whole number.

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

It is a polynomial because power of x is in whole numbers.

\frac{x - 1}{x + 2} \Rightarrow (x - 1) {(x + 2)}^{- 1}

It is also not a polynomial because power of x is negative.

Question:2

$\sqrt{2}$ is a polynomial of degree

(A) 2

(B) 0

(D) $\frac{1}{2}$

Answer:

[B]
Solution. Polynomial: It is an expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable (s)
Degree of polynomial: - Degree of polynomial is the highest degree of polynomial’s monomials with non-zero co-efficient.
Here we can write

\sqrt{2}

\sqrt{2} . x^{0}

because any variable having power zero is always one
Hence the degree of polynomial is zero.

Question:3

Degree of the polynomial 4x⁴ + 0x³ + 0x⁵ + 5x + 7 is :
(A) 4
(B) 5
(C) 3
(D) 7

Answer:

[A]
Solution: Polynomial:- It is an expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable(s)
Degree of a polynomial:- Degree of a polynomial is the highest power of the polynomial’s monomials with a non-zero coefficient.
Here given polynomial is 4x⁴ + 0x³+ 0x⁵ + 5x + 7 which can be write as 4x⁴ + 5x + 7
Here the highest power of the x-coefficient is 4 therefore the degree is 4.

Question:4

Degree of the zero polynomial is
(A) 0
(B) 1
(C) Any natural number
(D) Not defined

Answer:

(D) Not defined
Solution
Degree of polynomial:- Degree of a polynomial is the highest of the degree of polynomial’s monomials with the non-zero coefficient.
In zero polynomial, all the coefficients are zero. So, we cannot determine its degree therefore we say it is undefined

Question:5

If $p (x) = x^{2} - 2 \sqrt{2} x + 1$ , then $p (2 \sqrt{2})$ is equal to
(A) 0
(B) 1
(C) $(4 \sqrt{2})$
(D) $(8 \sqrt{2})$

Answer:

(B) 1
Solution:

p (x) = x^{2} - 2 \sqrt{2} x + 1

p (2 \sqrt{2})

means that the value of x is

2 \sqrt{2}

Put

x = 2 \sqrt{2}

in the given equation

p (2 \sqrt{2}) = {(2 \sqrt{2})}^{2} - (2 \sqrt{2}) (2 \sqrt{2}) + 1 p (2 \sqrt{2}) = {(2 \sqrt{2})}^{2} - {(2 \sqrt{2})}^{2} + 1 p (2 \sqrt{2}) = 1

Hence the value of

p (2 \sqrt{2})

is 1.

Question:6

The value of the polynomial 5x – 4x² + 3 when x = – 1 is
(A) –6
(B) 6
(C) 2
(D) –2

Answer:

[A]
Solution: Here the given polynomial is p(x) = 5x – 4x² + 3
Then putting x = – 1 in the above equation we get
p(-1) = 5(–1) – 4(–1)² + 3
= – 5 – 4(–1)²+ 3
= – 5 – 4 + 3
= – 9 + 3
= – 6
Hence the answer is -6

Question:7

If $p (x) = x + 3$ , then $p (x) + p (- x)$ is equal to
(A) 3 (B) 2x (C) 0 (D) 6

Answer:

(D) 6
Solution:
The given equation is p(x)=x+3 ….. (1)
Put x=-x we get
p(-x)=-x+3 …..(2)
Now add equation 1 and 2 we get
p(x)+p(-x)=x+3-x+3
p(x)+p(-x)=6
Hence the value of p(x)+p(-x) is 6.

Question:8

Zero of the zero polynomial is
(A) 0
(B) 1
(C) Any real number
(D) Not define

Answer:

(C) Any real number
Solution
Zero Polynomial: - In zero polynomial all the co-efficient are zero and the degree of a polynomial is the highest of the degrees of its monomials with non- zero co-efficient but in zero polynomial all the co-efficient are zero.
For example:- p(x) = 0.x³ + 0.x²+ 0.x¹ + 0
=> p(x) = 0
Hence we can put any real value of x therefore are of the zero polynomial is any real number.
Therefore option (C) is correct

Question:9

Zero of the polynomial $p (x) = 2 x + 5$ is
(A) $- \frac{2}{5}$ (B) $- \frac{5}{2}$ (C) $\frac{2}{5}$ (D) $\frac{5}{2}$

Answer:

(B)

- \frac{5}{2}

Solution
We know that for finding the zero of a polynomial, we need to find a value of x for which the polynomial will be zero
i.e., p(x)=0
Here it is given that p(x)=2x+5
Put p(x)=0 in above equation we get
0=2x+5
-5=2x
x=

- \frac{5}{2}

Hence option (B) is correct.

Question:10

One of the zeroes of the polynomial $2 x^{2} + 7 x - 4$ is
(A) 2
(B) $\frac{1}{2}$
(C) $- \frac{1}{2}$
(D) -2

Answer:

(B)

\frac{1}{2}

Solution
We know that for finding the zero of a polynomial, we need to find a value of x for which the polynomial will be zero
i.e., p(x)=0
Here the polynomial is

2 x^{2} + 7 x - 4

Now let us find zeroes of this polynomial

2 x^{2} + 7 x - 4 = 0 2 x^{2} + 8 x - x - 4 = 0 2 x (x + 4) - 1 (x + 4) = 0 (x + 4) (2 x - 1) = 0 x = - 4, x = \frac{1}{2}

The zeros of the given polynomial are

- 4, \frac{1}{2}

Among the given options, (B) is correct.

Question:11

If $x^{51} + 51$ is divided by x+1 , the remainder is
(A) 0
(B) 1
(C) 49
(D) 50

Answer:

(D) 50
Solution:
In this question, we can use remainder theorem.
According to the remainder theorem when p(x) is divided by (x+a) then the remainder is p(-a) .

∴

x^{51} + 51

is divided by (x+1) then the remainder is p(-1)

p (- 1) = (- 1)^{51} + 51 p (- 1) = - 1 + 51 p (- 1) = 50

Hence the remainder is 50
Hence option (D) is correct.

Question:12

If x+1 is a factor of the polynomial 2x²+kx , then the value of k is
(A) –3 (B) 4 (C) 2 (D) –2

Answer:

2 x^{2} + k x

Let

p (x) = 2 x^{2} + k x

It is given that

(x + 1)

is a factor of given polynomial.
Then according to factor theorem,

p (- 1) = 0

\Rightarrow 2 (- 1)^{2} + k (- 1) = 0 \Rightarrow 2 - k = 0 \Rightarrow k = 2

So the value of k is 2.
Hence option (C) is correct.

Question:13

x+1 is a factor of the polynomial
$(A) x^{3} + x^{2} - x + 1 (B) x^{3} + x^{2} + x + 1 (C) x^{4} + x^{3} + x^{2} + 1 (D) x^{4} + 3 x^{3} + 3 x^{2} + x + 1$

Answer:

(B)

x^{3} + x^{2} + x + 1

Solution:
We know that if

(x + a)

is factor of the polynomial f(x), then it always satisfies

f (- a) = 0

Hence (x+1) is factor of that polynomial which satisfies f(-1)=0 .

(A) f (x) = x^{3} + x^{2} - x + 1 f (- 1) = (- 1)^{3} + (- 1)^{2} - (- 1) + 1 = - 1 + 1 + 1 + 1 f (- 1) = 2

Not satisfied

(B) f (x) = x^{3} + x^{2} + x + 1 f (- 1) = (- 1)^{3} + (- 1)^{2} + (- 1) + 1 = - 1 + 1 - 1 + 1 f (- 1) = 0

Satisfied

(C) f (x) = x^{4} + x^{3} + x^{2} + 1 f (- 1) = (- 1)^{4} + (- 1)^{3} + (- 1)^{2} + 1 = 1 - 1 + 1 + 1 f (- 1) = 2

Not satisfied

(D) f (x) = x^{4} + 3 x^{3} + 3 x^{2} + x + 1 f (- 1) = (- 1)^{4} + 3 (- 1)^{3} + 3 (- 1)^{2} + (- 1) + 1 = 1 - 3 + 3 - 1 + 1 f (- 1) = 1

Not satisfied
Hence

(x + 1)

is a factor of

x^{3} + x^{2} + x + 1

Therefore option (B) is correct

Question:14

One of the factors of $(25 x^{2} - 1) + {(1 + 5 x)}^{2}$ is
(A) 5+x (B) 5-x (C) 5x-1 (D) 10x

Answer:

(D) 10x
Solution
Given,

(25 x^{2} - 1) + {(1 + 5 x)}^{2}

We know that,

(5 x + 1)^{2} = (5 x)^{2} + 2 \times (5 x) \times 1 + 1

{u s i n g (a + b)^{2} = a^{2} + b^{2} + 2 a b}

Now, using the above

(25 x^{2} - 1) + {(1 + 5 x)}^{2} = 25 x^{2} - 1 + 1^{2} + (5 x)^{2} + 2 \times 1 \times 5 x = 0 25 x^{2} - 1 + 1 + 25 x^{2} + 10 x = 0 50 x^{2} + 10 x = 0 10 x (5 x + 1) = 0

The factors of

(25 x^{2} - 1) + {(1 + 5 x)}^{2}

are 10x and (5x+1)
Hence in the above option, (D) is correct.

Question:15

The value of $249^{2} - 248^{2}$ is
(A) $1^{2}$ (B) 477 (C) 487 (D) 497

Answer:

(D) 497
Solution
Here we find the value of

249^{2} - 248^{2}

by using the identity

a^{2} - b^{2} = (a - b) (a + b)

249^{2} - 248^{2} = (249 - 248) (249 + 248) 249^{2} - 248^{2} = (1) (497) 249^{2} - 248^{2} = 497

Hence the answer is an option (D) 497.

Question:16

The factorization of $4 x^{2} + 8 x + 3$ is
(A) $(x + 1) (x + 3)$
(B) $(2 x + 1) (2 x + 3)$
(C) $(2 x + 2) (2 x + 5)$
(D) $(2 x - 1) (2 x - 3)$

Answer:

Let us factorize the given polynomial

4 x^{2} + 8 x + 3

To factorize ax² + bx + c, we have to distribute “bx” into “px” and “qx” such that
p + q = b and p.q = a.c
So,

4 x^{2} + 8 x + 3 = 0

can be written as

4 x^{2} + 6 x + 2 x + 3 = 0

{6+2 = 8; (6)(2) = (4)(3)}

2 x (2 x + 3) + 1 (2 x + 3) = 0 (2 x + 3) (2 x + 1) = 0

Hence option B is correct

Question:17

Which of the following is a factor of ${(x + y)}^{3} - (x^{3} + y^{3})$
$A . x^{2} + y^{2} + 2 x y B . x^{2} + y^{2} - 2 x y C . x y^{2} D . 3 x y$

Answer:

(D) 3xy
Solution
Given,

{(x + y)}^{3} - (x^{3} + y^{3})

We know that,

{(a + b)^{3} = a^{3} + b^{3} + 3 a^{2} b + 3 a b^{3}}

So,

(x + y)^{3} = x^{3} + y^{3} + 3 x^{2} y + 3 x y^{3}

{(x + y)}^{3} - (x^{3} + y^{3}) = x^{3} + y^{3} + 3 x^{2} y + 3 x y^{3} - x^{3} - y^{3} = 3 x^{2} + 3 x y^{2} = 3 x y (x + y)

The factors of the above polynomial are
3xy and (x+y)
Hence option D is correct

Question:18

The coefficient of x in the expansion of (x+3)³ is
(A) 1 (B) 9 (C) 18 (D) 27

Answer:

(D) 27
Solution
The given expansion is (x+3)³
Using

(a + b)^{3} = a^{3} + b^{3} + 3 a^{2} b + 3 a b^{3}

in above expansion we get

(x + 3)^{3} = x^{3} + 3^{3} + 3 x^{2} .3 + 3 x {.3}^{3} = x^{3} + 27 + 9 x^{2} + 27 x

Here the coefficient of x is 27
Therefore option (D) is correct

Question:19

If $\frac{x}{y} + \frac{y}{x} = - 1$ $(x, y \neq 0)$ the vlaue of $x^{3} - y^{3}$ is
(A) 1 (B) -1 (C)0 (D) $\frac{1}{2}$

Answer:

\frac{x}{y} + \frac{y}{x} = - 1

(x, y \neq 0)

Simplifying the above equation we have

\frac{x^{2} + y^{2}}{x y} = - 1 x^{2} + y^{2} = - x y x^{2} + y^{2} + x y = 0 \dots \dots \dots (1) x^{3} - y^{3} = (x - y) (x^{2} + y^{2} + x y) \dots \dots \dots (2) {∵ a^{3} - b^{3} = (a - b) (a^{2} + b^{2} + a b)}

Put the value of equation 1 in 2 we get

x^{3} - y^{3} = (x - y) (0) x^{3} - y^{3} = 0

Hence option C is correct.

Question:20

If $49 x^{2} - b = (7 x + \frac{1}{2}) (7 x - \frac{1}{2})$ then the valueof b is,
$(A) 0$ $(B) \frac{1}{\sqrt{2}}$ $(C) \frac{1}{4}$ $(D) \frac{1}{2}$

Answer:

Given:

49 x^{2} - b = (7 x + \frac{1}{2}) (7 x - \frac{1}{2})

(7 x + \frac{1}{2}) (7 x - \frac{1}{2}) = (7 x) - {(\frac{1}{2})}^{2}

{u s i n g (a + b) (a - b) = a^{2} - b^{2}}

Simplifying the given equation we have

49 x^{2} - b = (7 x)^{2} - {(\frac{1}{2})}^{2} 49 x^{2} - b = 49 x^{2} - \frac{1}{4} - b = - \frac{1}{4} b = \frac{1}{4}

Therefore option (C) is correct.

Question:21

If $a + b + c = 0$ , then $a^{3} + b^{3} + c^{3}$ is equal to
(A) 0 (B) abc (C) 3abc (D) 2abc

Answer:

a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a)

Given,

a + b + c = 0

…..(1)

a^{3} + b^{3} + c^{3} = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a) + 3 a b c

…..(2)
Using equation 1 in equation 2 we have

a^{3} + b^{3} + c^{3} = (0) (a^{2} + b^{2} + c^{2} - a b - b c - c a) + 3 a b c a^{3} + b^{3} + c^{3} = 3 a b c

Therefore option (C) is correct.

NCERT Exemplar Class 9 Maths Solutions Chapter 2-Exercise 2.2

Question:1

Which of the following expressions are polynomials? Justify your answer
(i) 8
(ii) $\sqrt{3} x^{2} - 2 x$
(iii) $1 - \sqrt{5 x}$
(iv) $\frac{1}{5 x^{- 2}} + 5 x + 7$
(v) $\frac{(x - 2) (x - 4)}{x}$
(vi) $\frac{x}{x + 1}$
(vii) $\frac{1}{7} a^{3} - \frac{2}{\sqrt{3}} a^{2} + 4 a - 7$
(viii) $\frac{1}{2 x}$

Answer:

(i, ii, iv, vii)
Solution
Polynomial:- It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s)
Its degree is always a whole number.
For example:

x^{0}, x + 2

etc.
(i) Here 8 is a polynomial because it can also be written as

8. x^{0}

i.e., multiply by

x^{0}

.
(ii)

\sqrt{3} x^{2} - 2 x

is also a polynomial having degree two.
(iii)

1 - \sqrt{5 x}

is not a polynomial because its exponent is in fraction.
(iv)

\frac{1}{5 x^{- 2}} + 5 x + 7

can be written as

5 x^{2} + 5 x + 7

and it is a polynomial having degree two.
(v)

\frac{(x - 2) (x - 4)}{x}

is not polynomial because it has negative exponent.
(vi)

\frac{1}{x + 1}

is not a polynomial because it have negative exponent.
(vii)

\frac{1}{7} a^{3} - \frac{2}{\sqrt{3}} a^{2} + 4 a - 7

is a polynomial of degree three.
(viii)

\frac{1}{2 x}

is not a polynomial because it have negative exponent.

Question:2

Write whether the following statements are True or False. Justify your answer.
i. A binomial can have at most two terms
ii. Every polynomial is a binomial.
iii. A binomial may have degree 5.
iv. Zero of a polynomial is always 0
v.A polynomial cannot have more than one zero.
vi. The degree of the sum of two polynomials each of degree 5 is always 5

Answer:

i. False
Solution :- Binomial: A binomial is an expression that has two numbers, terms or letters joined by the sign + or –.
Binomial necessarily means consisting of two terms only. These terms should not be like terms.
For example:

x^{2} + 1

is a binomial
x + 2x is not a binomial as these are like terms.
So the given statement is false, as a binomial has exactly two terms (not at most two terms).
ii. False
Solution:- Binomial: A binomial is an expression that has two numbers, terms or letters joined by the sign + or –.
Binomial necessarily means consisting of two terms only. These terms should not be like terms.
For example:

x^{2} + 1

is a binomial
x + 2x is not a binomial as these are like terms.
Polynomial: It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s)
Its degree is always a whole number.
For example:

x^{0}, x^{2} + 2

etc.
Because a binomial has exactly two terms but a polynomial can be monomial (single term), binomial (two terms), trinomial (three terms) etc.

∴

The given statement is False.
iii. True
Solution:- Binomial: A binomial is an expression that has two numbers, terms or letters joined by the sign + or –.
Binomial necessarily means consisting of two terms only. These terms should not be like terms.
For example:

x^{2} + 1

is a binomial
x + 2x is not a binomial as these are like terms.
Degree of polynomial: Degree of polynomial is the highest power of the polynomial’s monomials with non-zero coefficient.
For any binomial of the form

x^{5} + 2

, we can see that the degree is 5.
So, a binomial may have degree 5.
Therefore the given statement is True.
iv. False
Solution :- Polynomial:- It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s)
Its degree is always a whole number.
For example:

x^{0}, x^{2} + 2

etc.
We know that for finding the zero of a polynomial, we need to find a value of x for which the polynomial will be zero
i.e., p(x)=0
Let us consider an example:
p(x) = x - 2
Now to find the zero of this polynomial, we have:
x - 2 =0
x = 2 (which is not zero)
Hence the given statement is false because zero of a polynomial can be any real number.
v. False
Solution :- Polynomial: It is an expression of more than two algebraic terms, especially the sum of several terms that contains different powers of the same variable(s).
Its degree is always a whole number.
For example:

x^{0}, x + 2, x^{3} + 1, x^{4} + x^{3} + x^{2} + 1

etc.
We know that for finding the zero of a polynomial, we need to find a value of x for which the polynomial will be zero
i.e., p(x)=0
So, a polynomial can have any number of zeroes. It depends upon the degree of polynomial.

∴

The given statement is False.
vi. False
Solution :- Degree of polynomial: Degree of polynomial is the highest power of the polynomial’s monomials with non-zero coefficient.
The degree of the sum of two polynomials may be less than or equal to 5.
For example:

x^{5} + 1

and

- x^{5} + 2 x^{3} + 1

are two polynomials of degree 5 but the degree of the sum of the two polynomials

2 x^{3} + 2

is 3.
Hence the given statement is False.

NCERT Exemplar Class 9 Maths Solutions Chapter 2-Exercise 2.3

Question:1

Classify the following polynomials as polynomials in one variable, two variables etc.
(i) $x^{2} + x + 1$
(ii) $y^{2} - 5 y$
(iii) $x y + y z + z x$
(iv) $x^{2} - 2 x y + y^{2} + 1$

Answer:

(i) One Variable
(ii) One Variable
(iii) Three Variables
(iv) Two Variables
Solution
In mathematics, a variable is a symbol whose role is to act as a substitute for a changing quantity or expression. It is often used to represent an arbitrary element.
For example: - x, y, z etc.
(i) Here,

x^{2} + x + 1

is a one variable polynomial, because it contains only one variable,
i.e., x
(ii)

y^{2} - 5 y

Here,

y^{2} - 5 y

is a one variable polynomial, because it contains only one variable,
i.e., y
(iii)

x y + y z + z x

Here,

x y + y z + z x

is three variable polynomial, because it contains three variables
i.e., x, y and z
(iv)

x^{2} - 2 x y + y^{2} + 1

Here,

x^{2} - 2 x y + y^{2} + 1

is two variable polynomials, because it contains two variables,
i.e., x and y

Question:2

Determine the degree of each of the following polynomials:
(i) $2 x - 1$
(ii) $- 10$
(iii) $x^{3} - 9 x + 3 x^{5}$
(iv) $y^{3} (1 - y^{4})$

Answer:

(i) One
(ii) Zero
(iii) Five
(iv) Seven
Solution
Degree of polynomial: Degree of polynomial is the highest power of the polynomial’s monomial with non-zero coefficient
(i)

2 x - 1

Here highest power of x is one therefore degree is one.
(ii)

- 10

Here –10 can be written as

- 10. x^{0}

hence the degree of polynomial is zero or we can say that it is a constant polynomial.
(iii)

x^{3} - 9 x + 3 x^{5}

Here highest power of x is Five therefore degree is Five.
(iv)

y^{3} (1 - y^{4})

Here

y^{3} (1 - y^{4}) = y^{3} - y^{7}

Highest power of y is seven therefore degree is seven.

Question:3

For the polynomial $\frac{x^{3} + 2 x + 1}{5} - \frac{7}{2} x^{2} - x^{6}$ , write
(i) The degree of the polynomial
(ii) The coefficient of $x^{3}$
(iii) The coefficient of $x^{6}$
(iv) The constant term

Answer:

(i) 6
(ii)

\frac{1}{5}

(iii) -1
(iv)

\frac{1}{5}

Solution

\frac{x^{3} + 2 x + 1}{5} - \frac{7}{2} x^{2} - x^{6} = \frac{x^{3}}{5} + \frac{2 x}{5} + \frac{1}{5} - \frac{7 x^{2}}{2} - x^{6} = - x^{6} + \frac{x^{3}}{5} - \frac{7 x^{2}}{2} + \frac{2 x}{5} + \frac{1}{5} = (- 1) x^{6} + (\frac{1}{5}) x^{3} - (\frac{7}{2}) x^{2} + (\frac{2}{5}) x + \frac{1}{5}

(i) Degree of polynomial: Degree of polynomial is the highest power of the polynomial’s monomial with non-zero coefficient
Here the highest degree of x is 6 therefore the degree of the polynomial is six.
(ii) The coefficient of

x^{3}

Here

\frac{1}{5}

is multiplied by

x^{3}

therefore coefficient of

x^{3}

\frac{1}{5}

(iii) The coefficient of

x^{6}

Here –1 is multiplied by

x^{6}

therefore coefficient of

x^{6}

is –1.
(iv) The constant term
The constant term is the one without the variable x.
So,

\frac{1}{5}

is the constant term

Question:4

Write the coefficient of $x^{2}$ in each of the following
(i) $\frac{π}{6} x + x^{2} - 1$
(ii) $3 x - 5 + 0. x^{2}$
(iii) $(x - 1) (3 x - 4)$
(iv) $(2 x - 5) (2 x^{2} - 3 x + 1)$

Answer:

(i) 1
(ii) 0
(iii) 3
(iv) -16
Solution
The value which is multiplied by

x^{2}

is its coefficient.
(i)

\frac{π}{6} x + x^{2} - 1 = (\frac{π}{6}) x + (1) x^{2} - 1

Here

x^{2}

is multiplied with 1.
Hence the coefficient of

x^{2}

is one.
(ii)

3 x - 5 + 0. x^{2}

Here

x^{2}

is multiplied with 0.
Hence the coefficient of

x^{2}

is 0.
(iii)

(x - 1) (3 x - 4)

3 x^{2} - 4 x - 3 x + 4 3 x^{2} - 7 x + 4

Here

x^{2}

is multiplied with 3.
Hence the coefficient of

x^{2}

is 3.
(iv)

(2 x - 5) (2 x^{2} - 3 x + 1)

4 x^{3} - 6 x^{2} + 2 x - 10 x^{2} + 15 x - 5 4 x^{3} - 16 x^{2} + 17 x - 5

Here

x^{2}

is multiplied with -16.
Hence the coefficient of

x^{2}

is -16.

Question:5

Classify the following as a constant, linear, quadratic and cubic polynomials:
(i) $2 - x^{2} + x^{3}$
(ii) $x^{3}$
(iii) $5 t - \sqrt{7}$
(iv) $4 - 5 y^{2}$
(v) $3. x^{0}$
(vi) $2 + x$
(vii) $y^{3} - y$
(viii) $1 + x + x^{2}$
(ix) $t^{2}$
(x) $\sqrt{2} x - 1$

Answer:

(i) Cubic Polynomial
(ii) Cubic Polynomial
(iii) Linear Polynomial
(iv) Quadratic Polynomial
(v) Constant Polynomial
(vi) Linear Polynomial
(vii) Cubic Polynomial
(viii) Quadratic Polynomial
(ix) Quadratic Polynomial
(x) Linear Polynomial
Solution
Degree of a polynomial: Degree of a polynomial is the highest power of the polynomial’s monomials with non-zero coefficient
Depending on the degree we can classify functions as:

Constant – Degree is zero
Linear – Degree is one
Quadratic – Degree is two
Cubic – Degree is three

(i)

2 - x^{2} + x^{3}

Here highest power of x is three therefore degree is three.
Cubic polynomial.
(ii)

x^{3}

Here highest power of x is three therefore degree is three.
Cubic polynomial.
(iii)

5 t - \sqrt{7}

Here highest power of t is one therefore degree is one.
Linear polynomial.
(iv)

4 - 5 y^{2}

Here highest power of y is two therefore degree is two.
Quadratic polynomial.
(v)

3. x^{0}

Here highest power of x is zero therefore degree is zero.
Constant polynomial.
(vi)

2 + x

Here highest power of x is one therefore degree is one.
Linear polynomial.
(vii)

y^{3} - y

Here highest power of y is three therefore degree is three.
Cubic polynomial.
(viii)

1 + x + x^{2}

Here highest power of x is two therefore degree is two.
Quadratic polynomial.
(ix)

t^{2}

Here highest power of t is two therefore degree is two.
Quadratic polynomial.
(x)

\sqrt{2} x - 1

Here highest power of x is one therefore degree is one.
Linear polynomial.

Question:6

Give an example of a polynomial, which is:
(i) monomial of degree 1
(ii) binomial of degree 20
(iii) Trinomial of degree 2

Answer:

(i) x
(ii)

3 x^{20} + 4

(iii)

x^{2} + 2 x + 5

Solution
Monomial: It is an algebraic expression consisting of only one term.
Example: 2x
Binomial means consisting of exactly two terms. These terms should not be like terms.
For example:

x^{2} + 1

is a binomial
x + 2x is not a binomial as these are like terms.
Trinomial means consisting of exactly three terms. These terms should not be like terms.
For example:

x^{2} + 2 x + 1

is a trinomial
x + 2x is not a binomial as these are like terms.
Degree: Degree of an expression is the highest power of the monomial present in the expression with non-zero coefficient
(i) monomial of degree 1:
2x, x, y, 7y etc.
(ii) binomial of degree 20:

3 x^{20} + 4

(iii) Trinomial of degree 2:

x^{2} + 2 x + 5

Question:7

Find the value of the polynomial $3 x^{3} - 4 x^{2} + 7 x - 5$ , when x = 3 and also when x = – 3.

Answer:

[61, –143]
Solution.
Here the given polynomial is p(x) = 3x³ – 4x² + 7x – 5
Put x = 3 in given polynomial we get
p(3) = 3(3)³ – 4(3)² + 7(3) – 5
= 81 – 36 + 21 – 5
= 45 + 16
= 61
Put x = – 3 in given polynomial we get
p(-3) = 3(–3)³ – 4(–3)² + 7(–3) – 5
= –81 – 36 – 21 – 5
= – 117 – 26
= – 143

Question:8

If $p (x) = x^{2} - 4 x + 3$ evaluate : $p (2) - p (- 1) + p (\frac{1}{2})$

Answer:

- \frac{31}{4}

Solution
Given polynomial is

p (x) = x^{2} - 4 x + 3

Put

x = 2

we get

p (2) = (2)^{2} - 4 (2) + 3 = 4 - 8 + 3 = - 1

Put x=-1 we get

p (- 1) = (- 1)^{2} - 4 (- 1) + 3 = 1 + 4 + 3 = 8

Put

x = \frac{1}{2}

we get

p (\frac{1}{2}) = (\frac{1}{2})^{2} - 4 (\frac{1}{2}) + 3 = \frac{1}{4} - 2 + 3 = \frac{1}{4} + 1 = \frac{1 + 4}{4} = \frac{5}{4}

Now

p (2) - p (- 1) + p (\frac{1}{2})

= (- 1) - 8 + \frac{5}{4} = - 9 + \frac{5}{4} = \frac{- 36 + 5}{4} = - \frac{31}{4}

Hence the answer is

- \frac{31}{4}

Question:9

Find P(0), P(1), P(–2) for the following polynomials:
i. $P (x) = 10 x - 4 x^{2} - 3$
ii. $P (y) = (y + 2) (y - 2)$

Answer:

(i) P(0) = –3; P(1) = 3; P(–2) = –39
Solution: Given polynomial is P(x) = 10x – 4x² – 3
Put x = 0,
P(0) = 10(0) – 4(0) – 3
= 0 – 0 – 3
= – 3
Put x = 1,
P(1) = 10(1) – 4(1)2 – 3
= 10 – 4 – 3
= 10 – 7
= 3
Put x = –2,
P(–2) = 10(–2) –4(–2)2 – 3
= – 20 – 16 – 3
= – 39
Hence, P(0) = –3; P(1) = 3; P(–2) = –39
(ii)

P(0) = – 4; P(1) = – 3; P(–2) = 0

Solution.

Given polynomial is P(y) = (y + 2) (y – 2)

Put y = 0,

P(0)= (0 + 2) (0 – 2)

= (2) (–2)

= – 4

Put y = 1,

P(1) = (1 + 2) (1 – 2)

= (3) (–1)

= – 3

Put y = – 2,

P(–2) = (–2 + 2) (–2 – 2)

= (0) (–4)

= 0

Hence, P(0) = – 4; P(1) = – 3; P(–2) = 0.

Question:10

Verify whether the following are True or False :
i)-3 is a zero of x-3
ii) $- \frac{1}{3}$ is a zero of $3 x + 1$
iii) $- \frac{4}{5}$ is a zero of $4 - 5 y$
iv) 0 and 2 are the zeroes of $t^{2} - 2 t$
v) –3 is a zero of $y^{2} + y - 6$

Answer:

i) False
Solution:- Zeroes of a polynomial can be defined as points where the value of the polynomial becomes zero as a whole.
Let,
p(x) = x-3
Putting x = -3
p(x) =

- 3 - 3 = - 6 \neq 0

Hence -3 is not a zero of x - 3
Therefore given statement is False.
ii) True
Solution :- Zeroes of a polynomial can be defined as points where the value of the polynomial becomes zero as a whole.
Let,
p(x) = 3x+1
Putting x =

- \frac{1}{3}

p(x) =

3 (\frac{- 1}{3}) + 1 = - 1 + 1 = 0

Hence

- \frac{1}{3}

is a zero of 3x+1.
Therefore given statement is True.
iii) False
Solution :- Zeroes of a polynomial can be defined as points where the value of the polynomial becomes zero as a whole.
Let,
p(y) =

4 - 5 y

Putting y =

- \frac{4}{5}

p (y) = 4 - 5 (\frac{- 4}{5}) = 4 + 4 = 16 \neq 0

Hence

- \frac{4}{5}

is not a zero of 4-5y
Therefore given statement is False.
iv) True
Solution
Method 1: Zeroes of a polynomial can be defined as points where the value of the polynomial becomes zero as a whole.
Let,

p (t) = t^{2} - 2 t

Putting t = 0

p (t) = 0^{2} - 2 (0) = 0

Putting t = 2

p (t) = 2^{2} - 2 (2) = 0

Hence 0 and 2 are the zeroes of

t^{2} - 2 t

Therefore given statement is True.
Method 2:

t^{2} - 2 t = 0

The given expression can be factorized as:

t (t - 2) = 0

Hence, t = 0, t – 2 = 0
So we get the zeroes as t=0,t=2
Therefore given statement is True.
v) True
Solution
Method 1: Zeroes of a polynomial can be defined as points where the value of the polynomial becomes zero as a whole.
Let,

P (y) = y^{2} + y - 6

Putting y = -3

P (y) = (- 3)^{2} + (- 3) - 6 = 9 - 3 - 6 = 0

Hence -3 is the zero of

y^{2} + y - 6

Therefore given statement is True.
Method 2:

y^{2} + y - 6 = 0

The given expression can be factorized as follows:

y^{2} + 3 y - 2 y - 6 = 0 y (y + 3) - 2 (y + 3) = 0 (y + 3) (y - 2) = 0 y = - 3, y = 2

Hence zeroes are -3 and 2
Therefore the given statement is True.

Question:11

Find the zeroes of the polynomial in each of the following :
i) p(x) = x - 4
ii) g(x) = 3 - 6x
iii)q(x) = 2x - 7
iv) h(y) = 2y

Answer:

i) 4
Solution
p(x)=x-4
We know that for finding the zero of a polynomial, we need to find a value of x for which the polynomial will be zero
i.e., p(x)=0
x-4=0
x=4
Hence 4 is a zero of the given polynomial.
ii)

\frac{1}{2}

Solution

g (x) = 3 - 6 x

We know that for finding the zero of a polynomial, we need to find a value of x for which the polynomial will be zero
i.e., g(x)=0
3-6x=0
3=6x

\frac{3}{6} = x

x = \frac{1}{2}

Hence

\frac{1}{2}

is a zero of the given polynomial.
iii)

\frac{7}{2}

Solution
q(x)=2x-7
We know that for finding the zero of a polynomial, we need to find a value of x for which the polynomial will be zero
i.e., q(x)=0
2x-7=0
2x=7

x = \frac{7}{2}

Hence

\frac{7}{2}

is a zero of the given polynomial.

iv) 0
Solution
We know that for finding the zero of a polynomial, we need to find a value of x for which the polynomial will be zero
i.e., h(x)=0
2y=0

y = \frac{0}{2}

y=0
Hence 0 is a zero of the given polynomial.

Question:12

Find the zeroes of the polynomial : $p (x) = (x - 2)^{2} - (x + 2)^{2}$

Answer:

[0]
Solution:

p (x) = (x - 2)^{2} - (x + 2)^{2}

We know that for finding the zero of a polynomial, we need to find a value of x for which the polynomial will be zero
i.e., p(x) = 0
That is,

(x - 2)^{2} - (x + 2)^{2} = 0 (x^{2} + 2^{2} - 2 \times x \times 2) - (x^{2} + 2^{2} + 2 \times x \times 2) = 0 ∵ (a - b)^{2} = a^{2} + b^{2} - 2 a b ∵ (a + b)^{2} = a^{2} + b^{2} + 2 a b S o, x^{2} + 4 - 4 x - x^{2} - 4 - 4 x = 0 - 4 x - 4 x = 0 - 8 x = 0 x = 0

Hence 0 is the zero of the given polynomial

Question:13

By actual division, find the quotient and the remainder when the first polynomial isdivided by the second polynomial : $x^{2} + 4$ :x-1

Answer:

Quotient

= x^{3} + x^{2} + x + 1

Remainder =2
Solution
Let

p (x) = x^{4} + 1

g (x) = x - 1

Divide p(x) and g(x)

Quotient

= x^{3} + x^{2} + x + 1

Remainder =2

Question:14

By Remainder Theorem find the remainder, when p(x) is divided by g(x) , where
$(i) p (x) = x^{3} - 2 x^{2} - 4 x - 1; g (x) = x + 1$
$(i i) p (x) = x^{3} - 3 x^{2} + 4 x + 50; g (x) = x - 3$
$(i i i) p (x) = 4 x^{3} - 12 x^{2} + 14 x - 3; g (x) = 2 x - 1$
$(i v) p (x) = x^{3} - 6 x^{2} + 2 x - 4; g (x) = 1 - \frac{3}{2} x$

Answer:

(i) 0
Solution:- According to remainder theorem when p(x) is divided by (x+a) then the remainder is p(-a) .

p (x) = x^{3} - 2 x^{2} - 4 x - 1; g (x) = x + 1

So, when p(x) is divided by g(x) then remainder will be p(-1) .

p (- 1) = (- 1)^{3} - 2 (- 1)^{2} - 4 (- 1) - 1 = - 1 - 2 + 4 - 1 = - 4 + 4 = 0

Hence the remainder is zero
(ii)62
Solution:- According to remainder theorem when p(x) is divided by (x+a) then the remainder is p(-a) .

p (x) = x^{3} - 3 x^{2} + 4 x + 50; g (x) = x - 3

So, when p(x) is divided by g(x) then remainder will be p(3) .

p (3) = 3^{3} - 3 (3)^{2} + 4 (3) + 50 p (3) = 27 - 27 + 12 + 50 p (3) = 62

Hence the remainder is 62.
(iii)

\frac{3}{2}

Solution:- According to remainder theorem when p(x) is divided by (x+a) then the remainder is p(-a) .

p (x) = 4 x^{3} - 12 x^{2} + 14 x - 3; g (x) = 2 x - 1

So, when p(x) is divided by g(x) then remainder will be

p (\frac{1}{2})

p (\frac{1}{2}) = 4 {(\frac{1}{2})}^{3} - 12 {(\frac{1}{2})}^{2} + 14 (\frac{1}{2}) - 3 = 4 \times \frac{1}{8} - 12 \times \frac{1}{4} + 7 - 3 = \frac{1}{2} - 3 + 7 - 3 = \frac{1}{2} + 1 = \frac{1 + 2}{2} = \frac{3}{2}

Hence the remainder is

\frac{3}{2}

(iv)

- \frac{136}{27}

Solution:- According to remainder theorem when p(x) is divided by (x+a) then the remainder is p(-a) .

p (x) = x^{3} - 6 x^{2} + 2 x - 4; g (x) = 1 - \frac{3}{2} x

So, when p(x) is divided by g(x) then remainder will be

p (\frac{2}{3})

p (\frac{2}{3}) = {(\frac{2}{3})}^{3} - 6 {(\frac{2}{3})}^{2} + 2 (\frac{2}{3}) - 4 = \frac{8}{27} - 6 (\frac{4}{9}) + \frac{4}{3} - 4 = \frac{8 - 72 + 36 - 108}{27} = - \frac{136}{27}

Hence the remainder is

- \frac{136}{27}

Question:15

Check whether p(x) is multiple of g(x) or not
$(i) p (x) = x^{3} - 5 x^{2} + 4 x - 3, g (x) = x - 2 (i i) p (x) = 2 x^{3} - 11 x^{2} - 4 x + 5, g (x) = 2 x + 1$

Answer:

(i) No
Solution
Given

p (x) = x^{3} - 5 x^{2} + 4 x - 3, g (x) = x - 2

According to remainder theorem if p(x) is a multiple of (x+a) then p(-a)=0
Here p(x) is

x^{3} - 5 x^{2} + 4 x - 3

and g(x) is

x - 2

If p(x) is a multiple of g(x) then p(2)=0

p (2) = (2)^{3} - 5 (2)^{2} + 4 (2) - 3 = 8 - 20 + 8 - 3 = 16 - 23 = - 7 \neq 0

p (2) \neq 0

Hence p(x) is not a multiple of g(x) .
(ii) No
Solution
Given

p (x) = 2 x^{3} - 11 x^{2} - 4 x + 5, g (x) = 2 x + 1

According to remainder theorem if p(x) is a multiple of (x+a) then p(-a)=0
If p(x) is a multiple of g(x) then

p (- \frac{1}{2}) = 0

p (- \frac{1}{2}) = 2 {(- \frac{1}{2})}^{3} - 11 {(- \frac{1}{2})}^{2} - 4 (- \frac{1}{2}) + 5 = 2 \times \frac{- 1}{8} - 11 \times \frac{1}{4} + 2 + 5 = - \frac{1}{4} - \frac{11}{4} + 7 = \frac{- 1 - 11 + 28}{4} = \frac{16}{4} = 4 p (- \frac{1}{2}) \neq 0

Hence p(x) is not a multiple of g(x) .

Question:16

Show that
(i) $x + 3$ is a factor of $69 + 11 x - x^{2} + x^{3}$
(ii) $2 x - 3$ is a factor of $x + 2 x^{3} - 9 x^{2} + 12$

Answer:

(i) Here

g (x) = x + 3

and

p (x) = 69 + 11 x - x^{2} + x^{3}

According to remainder theorem if x+a is a factor of p(x) then p(-a)=0
So, if g(x) is a factor of p(x) then p(-3)=0

p (- 3) = 69 + 11 (- 3) - (- 3)^{2} + (- 3)^{3} = 69 - 33 - 9 - 27 = 69 - 69 = 0 p (- 3) = 0

Therefore x+3 is a factor of

69 + 11 x - x^{2} + x^{3}

Hence proved
(ii) Here

g (x) = 2 x - 3

and

p (x) = x + 2 x^{3} - 9 x^{2} + 12

According to remainder theorem if x+a is a factor of p(x) then p(-a)=0
So, if g(x) is a factor of p(x) then

p (\frac{3}{2}) = 0

p (\frac{3}{2}) = (\frac{3}{2}) + 2 {(\frac{3}{2})}^{3} - 9 {(\frac{3}{2})}^{2} + 12 = \frac{3}{2} + 2 \times \frac{27}{8} - 9 \times \frac{9}{4} + 12 = \frac{3}{2} + \frac{27}{4} - \frac{81}{4} + 12 = \frac{6 + 27 - 81 + 48}{4} = \frac{0}{4} = 0 p (\frac{3}{2}) = 0

Hence 2x-3 is a factor of

x + 2 x^{3} - 9 x^{2} + 12

Hence proved

Question:17

Determine which of the following polynomials has $x - 2 a$ factor
(i) $3 x^{2} + 6 x - 24$
(ii) $4 x^{2} + x - 2$

Answer:

(i)

3 x^{2} + 6 x - 24

only
Solution
We know that if

(x + a)

is factor of the polynomial f(x), then it always satisfies f(-a)=0
(i) Here polynomial is

3 x^{2} + 6 x - 24

p (x) = 3 x^{2} + 6 x - 24

According to remainder theorem if x-2 is a factor of p(x) then p(2)=0

p (2) = 3 (2)^{2} + 6 (2) - 24 = 3 (4) + 12 - 24 = 12 - 12 = 0 p (2) = 0

Hence x-2 is a factor of

3 x^{2} + 6 x - 24

(ii) Here

p (x) = 4 x^{2} + x - 2

According to remainder theorem if x-2 is a factor of p(x) then p(2)=0

p (2) = 4 (2)^{2} + 2 - 2 = 16 p (2) \neq 0

Hence x-2 is not a factor of

4 x^{2} + x - 2

Question:18

Show that p - 1 is a factor of p¹⁰ - 1 and also of p¹¹ - 1 .

Answer:

To prove : Here we have to prove that

p - 1

is a factor of

p^{10} - 1

and also of

p^{11} - 1

.
We know that if (x+a) is factor of the polynomial f(x), then it always satisfies f(-a)=0
If

p - 1

is a factor of

p^{11} - 1

and

p^{10} - 1

then by putting the value of p = 1, the given polynomials should be equal to zero.
Let

f (p) = p^{10} - 1 f (1) = (1)^{10} - 1 = 1 - 1 = 0 f (1) = 0

Hence p-1 is a factor of

p^{10} - 1

Let

g (p) = p^{11} - 1 g (1) = (1)^{11} - 1 = 1 - 1 = 0 g (1) = 0

Hence p-1 is a factor of

p^{11} - 1

Hence proved

Question:19

For what value of m is $x^{3} - 2 m x^{2} + 16$ divisible by $x + 2$ ?

Answer:

m = 1
Solution
Given :

x^{3} - 2 m x^{2} + 16

is divisible by

x + 2

We know that if (x+a) is factor of the polynomial f(x), then it always satisfies f(-a)=0
If

x^{3} - 2 m x^{2} + 16

is divisible by

x + 2

then according to remainder theorem if we put x=-2 in the polynomial

x^{3} - 2 m x^{2} + 16

then the output must be equal to 0.
Let

x^{3} - 2 m x^{2} + 16

Put x=-2

p (- 2) = (- 2)^{3} - 2 m (- 2)^{2} + 16 0 = - 8 - 8 m + 16 0 = - 8 m + 8 8 m = 8 m = \frac{8}{8} = 1

Hence, m=1

Question:20

If $x + 2 a$ is a factor of $x^{5} - 4 a^{2} x^{3} + 2 x + 2 a + 3$ , find a.

Answer:

a = \frac{3}{2}

Solution
We know that if (x+a) is factor of the polynomial p(x), then it always satisfies p(-a)=0
Given

x + 2 a

is a factor of

x^{5} - 4 a^{2} x^{3} + 2 x + 2 a + 3

Let

p (x) = x^{5} - 4 a^{2} x^{3} + 2 x + 2 a + 3

g (x) = x + 2 a

If g(x) is a factor of p(x), then p(-2a)=0

p (- 2 a) = (- 2 a)^{5} - 4 a^{2} (- 2 a)^{3} + 2 (- 2 a) + 2 a + 3 0 = - 32 a^{5} + 32 a^{5} - 4 a + 2 a + 3 0 = - 2 a + 3 2 a = 3

So,

a = \frac{3}{2}

Question:21

Find the value of m so that $2 x - 1$ be a factor of $8 x^{4} + 4 x^{3} - 16 x^{2} + 10 x + m$ .

Answer:

m=-2
Solution
We know that if (x+a) is factor of the polynomial f(x), then it always satisfies f(-a)=0
Given,

2 x - 1

is a factor of

8 x^{4} + 4 x^{3} - 16 x^{2} + 10 x + m

Let

p (x) = 8 x^{4} + 4 x^{3} - 16 x^{2} + 10 x + m

g (x) = 2 x - 1

According to remainder theorem if g(x) is a factor of p(x) then

p (\frac{1}{2}) = 0

p (\frac{1}{2}) = 8 {(\frac{1}{2})}^{4} + 4 {(\frac{1}{2})}^{3} - 16 {(\frac{1}{2})}^{2} + 10 (\frac{1}{2}) + m 0 = 8 \times \frac{1}{16} + 4 \times \frac{1}{8} - 16 \times \frac{1}{4} + 5 + m 0 = \frac{1}{2} + \frac{1}{2} - 4 + 5 + m 0 = 1 + 1 + m m = - 2

Question:22

If $x + 1$ is a factor of $a x^{3} + x^{2} - 2 x + 4 a - 9$ , find the value of a.

Answer:

a = 2
Solution:
We know that if (x+a) is factor of the polynomial f(x), then it always satisfies f(-a)=0
Given

x + 1

is a factor of

a x^{3} + x^{2} - 2 x + 4 a - 9

Let

p (x) = a x^{3} + x^{2} - 2 x + 4 a - 9

g (x) = x + 1

According to remainder theorem if g(x) is a factor of p(x) then p(-1)=0

p (- 1) = a (- 1)^{3} + (- 1)^{2} - 2 (- 1) + 4 a - 9 0 = - a + 1 + 2 + 4 a - 9 0 = 3 a - 6 3 a = 6 a = \frac{6}{3} = 2

Hence a = 2.

Question:23

Factorize :
(i) $x^{2} + 9 x + 18$
(ii) $6 x^{2} + 7 x - 3$
(iii) $2 x^{2} - 7 x - 15$
(iv) $84 - 2 r - 2 r^{2}$

Answer:

(i) (x + 6) (x + 3)

Solution
Given,

x^{2} + 9 x + 18

To factorize

a x^{2} + b x + c

, we have to distribute “bx” into “px” and “qx” such that
p + q = b and p.q = a.c
The given equation can be written as:

x^{2} + 9 x + 18 \Rightarrow x^{2} + 3 x + 6 x + 18 {6 + 3 = 9, (6) (3) = 18} \Rightarrow x (x + 3) + 6 (x + 3) \Rightarrow (x + 3) (x + 6) .

(i i) (2 x + 3) (3 x - 1)

Solution
Given,

6 x^{2} + 7 x - 3

To factorize

a x^{2} + b x + c

, we have to distribute “bx” into “px” and “qx” such that
p + q = b and p.q = a.c
The given equation can be written as:

6 x^{2} + 7 x - 3 \Rightarrow 6 x^{2} + 9 x - 2 x - 3 {9 - 2 = 7, (9) (- 2) = (6) (- 3)} \Rightarrow (6 x^{2} + 9 x) - (2 x + 3) \Rightarrow 3 x (2 x + 3) - 1 (2 x + 3) \Rightarrow (2 x + 3) (3 x - 1) .

(i i i) (x - 5) (2 x - 3)

Solution
Given,

2 x^{2} - 7 x - 15

To factorize

a x^{2} + b x + c

, we have to distribute “bx” into “px” and “qx” such that
p + q = b and p.q = a.c
The given equation can be written as:

2 x^{2} - 7 x - 15 \Rightarrow 2 x^{2} - 10 x + 3 x - 15 {- 10 + 3 = - 7, (- 10) (3) = (2) (- 15)} \Rightarrow (2 x^{2} - 10 x) + (3 x - 15) \Rightarrow 2 x (x - 5) + 3 (x - 5) \Rightarrow (x - 5) (2 x + 3) .

(i v) (6 - r) (14 - 2 r)

Solution
Given,

84 - 2 r - 2 r^{2}

To factorize

a x^{2} + b x + c

, we have to distribute “bx” into “px” and “qx” such that
p + q = b and p.q = a.c
The given equation can be written as:

84 - 2 r - 2 r^{2} \Rightarrow 84 - 14 r + 12 r - 2 r^{2} {- 14 + 12 = - 2, (- 14) (12) = (84) (- 2)} \Rightarrow (84 - 14 r) + (12 r - 2 r^{2}) \Rightarrow 14 (6 - r) + 2 r (6 - r) \Rightarrow (6 - r) (14 + 2 r) .

Question:24

Factorize :
$(i) 2 x^{3} - 3 x^{2} - 17 x + 30$
$(i i) x^{3} - 6 x^{2} + 11 x - 6$
$(i i i) x^{3} + x^{2} - 4 x - 4$
$(i v) 3 x^{3} - x^{2} - 3 x + 1$

Answer:

(i) (x - 2) (x + 3) (2 x - 5)

Solution
Let

p (x) = 2 x^{3} - 3 x^{2} - 17 x + 30

By trail, we find that

p (2) = 0

p (2) = 2 (2)^{3} - 3 (2)^{2} - 17 (2) + 30 = 16 - 12 - 34 + 30 = 0

Hence

x - 2

is a factor of

p (x)

2 x^{3} - 3 x^{2} - 17 x + 30 = (x - 2) (2 x^{2} + x - 15) = (x - 2) (2 x^{2} + 6 x - 5 x - 15) = (x - 2) (2 x (x + 3) - 5 (x + 3)) 2 x^{3} - 3 x^{2} - 17 x + 30 = (x - 2) (x + 3) (2 x - 5)

Hence the factorized form is

(x - 2) (x + 3) (2 x - 5)

(i i) (x - 1) (x - 2) (x - 3)

Solution
Let

p (x) = x^{3} - 6 x^{2} + 11 x - 6

By trail, we find that

p (1) = 0

p (1) = 1 - 6 + 11 - 6 = 0

Hence

x - 1

is a factor of

p (x)

p (x) = (x - 1) (x^{2} - 5 x + 6) = (x - 1) (x^{2} - 3 x - 2 x + 6)) = (x - 1) (x (x - 3) - 2 (x - 3)) = (x - 1) (x - 2) (x - 3)

Hence the factorized form is

(x - 1) (x - 2) (x - 3)

(i i i) (x + 1) (x - 2) (x + 2)

Solution
Let

p (x) = x^{3} + x^{2} - 4 x - 4

By trail, we find that

p (- 1) = 0

p (- 1) = - 1 + 1 + 4 - 4 = 0

Hence

x + 1

is a factor of

p (x)

p (x) = (x + 1) (x^{2} - 4) = (x + 1) (x^{2} - (2)^{2})) = (x + 1) (x - 2) (x + 2) (u s i n g a^{2} - b^{2} = (a - b) (a + b)) = (x + 1) (x - 2) (x + 2)

Hence the factorized form is

(x + 1) (x - 2) (x + 2)

(i v) (x - 1) (x + 1) (3 x - 1)

Solution
Let

p (x) = 3 x^{3} - x^{2} - 3 x - 1

By trail, we find that

p (1) = 0

p (1) = 3 (1)^{3} - (1)^{2} + 3 (1) + 1 = 3 - 1 - 3 + 1 = 0

Hence

x - 1

is a factor of

p (x)

p (x) = (x - 1) (3 x^{3} + 2 x - 1) = (x - 1) (3 x^{2} + 3 x - x - 1) = (x - 1) (3 x (x + 1) - 1 (x + 1)) = (x - 1) (x + 1) (3 x - 1)

Hence the factorized form is

(x - 1) (x + 1) (3 x - 1)

Question:25

Using suitable identity, evaluate the following
$(i) 103^{3} (i i) 101 \times 102 (i i i) 999^{2}$

Answer:

(i) 1092727
Solution
Given,

(103)^{3} = (100 + 3)^{3} = 100^{3} + 3^{3} + 3 (100)^{2} (3) + 3 (100) (3)^{2} (u s i n g (a + b)^{3} = a^{3} + b^{3} + 3 a^{2} b + 3 a b^{2}) = 1000000 + 27 + 90000 + 2700 = 1092727

Hence the answer is 1092727
(ii) 10302
Solution
Given,

101 \times 102 = (100 + 1) (100 + 2)

Using

(x + a) (x + b) = x^{2} + (a + b) x + a b

Put x = 100, a = 1, b = 2

(100 + 1) (100 + 2) = 100^{2} + (1 + 2) 100 + (1) (2) = 10000 + 300 + 2 = 10302

Hence the answer is 10302
(iii)998001
Solution
Given,

(999)^{2} = (1000 - 1)^{2}

Using

(a - b)^{2} = a^{2} + b^{2} - 2 a b

Putting a = 1000, b = 1

(1000 - 1)^{2} = (1000)^{2} + (1)^{2} - 2 (1000) (1) = 1000000 + 1 - 2000 = 998001

Hence the answer is 998001.

Question:26

Factorize the following
$(i) 4 x^{2} + 20 x + 25$
$(i i) 9 y^{2} - 66 y z + 121 z^{2}$
$(i i i) {(2 x + \frac{1}{3})}^{2} - {(x - \frac{1}{2})}^{2}$

Answer:

(i) (2 x + 5) (2 x + 5)

Solution
Given,

4 x^{2} + 20 x + 25

We can write the given expression as:

(2 x)^{2} + (5)^{2} + 2 (2 x) (5) \dots (i)

Now, we know that

(a + b)^{2} = a^{2} + b^{2} + 2 a b

Comparing equation (i) with the above identity, we have:
a = 2x, b = 5
So,

(2 x)^{2} + (5)^{2} + 2 (2 x) (5) = (2 x + 5)^{2} = (2 x + 5) (2 x + 5)

(i i) (3 y - 11 z) (3 y - 11 z)

Solution
Given,

9 y^{2} - 66 y z + 121 z^{2}

We can write the given expression as:

(3 y)^{2} + (11 z)^{2} - 2 (3 y) (11 z) \dots (i)

Now, we know that

(a - b)^{2} = a^{2} + b^{2} - 2 a b

Comparing equation (i) with the above identity, we have:
a = 3y, b = 11z
So,

= (3 y)^{2} + (11 z)^{2} - 2 (3 y) (11 z) = (3 y - 11 z)^{2} = (3 y - 11 z) (3 y - 11 z)

(i i i) (x + \frac{5}{6}) (3 x - \frac{1}{6})

Solution
Given,

{(2 x + \frac{1}{3})}^{2} - {(x - \frac{1}{2})}^{2}

Now, we know that

a^{2} - b^{2} = (a - b) (a + b)

So,

{(2 x + \frac{1}{3})}^{2} - {(x - \frac{1}{2})}^{2} = (2 x + \frac{1}{3} - (x - \frac{1}{2})) . (2 x + \frac{1}{3} + (x - \frac{1}{2})) = (2 x + \frac{1}{3} - x + \frac{1}{2}) . (3 x + \frac{2 - 3}{6}) = (x + \frac{2 + 3}{6}) (3 x - \frac{1}{6}) = (x + \frac{5}{6}) (3 x - \frac{1}{6})

Hence the factorized form is:

(x + \frac{5}{6}) (3 x - \frac{1}{6})

Question:27

Factorize the following :-
$(i) 9 x^{2} - 12 x + 3$
$(i i) 9 x^{2} - 12 x + 4$

Answer:

(i) (x - 1) (9 x - 3)

Given,

9 x^{2} - 12 x + 3

To factorize ax² + bx + c, we have to distribute “bx” into “px” and “qx” such that
p + q = b and p.q = a.c
The given equation can be written as:

9 x^{2} - 9 x - 3 x + 3

{-9 - 3 = -12, (-9) (-3) = (9) (3)}

= (9 x^{2} - 9 x) - (3 x - 3) = 9 x (x - 1) - 3 (x - 1) = (x - 1) (9 x - 3)

(i i) (3 x - 2) (3 x - 2)

Given,

9 x^{2} - 12 x + 4

To factorize ax² + bx + c, we have to distribute “bx” into “px” and “qx” such that
p + q = b and p.q = a.c
The given equation can be written as:

9 x^{2} - 6 x - 6 x + 4

{-6 - 6 = -12, (-6) (-6) = (9) (4)}

= (9 x^{2} - 6 x) - (6 x - 4) = 3 x (3 x - 2) - 2 (3 x - 2) = (3 x - 2) (3 x - 2)

Question:28

Expand the following
$(i) (4 a - b + 2 c)^{2} (i i) (3 a - 5 b - c)^{2} (i i i) (- x + 2 y - 3 z)^{2}$

Answer:

(i)

16 a^{2} + b^{2} + 4 c^{2} - 8 a b - 4 b c + 16 a c

Solution

(4 a - b + 2 c)^{2} = (4 a + (- b) + 2 c)^{2} \dots \dots (i)

Now, we know that

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 a c

Comparing the equation (i) with the above identity, we get:

(4 a - b + 2 c)^{2} = (4 a)^{2} + (- b)^{2} + (2 c)^{2} + 2 (4 a) (- b) + 2 (- b) (2 c) + 2 (4 a) (2 c) = 16 a^{2} + b^{2} + 4 c^{2} - 8 a b - 4 b c + 16 a c

Hence the expanded form is

16 a^{2} + b^{2} + 4 c^{2} - 8 a b - 4 b c + 16 a c

(ii)

9 a^{2} + 25 b^{2} + c^{2} - 30 a b + 10 b c - 6 a c

Solution

(3 a - 5 b - c)^{2} = (3 a + (- 5 b) + (- c))^{2} \dots \dots (i)

Now, we know that

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 a c

Comparing the equation (i) with the above identity, we get:

(3 a - 5 b - c)^{2} = (3 a)^{2} + (- 5 b)^{2} + (- c)^{2} + 2 (3 a) (- 5 b) + 2 (- 5 b) (- c) + 2 (3 a) (- c) = 9 a^{2} + 25 b^{2} + c^{2} - 30 a b + 10 b c - 6 a c

Hence the expanded form is

9 a^{2} + 25 b^{2} + c^{2} - 30 a b + 10 b c - 6 a c

(iii)

x^{2} + 4 y^{2} + 9 z^{2} - 4 x y - 12 y z + 6 x z

Solution
Now, we know that

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 a c

Comparing the equation (i) with the above identity, we get:

(- x + 2 y - 3 z)^{2} = (- x)^{2} + (2 y)^{2} + (- 3 z)^{2} + 2 (- x) (2 y) + 2 (2 y) (- 3 z) + 2 (- 3 z) (- x) = x^{2} + 4 y^{2} + 9 z^{2} - 4 x y - 12 y z + 6 x z

Hence the expanded form is

x^{2} + 4 y^{2} + 9 z^{2} - 4 x y - 12 y z + 6 x z

Question:29

Factorize the following
$(i) 9 x^{2} + 4 y^{2} + 16 z^{2} + 12 x y - 16 y z - 24 x z$
$(i i) 25 x^{2} + 16 y^{2} + 4 z^{2} - 40 x y + 16 y z - 20 x z$
$(i i i) 16 x^{2} + 4 y^{2} + 9 z^{2} - 16 x y - 12 y z + 24 x z$

Answer:

(i) (3 x + 2 y - 4 z) (3 x + 2 y - 4 z)

Solution
Given,

9 x^{2} + 4 y^{2} + 16 z^{2} + 12 x y - 16 y z - 24 x z

This can be written as:

(3 x)^{2} + (2 y)^{2} + (4 z)^{2} + 2 (3 x) (2 y) + 2 (2 y) (- 4 z) + 2 (3 x) (- 4 z)

…(i)
We know that,

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a

Comparing equation (i) with RHS of above identity, we get:
a = 3x, b = 2y, c = -4z
So, we get:

(3 x)^{2} + (2 y)^{2} + (- 4 z)^{2} + 2 (3 x) (2 y) + 2 (2 y) (- 4 z) + 2 (3 x) (- 4 z)

= (3 x + 2 y - 4 z)^{2}

Hence the factorized form is

(3 x + 2 y - 4 z) (3 x + 2 y - 4 z)

(i i) (- 5 x + 4 y + 2 z) (- 5 x + 4 y + 2 z)

Solution
Given,

25 x^{2} + 16 y^{2} + 4 z^{2} - 40 x y + 16 y z - 20 x z

This can be written as:

(- 5 x)^{2} + (4 y)^{2} + (2 z)^{2} + 2 (- 5 x) (4 y) + 2 (4 y) (2 z) + 2 (- 5 x) (2 z)

…(i)
We know that,

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a

Comparing equation (i) with RHS of above identity, we get:
a = -5x, b = 4y, c = 2z
So, we get:

(- 5 x)^{2} + (4 y)^{2} + (2 z)^{2} + 2 (- 5 x) (4 y) + 2 (4 y) (2 z) + 2 (- 5 x) (2 z)

= (- 5 x + 4 y + 2 z)^{2} = (- 5 x + 4 y + 2 z) (- 5 x + 4 y + 2 z)

Hence the factorized form is

(- 5 x + 4 y + 2 z) (- 5 x + 4 y + 2 z)

(i i i) (4 x - 2 y + 3 z) (4 x - 2 y + 3 z)

Solution
Given,

16 x^{2} + 4 y^{2} + 9 z^{2} - 16 x y - 12 y z + 24 x z

This can be written as:

(4 x)^{2} + (- 2 y)^{2} + (3 z)^{2} + 2 (4 x) (- 2 y) + 2 (- 2 y) (3 z) + 2 (4 x) (3 z)

…(i)
We know that,

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a

Comparing equation (i) with RHS of above identity, we get:
a = 4x, b = -2y, c = 3z
So, we get:

(4 x)^{2} + (- 2 y)^{2} + (3 z)^{2} + 2 (4 x) (- 2 y) + 2 (- 2 y) (3 z) + 2 (4 x) (3 z)

= (4 x - 2 y + 3 z)^{2} = (4 x - 2 y + 3 z) (4 x - 2 y + 3 z)

Hence the factorized form is

(4 x - 2 y + 3 z) (4 x - 2 y + 3 z)

Question:30

If $a + b + c = 9$ and $a b + b c + c a = 26$ , find $a^{2} + b^{2} + c^{2}$ .

Answer:

29
Solution
Given ab+bc+ac=26, a+b+c=9
As we know that

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a)

…..(1)
Put a+b+c=9 , ab+bc+ac=26 in (1)

(9)^{2} = a^{2} + b^{2} + c^{2} + 2 \times 26 a^{2} + b^{2} + c^{2} = 81 - 52 a^{2} + b^{2} + c^{2} = 29

Hence the answer is 29.

Question:31

Expand the following :
$(i) (3 a - 2 b)^{3}$
$(i i) {(\frac{1}{x} + \frac{y}{3})}^{3}$
$(i i i) {(4 - \frac{1}{3 x})}^{3}$

Answer:

(i) 27 a^{3} - 8 b^{2} - 54 a^{2} b + 36 a b^{2}

Solution: Given

(3 a - 2 b)^{3}

We know that

(a - b)^{3} = a^{3} - b^{3} - 3 a^{2} b + 3 a b^{2}

So we get:

(3 a - 2 b)^{3} = (3 a)^{3} - (2 b)^{3} - 3 (3 a)^{2} (2 b) + 3 (3 a) (2 b)^{2}

(3 a - 2 b)^{3} = 27 a^{3} - 8 b^{2} - 54 a^{2} b + 36 a b^{2}

Hence the expanded form is

27 a^{3} - 8 b^{2} - 54 a^{2} b + 36 a b^{2}

(i i) \frac{1}{x^{3}} + \frac{y^{3}}{27} + \frac{y}{x^{2}} + \frac{y^{2}}{3 x}

Solution: Given

{(\frac{1}{x} + \frac{y}{3})}^{3}

We know that

(a + b)^{3} = a^{3} + b^{3} + 3 a^{2} b + 3 a b^{2}

So we get:

{(\frac{1}{x} + \frac{y}{3})}^{3} = {(\frac{1}{x})}^{3} + {(\frac{y}{3})}^{3} + 3 {(\frac{1}{x})}^{2} (\frac{y}{3}) + 3 (\frac{1}{x}) {(\frac{y}{3})}^{2}

= \frac{1}{x^{3}} + \frac{y^{3}}{27} + \frac{3 y}{3 x^{2}} + \frac{3 y^{2}}{9 x}

= \frac{1}{x^{3}} + \frac{y^{3}}{27} + \frac{y}{x^{2}} + \frac{y^{2}}{3 x}

Hence the expanded form is

= \frac{1}{x^{3}} + \frac{y^{3}}{27} + \frac{y}{x^{2}} + \frac{y^{2}}{3 x}

(i i i) 64 - \frac{1}{27 x^{3}} - \frac{16}{x} + \frac{4}{3 x^{2}}

Solution: Given

{(4 - \frac{1}{3 x})}^{3}

We know that

(a - b)^{3} = a^{3} - b^{3} - 3 a^{2} b + 3 a b^{2}

So we get:

{(4 - \frac{1}{3 x})}^{3} = (4)^{3} - {(\frac{1}{3 x})}^{3} - 3 (4)^{2} (\frac{1}{3 x}) + 3 (4) {(\frac{1}{3 x})}^{2}

= 64 - \frac{1}{27 x^{3}} - \frac{48}{3 x} + \frac{12}{9 x^{2}}

= 64 - \frac{1}{27 x^{3}} - \frac{16}{x} + \frac{4}{3 x^{2}}

Hence the expanded form is

64 - \frac{1}{27 x^{3}} - \frac{16}{x} + \frac{4}{3 x^{2}}

Question:32

Factorize the following :
$(i) 1 - 64 a^{3} - 12 a + 48 a^{2} (i i) 8 p^{3} + \frac{12}{5} p^{2} + \frac{6}{25} p + \frac{1}{125}$

Answer:

(i)(1−4a)(1−4a)(1−4a)
Given, 1−64a³−12a+48a²
The given equation can be written as:
=(1)³−(4a)³−3(1)²(4a)+3(1)(4a)² ..........(i)
Now we know that:
a³−b³−3a²b+3ab²=(a−b)³
Comparing equation (i) with the above identity, we get:
∴(1)³−(4a)³−3(1)²(4a)+3(1)(4a)²
=(1−4a)³
Hence the factorized form is (1−4a)(1−4a)(1−4a)

(ii)

(2 p + \frac{1}{5}) (2 p + \frac{1}{5}) (2 p + \frac{1}{5})

Given,

8 p^{3} + \frac{12}{5} p^{2} + \frac{6}{25} p + \frac{1}{125}

The given equation can be written as:

(2 p)^{3} + (3 \times (2 p)^{2} \times \frac{1}{5}) + (3 (2 p) {(\frac{1}{5})}^{2}) + {(\frac{1}{5})}^{3} = (2 p)^{3} + {(\frac{1}{5})}^{3} + (3 \times (2 p)^{2} \times \frac{1}{5}) + (3 (2 p) {(\frac{1}{5})}^{2})

…(i)
Now we know that:

a^{3} + b^{3} + 3 a^{2} b + 3 a b^{2} = (a + b)^{3}

Comparing equation (i) with the above identity, we get:

(2 p)^{3} + {(\frac{1}{5})}^{3} + (3 \times (2 p)^{2} \times \frac{1}{5}) + (3 (2 p) {(\frac{1}{5})}^{2}) = {(2 p + \frac{1}{5})}^{3}

Hence the factorized form is

(2 p + \frac{1}{5}) (2 p + \frac{1}{5}) (2 p + \frac{1}{5})

Question:33

Find the following product
(i) $(\frac{x}{2} + 2 y) (\frac{x^{2}}{4} - x y + 4 y^{2})$
(ii) $(x^{2} - 1) (x^{4} + x^{2} + 1)$

Answer:

(i)

\frac{x^{3}}{8} + 8 y^{3}

Given:

(\frac{x}{2} + 2 y) (\frac{x^{2}}{4} - x y + 4 y^{2})

= \frac{x}{2} (\frac{x^{2}}{4} - x y + 4 y^{2}) + 2 y (\frac{x^{2}}{4} - x y + 4 y^{2}) = \frac{x^{3}}{8} - \frac{x^{2} y}{2} + \frac{4 x y^{2}}{2} + \frac{2 y x^{2}}{4} - 2 x y^{2} + 8 y^{3} = \frac{x^{3}}{8} - \frac{x^{2} y}{2} + \frac{x^{2} y}{2} + 2 x y^{2} - 2 x y^{2} + 8 y^{3} = \frac{x^{3}}{8} + 8 y^{3}

(ii)

x^{6} - 1

Given

(x^{2} - 1) (x^{4} + x^{2} + 1)

= x^{2} (x^{4} + x^{2} + 1) - 1 (x^{4} + x^{2} + 1) = x^{6} + x^{4} + x^{2} - x^{2} - x^{4} - x^{2} - 1 = x^{6} - 1

Question:34

Factorise:
(i) $1 + 64 x^{3}$
(ii) $a^{3} - 2 \sqrt{2} b^{3}$

Answer:

(i)

(1 + 4 x) (1 + 16 x^{2} - 4 x)

Given

1 + 64^{3}

(1)^{3} + (4 x)^{3}

we know that

a^{3} + b^{3} = (a + b) (a^{2} + b^{2} - a b)

(1)^{3} + {(4 x)}^{3} = (1 + 4 x) ((1)^{2} + {(4 x)}^{2} - (1) (4 x))

(1 + 4 x) (1 + 16 x^{2} - 4 x)

(ii)

(a - \sqrt{2} b) (a^{2} + 2 b^{2} + \sqrt{2} a b)

Given

a^{3} - 2 \sqrt{2} b^{3}

This can be written as

(a)^{3} - (\sqrt{2} b)^{3}

we know that

a^{3} - b^{3} = (a + b) (a^{2} + b^{2} + a b)

(a)^{3} - (\sqrt{2} b)^{3} = (a - \sqrt{2} b) (a^{2} + (\sqrt{2} b)^{2} + \sqrt{2} a b)

(a - \sqrt{2} b) (a^{2} + 2 b^{2} + \sqrt{2} a b)

Question:35

Find the following product $(2 x - y + 3 z) (4 x^{2} + y^{2} + 9 z^{2} + 2 x y + 3 y z - 6 x z)$

Answer:

8 x^{3} - y^{3} + 27 x^{3}

Solution
Given:

(2 x - y + 3 z) (4 x^{2} + y^{2} + 9 z^{2} + 2 x y + 3 y z - 6 x z)

This can be written as

(2 x - y + 3 z) ((2 x)^{2} + (y)^{2} + (3 z)^{2} - (2 x) (- y) - (- y) (3 z) - (2 x) (3 z))

We know that

a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a)

Comparing equation (i) with the above identity we get:
a = 2x, b = -y, c = 3z
Hence,

(2 x - y + 3 z) ((2 x)^{2} + (y)^{2} + (3 z)^{2} - (2 x) (- y) - (- y) (3 z) - (2 x) (3 z)) = (2 x)^{3} + (- y)^{3} + (3 z)^{3} - 3 (2 x) (- y) (3 z) = 8 x^{3} - y^{3} + 27 z^{3} + 18 x y z

Question:36

Factorize
(i) $a^{3} - 8 b^{3} - 64 c^{3} - 24 a b c$
(ii) $2 \sqrt{2} a^{3} + 8 b^{3} - 27 c^{3} + 18 \sqrt{2} a b c$

Answer:

(i)

(a - 2 b - 4 c) (a^{2} + 4 b^{2} + 16 c^{2} + 2 a b - 8 b c + 4 a c)

Solution:
Given:

a^{3} - 8 b^{3} - 64 c^{3} - 24 a b c

This can be written as:

a^{3} + (- 2 b)^{3} + (- 4 c)^{3} - 3 (a) (- 2 b) (- 4 c)

We know that

a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - a c)

Using this identity we get

a^{3} + (- 2 b)^{3} + (- 4 c)^{3} - 3 (a) (- 2 b) (- 4 c)

= (a + (- 2 b) + (- 4 c)) (a^{2} - (- 2 b)^{2} + (- 4 c)^{2} - (a) (- 2 b) - (- 2 b) (- 4 c) - (a) (- 4 c))

= (a - 2 b - 4 c) (a^{2} + 4 b^{2} + 16 c^{2} + 2 a b - 8 b c + 4 a c)

(i)

(\sqrt{2} a + 2 b - 3 c) (2 a^{2} + 4 b^{2} + 9 c^{2} - 2 \sqrt{2} a b + 6 b c + 3 \sqrt{2} c a)

Solution:
Given:

2 \sqrt{2} a^{3} + 8 b^{3} - 27 c^{3} + 18 \sqrt{2} a b c

This can be written as:

(\sqrt{2} a)^{3} + (2 b)^{3} - (3 c)^{3} - 3 (\sqrt{2} a) (2 b) (3 c)

We know that

a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - a c)

Using this identity we get

(\sqrt{2} a)^{3} + (2 b)^{3} - (3 c)^{3} - 3 (\sqrt{2} a) (2 b) (3 c)

= ((\sqrt{2} a) + (2 b) + (- 3 c)) ((\sqrt{2} a)^{2} + (2 b)^{2} + (- 3 c)^{2} - (\sqrt{2} a) (2 b) - (2 b) (- 3 c) - (\sqrt{2} a) (- 3 c))

(\sqrt{2} a + 2 b - 3 c) (2 a^{2} + 4 b^{2} + 9 c^{2} - 2 \sqrt{2} a b + 6 b c + 3 \sqrt{2} c a)

Question:37

Without actually calculating the cubes, find the value of
(i) ${(\frac{1}{2})}^{3} + {(\frac{1}{3})}^{3} - {(\frac{5}{6})}^{3}$
(ii) $(0.2)^{3} - (0.3)^{3} + (0.1)^{3}$

Answer:

(i)

\frac{- 5}{12}

Given

{(\frac{1}{2})}^{3} + {(\frac{1}{3})}^{3} - {(\frac{5}{6})}^{3}

We know that

a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a)

When a + b + c = 0, we get

a^{3} + b^{3} + c^{3} = 3 a b c

Here,

\frac{1}{2} + \frac{1}{3} - \frac{5}{6} = \frac{3 + 2 - 5}{6} = \frac{0}{6} = 0

So using the above identity, we get:

∴ {(\frac{1}{2})}^{3} + {(\frac{1}{3})}^{3} + {(\frac{- 5}{6})}^{3} = 3 \times \frac{1}{2} \times \frac{1}{3} \times \frac{- 5}{6} = \frac{- 5}{12}

Hence the answer is

\frac{- 5}{12}

(ii)-0.018
Solution
Given:

(0.2)^{3} - (0.3)^{3} + (0.1)^{3}

We know that

a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a)

When a + b + c = 0, we get

a^{3} + b^{3} + c^{3} = 3 a b c

Here,

0.2 + (- 0.3) + 0.1 = 0

So,

(0.2)^{3} - (0.3)^{3} + (0.1)^{3} = 3 (0.2) (- 0.3) (0.1) = - 0.018

Hence the answer is -0.018.

Question:38

Without finding the cubes, factorize $(x - 2 y)^{3} + (2 y - 3 z)^{3} + (3 z - x)^{3}$

Answer:

3 (x - 2 y) (2 y - 3 z) (3 z - x)

solution :
given

(x - 2 y)^{3} + (2 y - 3 z)^{3} + (3 z - x)^{3}

We know that

a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a)

When a + b + c = 0, we get

a^{3} + b^{3} + c^{3} = 3 a b c

Here,

(x - 2 y) + (2 y - 3 z) + (3 z - x) = 0

So using the above identity, we get:

(x - 2 y)^{3} + (2 y - 3 z)^{3} + (3 z - x)^{3}

= 3 (x - 2 y) (2 y - 3 z) (3 z - x)

Hence the answer is

3 (x - 2 y) (2 y - 3 z) (3 z - x)

Question:39

Find the value of
(i) $x^{3} + y^{3} - 12 x y - 64$ when $x + y = - 4$
(ii) $x^{3} - 8 y^{3} - 36 x y - 216$ when $x = 2 y + 6$

Answer:

(i) 0
Solution :-
Here

x + y = - 4

We know that

a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a)

When a + b + c = 0, we get

a^{3} + b^{3} + c^{3} = 3 a b c

Here,

x + y + 4 = 0

So using the above identity, we get:

x^{3} + y^{3} + 4^{3} = 3 \times x \times y \times 4 = 12 x y \dots \dots (i)

Now

x^{3} + y^{3} - 12 x y + 64 = x^{3} + y^{3} + (4)^{3} - 12 x y 12 x y - 12 x y = 0

{from equation i}
Hence the answer is 0
(ii) 0
Solution :-
We know that

a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a)

When a + b + c = 0, we get

a^{3} + b^{3} + c^{3} = 3 a b c

Here,

x - 2 y - 6 = 0

So using the above identity, we get:

x^{3} + (- 2 y)^{3} - (- 6)^{3} = 3 x (- 2 y) (- 6) = 36 x y

x^{3} + 8 y^{3} - 216 = 36 x y

....................(i)
Now

x^{3} - 8 y^{3} - 216 - 36 x y = 36 x y - 36 x y = 0

Hence the answer is 0.

Question:40

Give possible expressions for the length and breadth of the rectangle whose area is given by $4 a^{2} + 4 a - 3$

Answer:

Length (2a+3)
Breadth (2a-1)
Or
Length = (2a-1)
Breadth =(2a+3)
Solution:
Given : Area of rectangle is 4a²+4a-3 …..(1)
Factorize equation 1, we get
=4a²+6a-2a-3
=2a(2a+3)-1(2a+3)
=(2a+3)(2a-1)
We know that area of rectangle is length × breadth
Hence
Length (2a+3)
Breadth (2a-1)
Or
Length = (2a-1)
Breadth =(2a+3)

NCERT Exemplar Class 9 Maths Solutions Chapter 2-Exercise 2.4

Question:1

If the polynomials az³ + 4z² + 3z – 4 and z³ – 4z + a leave the same remainder when divided by z – 3, find the value of a.

Answer:

a = –1
Solution
Let

p (z) = a z^{3} + 4 z^{2} + 3 z - 4

q (z) = z^{3} - 4 z + a

According to remainder theorem when p(x) is divided by (x+a) then the remainder is (-a) .
So when p(z) is divided by z – 3 then remainder is given by p(3).

p (3) = a \times 3^{3} + 4 \times 3^{2} + 3 \times 3 - 4 p (3) = 27 a + 36 + 9 - 4 p (3) = 27 a + 41 \dots \dots (1)

Similarly

q (3) = 3^{3} - 4 \times 3 + a q (3) = 27 - 12 + a q (3) = 15 + a \dots \dots (2)

According to question p(3) = q(3)

27 a + 41 = 15 + a 27 a - a = 15 - 41 26 a = - 26 a = - \frac{26}{26} a = - 1

Hence the answer is a=-1.

Question:2

The polynomial p(x) = x⁴ – 2x³ + 3x² – ax + 3a – 7 when divided by x + 1 leaves the remainder 19. Find the values of a. Also find the remainder when p(x) is divided by x + 2.

Answer:

Value of a is 5
Remainder is 62
Solution
Given: p(x) = x⁴– 2x³ + 3x²– ax + 3a – 7
According to remainder theorem when p(x) is divided by (x+a) then the remainder is p(-a) .
When we divide p(x) by (x + 1) then according to remainder theorem remainder is p(–1)

p (- 1) = (- 1)^{4} - 2 (- 1)^{3} + 3 (- 1)^{2} - a (- 1) + 3 a - 7 = 1 + 2 + 3 + a + 3 a - 7 = 4 a - 1

According to question p(-1)=19

\Rightarrow 4 a - 1 = 19 \Rightarrow 4 a = 19 + 1 \Rightarrow 4 a = 20 \Rightarrow a = \frac{20}{4} = 5 ∴ p (x) = x^{4} - 2 x^{3} + 3 x^{2} - 5 x + 3 (5) - 7 = x^{4} - 2 x^{3} + 3 x^{2} - 5 x + 8

When we divide p(x) by x + 2 then we get the remainder p(–2)

p (- 2) = (- 2)^{4} - 2 (- 2)^{3} + 3 (- 2)^{2} - 5 (- 2) + 8 = 16 + 16 + 12 + 10 + 8 = 62

Hence, value of a is 5.
Remainder is 62.

Question:3

If both x – 2 and $x - \frac{1}{2}$ are factors of px² + 5x + r, show that p = r.

Answer:

Let

f (x) = p x^{2} + 5 x + r

We know that if

(x + a)

is factor of the polynomial f(x), then it always satisfies

f (- a) = 0

(x-2) is a factor of f(x) then f(2) = 0
So,

p (2)^{2} + 5 (2) + r = 0 4 p + 10 + r = 0 \dots \dots (1)

…..(1)
Also

x - \frac{1}{2}

is a factor of f(x) then

f (\frac{1}{2}) = 0

p {(\frac{1}{2})}^{2} + 5 (\frac{1}{2}) + r = 0 \frac{p}{4} + \frac{5}{2} + r = 0 p + 10 + 4 r = 0 \dots (2)

Since (x – 2) and

x - \frac{1}{2}

are factors of f(x) then equation 1 and 2 are equal

4 p + 10 + r = 9 + 10 + 4 r 4 p - p = 4 r - r 3 p = 3 r p = r

Hence proved.

Question:4

Without actual division, prove that 2x⁴ – 5x³ + 2x² – x + 2 is divisible by x² – 3x + 2. [Hint: Factorize x²–3x + 2]

Answer:

First of all, factorize x² – 3x + 2
We get
x² – 2x-3x + 2
= x(x-2)-1(x-2)
= (x-2)(x-1)
According to remainder theorem when p(x) is divided by (x+a) then the remainder is p(-a) .
So if p(x) is divide by x² – 3x + 2 then p(2) and p(1) must be zero
p(2)=2(2)⁴ -5(2)³ +2(2)² -2+2
=32-40+8
=0
p(1)=2(1)⁴ -5(1)³ +2(1)² -1+2
=2-5+2-1+2
=6-6
=0
Hence both p(1) and p(2) are zero therefore p(x) is divisible by x² – 3x + 2
Hence proved.

Question:5

Simplify: $(2 x - 5 y)^{3} - (2 x + 5 y)^{3}$

Answer:

- 250 y^{2} - 120 x^{2} y

Solution
We know that

(a - b)^{3} = a^{3} - b^{3} - 3 a b (a - b) (a + b)^{3} = a^{3} + b^{3} + 3 a b (a + b) (2 x - 5 y)^{3} - (2 x + 5 y)^{3} = (18 x^{3} - 125 y^{3} - 30 x y (2 x - 5 y)) - (18 x^{3} + 125 y^{3} + 30 x y (2 x + 5 y)) = - 250 y^{3} - 60 x^{2} y + 150 x y^{2} - 60 x^{2} y - 150 x y^{3} = - 250 y^{3} - 120 x^{2} y

Question:6

Multiply: $x^{2} + 4 y^{2} + z^{2} + 2 x y + x z - 2 y z$ by $(- z + x - 2 y)$

Answer:

x^{3} - 8 y^{3} - z^{3} - 6 x y z

Solution
We have,

(x^{2} + 4 y^{2} + z^{2} + 2 x y + x z - 2 y z) (- z + x - 2 y)

This can be written as:

(x + (- 2 y) + (- z)) ((x)^{2} + (- 2 y)^{2} + (z)^{2} - (x) (- z) - (- z) (- 2 y)) \dots \dots (i)

We know that

a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a)

So comparing the RHS of equation (i) with the above identity:
a = x
b = -2y
c = -z
We get:

(x + (- 2 y) + (- z)) ((x)^{2} + (- 2 y)^{2} + (z)^{2} - (x) (- z) - (- z) (- 2 y)) = (x)^{3} + (- 2 y)^{3} + (- z)^{3} - 3 (x) (- 2 y) (- z) = x^{3} - 8 y^{3} - z^{3} - 6 x y z

Hence the answer is

x^{3} - 8 y^{3} - z^{3} - 6 x y z

Question:7

If a, b, c are all non-zero and a + b + c = 0, prove that $\frac{a^{2}}{b c} + \frac{b^{2}}{c a} + \frac{c^{2}}{a b} = 3$

Answer:

Given,

\frac{a^{2}}{b c} + \frac{b^{2}}{c a} + \frac{c^{2}}{a b} = 3

L.H.S.

\frac{a^{2}}{b c} + \frac{b^{2}}{c a} + \frac{c^{2}}{a b}

Taking LCM of denominators, we get
L.H.S

\frac{a^{3} + b^{3} + c^{3}}{a b c}

Now we know that if a + b + c = 0 then

a^{3} + b^{3} + c^{3} = 3 a b c

Putting the value in above equation:
L.H.S =

\frac{3 a b c}{a b c}

= 3 = RHS
Hence proved.

Question:8

If a + b + c = 5 and ab + bc + ca = 10, then prove that a³ + b³ + c³ –3abc = – 25

Answer:

Given: (a + b + c) = 5, ab + bc + ca = 10
To Prove: a³ + b³ + c³ –3abc = – 25
We know that

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a) (5)^{2} = a^{2} + b^{2} + c^{2} + 2 (10) 25 = a^{2} + b^{2} + c^{2} + 20 25 - 20 = a^{2} + b^{2} + c^{2} 5 = a^{2} + b^{2} + c^{2}

Now,

a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a)

Putting the values, we get:

a^{3} + b^{3} + c^{3} - 3 a b c = 5 [5 - (a b + b c + c a)] = 5 (5 - 10) = 5 (- 5) = - 25 = R . H . S

Hence proved.

Question:9

prove that: $(a + b + c)^{3} - a^{3} - b^{3} - c^{3} = 3 (a + b) (b + c) (c + a)$

Answer:

prove that:

(a + b + c)^{3} - a^{3} - b^{3} - c^{3} = 3 (a + b) (b + c) (c + a)

Taking left hand side

(a + b + c)^{3} - a^{3} - b^{3} - c^{3} = ((a + b + c)^{3} - a^{3}) (b^{3} + c^{3}) \dots \dots \dots \dots (i)

Now using the identity

x^{3} - y^{3} = (x - y) (x^{2} + y^{2} + x y)

So,

((a + b + c)^{3} - a^{3}) = (a + b + c - a) [(a + b + c)^{2} + a^{2} + a (a + b + c)]

Now using

x^{3} - y^{3} = (x - y) (x^{2} + y^{2} + x y)

(b^{3} + c^{3}) = [(b + c) (b^{2} + c^{2} - b c)]

So equation (i) becomes:

(a + b + c)^{3} - a^{3} - b^{3} - c^{3} = (a + b + c - a) [(a + b + c)^{2} + a^{2} + a (a + b + c)] - [(b + c) (b^{2} + c^{2} - b c)] = (b + c) [(a + b + c)^{2} + a^{2} + a (a + b + c)] [(b + c) (b^{2} + c^{2} - b c)] \dots \dots \dots (i i)

Now,

(a + b + c)^{3} - a^{3} - b^{3} - c^{3} = (b + c) [a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a + a^{2} + a^{2} + a b + a c - (b^{2} + c^{2} - b c)] = (b + c) [b^{2} + c^{2} + 3 a^{2} + 3 a b + 3 a c - b^{2} - c^{2} + 3 b c] = (b + c) [3 (a^{2}) + a b + a c + b c] = 3 (b + c) [a (a + b) + c (a + b)] = 3 (b + c) [(a + b) (b + c)] = 3 (a + b) (b + c) (c + a) = R . H . S

Hence proved.

NCERT Exemplar Class 9 Maths Solutions Chapter 2 Contains the Solutions of the Questions Based on the Mentioned Topics:

◊ Polynomial, monomial, and binomial as an algebraic expression.

◊ Degree of a polynomial: Maximum summation of exponents of variables in any term.

◊ Monomial and binomials: Expression of one-term and two-term, respectively.

◊ Coefficients of polynomials: The constant or number multiplied with variables in each term.

◊ Zeros of a polynomial of one variable: The value of a variable that will make polynomial zero value.

◊ Remainder theorem of polynomials; which will be the useful division of one polynomial by another

◊ Factor theorem of polynomials; we will use it to factorise the polynomials as a product of two or more than two polynomials of less degree.

◊ NCERT exemplar Class 9 Maths solutions chapter 2 includes applying identities for factorisation and division of polynomials and use of some algebraic identities.

NCERT Class 9 Exemplar Solutions for Other Subjects

NCERT Class 9 Maths Exemplar Solutions for Other Chapters:

Chapter 1 Number System

Chapter 3 Coordinate geometry

Chapter 4 Linear equations in Two Variable

Chapter 5 Introduction to Euclid’s Geometry

Chapter 6 Lines and Angles

Chapter 7 Triangles

Chapter 8 Quadrilaterals

Chapter 9 Area of Parallelograms and Triangles

Chapter 10 Circles

Chapter 11 Constructions

Chapter 12 Heron’s Formula

Chapter 13 Surface Areas and Volumes

Chapter 14 Statistics and Probability

Features of NCERT Exemplar Class 9 Maths Solutions Chapter 2:

These Class 9 Maths NCERT exemplar chapter 2 solutions will help the students understand the polynomials' concept and knowledge.
Polynomial is an algebraic expression of a single or many variables such that each term will have a different power of the variables.
These Class 9 Maths NCERT exemplar solutions chapter 2 Polynomials can be used strategically to learn and practice the concepts of polynomials and are appropriate to prepare a student to take on other books such as NCERT Class 9 Maths, RD Sharma Class 9 Maths or RS Aggarwal Class 9 Maths.

Check the solutions of questions given in the book

Chapter No.	Chapter Name
Chapter 1	Number Systems
Chapter 2	Polynomials
Chapter 3	Coordinate Geometry
Chapter 4	Linear Equations In Two Variables
Chapter 5	Introduction to Euclid's Geometry
Chapter 6	Lines And Angles
Chapter 7	Triangles
Chapter 8	Quadrilaterals
Chapter 9	Areas of Parallelograms and Triangles
Chapter 10	Circles
Chapter 11	Constructions
Chapter 12	Heron’s Formula
Chapter 13	Surface Area and Volumes
Chapter 14	Statistics
Chapter 15	Probability

Also, Read NCERT Solution Subject Wise

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Frequently Asked Questions (FAQs)

1. Can we solve the zeros of a polynomial of any degree?

A student can always solve the zeros of a polynomial with degree 2. It is known as the quadratic equation. Sometimes we can solve cubic equations or zeros of higher degree polynomials. With the help of a computer, we can draw the graph of a polynomial of any degree and then locate it's zero.

2. What is the binomial expansion taught in higher classes?

NCERT exemplar Class 9 Maths solutions chapter 2 states that it is the expansion of the nth power of any binomial such as (a+b)2; this is also known as a binomial theorem.

3. What is the difference between expression and equation?

In an equation, two expressions are equated, whereas expression is the mathematical representation of different terms of any variable.

4. What is the weightage of Polynomials in JEE Main and JEE Advanced?

A clear understanding of Polynomials can prepare a student to solve problems based on Algebra, which ranges up to 3% of the whole paper.

Also Read

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables

NCERT Solutions for Class 9 Maths Chapter 3 Coordinate Geometry

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials

NCERT Solutions for Class 9 Maths Chapter 1 Number Systems

NCERT Solutions for Class 9 - Download Subject-wise Solution PDF

NCERT Solutions for Class 9 Maths Chapter 14 Statistics

NCERT Exemplar Class 9 Maths Solutions Chapter 10 Circles

NCERT Exemplar Class 9 Maths Solutions Chapter 9 Areas of Parallelograms and Triangles

NCERT Exemplar Class 9 Science Solutions Chapter 9 Force and Laws of Motion

NCERT Exemplar Class 9 Science Solutions chapter 8 Motion

NCERT Exemplar Class 9 Solutions - Subject Wise Problems with Solutions

NCERT Exemplar Class 9 Science Solutions Chapter 11 Work and Energy

NCERT Books for Class 9 Maths 2025-26; Download Free PDF

NCERT Books for Class 9 All Subjects - Download Free PDF

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NCERT Books for Class 9 Hindi 2025-26: Download PDF

NCERT Books for Class 9 Science 2025-26; Download PDF (All Chapters)

NCERT Class 9 Maths Chapter 6 Notes Lines and Angles - Download PDF

Surface Areas And Volumes Class 9th Notes - Free NCERT Class 9 Maths Chapter 13 Notes - Download PDF

NCERT Class 9 Maths Notes - Download Chapter Wise PDFs

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Triangles Class 9th Notes - Free NCERT Class 9th Maths Chapter 7 Notes - Download PDF

NCERT Class 9 Maths Chapter 1 Notes Number System - Download Free PDF

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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NCERT Exemplar Class 9 Maths Solutions Chapter 2 Polynomials

NCERT Exemplar Class 9 Maths Solutions Chapter 2-Exercise 2.1

NCERT Exemplar Class 9 Maths Solutions Chapter 2-Exercise 2.2

NCERT Exemplar Class 9 Maths Solutions Chapter 2-Exercise 2.3

NCERT Exemplar Class 9 Maths Solutions Chapter 2-Exercise 2.4

NCERT Exemplar Class 9 Maths Solutions Chapter 2 Contains the Solutions of the Questions Based on the Mentioned Topics:

NCERT Class 9 Exemplar Solutions for Other Subjects

NCERT Class 9 Maths Exemplar Solutions for Other Chapters:

Features of NCERT Exemplar Class 9 Maths Solutions Chapter 2:

Check the solutions of questions given in the book

Also, Read NCERT Solution Subject Wise

Check NCERT Notes Subject Wise

Also Check NCERT Books and NCERT Syllabus here

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