NCERT Exemplar Class 9 Maths Solutions Chapter 2 Polynomials

NCERT Exemplar Class 9 Maths Solutions Chapter 2-Exercise 2.1

Question:1

Which one of the following is a polynomial?

Answer:

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 is not a polynomial because power of x must be a non-negative integer.
is also not a polynomial because degree of variable should always be a whole number.

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

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

Question:2

is a polynomial of degree

(A) 2

(B) 0

(C) 1

(D)

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 as 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 , then is equal to
(A) 0
(B) 1
(C)
(D)

Answer:

(B) 1
Solution:

means that the value of x is
Put in the given equation

Hence the value of 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 , then 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 is
(A) (B) (C) (D)

Answer:

(B)
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=
Hence option (B) is correct.

Question:10

One of the zeroes of the polynomial is
(A) 2
(B)
(C)
(D) -2

Answer:

(B)
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
Now let us find zeroes of this polynomial

The zeros of the given polynomial are
Among the given options, (B) is correct.

Question:11

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

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:

(C) 2
Solution:
The given polynomial is
Let
It is given that is a factor of given polynomial.
Then according to factor theorem,

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

Question:13

x+1 is a factor of the polynomial

Answer:

(B)
Solution:
We know that if is factor of the polynomial f(x), then it always satisfies
Hence (x+1) is factor of that polynomial which satisfies f(-1)=0 .
Not satisfied

Satisfied

Not satisfied
Not satisfied
Hence is a factor of
Therefore option (B) is correct

Question:14

One of the factors of is
(A) 5+x (B) 5-x (C) 5x-1 (D) 10x

Answer:

(D) 10x
Solution
Given,

We know that,

Now, using the above

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

Question:15

The value of is
(A) (B) 477 (C) 487 (D) 497

Answer:

(D) 497
Solution
Here we find the value of by using the identity

Hence the answer is an option (D) 497.

Question:16

The factorization of is
(A)
(B)
(C)
(D)

Answer:

Let us factorize the given polynomial
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,
can be written as
{6+2 = 8; (6)(2) = (4)(3)}

Hence option B is correct

Question:17

Which of the following is a factor of

Answer:

(D) 3xy
Solution
Given,

We know that,

So,

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 in above expansion we get

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

Question:19

If the vlaue of is
(A) 1 (B) -1 (C)0 (D)

Answer:

(C) 0
Solution
Given
Simplifying the above equation we have

Put the value of equation 1 in 2 we get

Hence option C is correct.

Question:20

If then the valueof b is,

Answer:

Given:

Simplifying the given equation we have

Therefore option (C) is correct.

Question:21

If , then is equal to
(A) 0 (B) abc (C) 3abc (D) 2abc

Answer:

(C) 3abc
Solution
We know that

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

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)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)

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: etc.
(i) Here 8 is a polynomial because it can also be written as i.e., multiply by .
(ii) is also a polynomial having degree two.
(iii) is not a polynomial because its exponent is in fraction.
(iv) can be written as and it is a polynomial having degree two.
(v) is not polynomial because it has negative exponent.
(vi) is not a polynomial because it have negative exponent.
(vii) is a polynomial of degree three.
(viii) 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: 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: 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: 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: 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 , 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: 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: 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: and are two polynomials of degree 5 but the degree of the sum of the two polynomials 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)
(ii)
(iii)
(iv)

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, is a one variable polynomial, because it contains only one variable,
i.e., x
(ii)
Here, is a one variable polynomial, because it contains only one variable,
i.e., y
(iii)
Here, is three variable polynomial, because it contains three variables
i.e., x, y and z
(iv)
Here, 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)
(ii)
(iii)
(iv)

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) Here highest power of x is one therefore degree is one.
(ii) Here –10 can be written as hence the degree of polynomial is zero or we can say that it is a constant polynomial.
(iii) Here highest power of x is Five therefore degree is Five.
(iv) Here Highest power of y is seven therefore degree is seven.

Question:3

For the polynomial , write
(i) The degree of the polynomial
(ii) The coefficient of
(iii) The coefficient of
(iv) The constant term

Answer:

(i) 6
(ii)
(iii) -1
(iv)
Solution

(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
Here is multiplied by therefore coefficient of is
(iii) The coefficient of
Here –1 is multiplied by therefore coefficient of is –1.
(iv) The constant term
The constant term is the one without the variable x.
So, is the constant term

Question:4

Write the coefficient of in each of the following
(i)
(ii)
(iii)
(iv)

Answer:

(i) 1
(ii) 0
(iii) 3
(iv) -16
Solution
The value which is multiplied by is its coefficient.
(i)
Here is multiplied with 1.
Hence the coefficient of is one.
(ii)
Here is multiplied with 0.
Hence the coefficient of is 0.
(iii)

Here is multiplied with 3.
Hence the coefficient of is 3.
(iv)

Here is multiplied with -16.
Hence the coefficient of is -16.

Question:5

Classify the following as a constant, linear, quadratic and cubic polynomials:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
(x)

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)
Here highest power of x is three therefore degree is three.
Cubic polynomial.
(ii)
Here highest power of x is three therefore degree is three.
Cubic polynomial.
(iii)
Here highest power of t is one therefore degree is one.
Linear polynomial.
(iv)
Here highest power of y is two therefore degree is two.
Quadratic polynomial.
(v)
Here highest power of x is zero therefore degree is zero.
Constant polynomial.
(vi)
Here highest power of x is one therefore degree is one.
Linear polynomial.
(vii)
Here highest power of y is three therefore degree is three.
Cubic polynomial.
(viii)
Here highest power of x is two therefore degree is two.
Quadratic polynomial.
(ix)
Here highest power of t is two therefore degree is two.
Quadratic polynomial.
(x)
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)
(iii)
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: 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: 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:

(iii) Trinomial of degree 2:

Question:7

Find the value of the polynomial , 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 evaluate :

Answer:

Solution
Given polynomial is
Put we get

Put x=-1 we get

Put we get

Now is

Hence the answer is .

Question:9

Find P(0), P(1), P(–2) for the following polynomials:
i.
ii.

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) is a zero of
iii) is a zero of
iv) 0 and 2 are the zeroes of
v) –3 is a zero of

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) =
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 =
p(x) =
Hence 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) =
Putting y =

Hence 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,

Putting t = 0

Putting t = 2

Hence 0 and 2 are the zeroes of
Therefore given statement is True.
Method 2:

The given expression can be factorized as:

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,

Putting y = -3

Hence -3 is the zero of
Therefore given statement is True.
Method 2:

The given expression can be factorized as follows:

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)
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., g(x)=0
3-6x=0
3=6x


Hence is a zero of the given polynomial.
iii)
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

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

Question:12

Find the zeroes of the polynomial :

Answer:

[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., p(x) = 0
That is,

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-1

Answer:

Quotient
Remainder =2
Solution
Let

Divide p(x) and g(x)

Quotient
Remainder =2

Question:14

By Remainder Theorem find the remainder, when p(x) is divided by g(x) , where

Answer:

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

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

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) .

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

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

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

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

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

Hence the remainder is .

Question:15

Check whether p(x) is multiple of g(x) or not

Answer:

(i) No
Solution
Given
According to remainder theorem if p(x) is a multiple of (x+a) then p(-a)=0
Here p(x) is and g(x) is
If p(x) is a multiple of g(x) then p(2)=0


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

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

Question:16

Show that
(i) is a factor of
(ii) is a factor of

Answer:

(i) Here and
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

Therefore x+3 is a factor of
Hence proved
(ii) Here and
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

Hence 2x-3 is a factor of
Hence proved

Question:17

Determine which of the following polynomials has factor
(i)
(ii)

Answer:

(i) only
Solution
We know that if is factor of the polynomial f(x), then it always satisfies f(-a)=0
(i) Here polynomial is

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

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

Hence x-2 is not a factor of .

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 is a factor of and also of .
We know that if (x+a) is factor of the polynomial f(x), then it always satisfies f(-a)=0
If is a factor of and then by putting the value of p = 1, the given polynomials should be equal to zero.
Let

Hence p-1 is a factor of
Let

Hence p-1 is a factor of
Hence proved

Question:19

For what value of m is divisible by ?

Answer:

m = 1
Solution
Given : is divisible by
We know that if (x+a) is factor of the polynomial f(x), then it always satisfies f(-a)=0
If is divisible by then according to remainder theorem if we put x=-2 in the polynomial then the output must be equal to 0.
Let
Put x=-2

Hence, m=1

Question:20

If is a factor of , find a.

Answer:

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


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

So,

Question:21

Find the value of m so that be a factor of .

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, is a factor of
Let

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

Question:22

If is a factor of , 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 is a factor of
Let

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

Hence a = 2.

Question:23

Factorize :
(i)
(ii)
(iii)
(iv)

Answer:

Solution
Given,
To factorize , 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:


Solution
Given,
To factorize , 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:


Solution
Given,
To factorize , 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:


Solution
Given,
To factorize , 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:

Question:24

Factorize :

Answer:

Solution
Let
By trail, we find that

Hence is a factor of


Hence the factorized form is

Solution
Let
By trail, we find that

Hence is a factor of


Hence the factorized form is

Solution
Let
By trail, we find that

Hence is a factor of


Hence the factorized form is

Solution
Let
By trail, we find that

Hence is a factor of


Hence the factorized form is .

Question:25

Using suitable identity, evaluate the following

Answer:

(i) 1092727
Solution
Given,

Hence the answer is 1092727
(ii) 10302
Solution
Given,
Using
Put x = 100, a = 1, b = 2

Hence the answer is 10302
(iii)998001
Solution
Given,
Using
Putting a = 1000, b = 1

Hence the answer is 998001.

Question:26

Factorize the following

Answer:

Solution
Given,
We can write the given expression as:

Now, we know that

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


Solution
Given,
We can write the given expression as:

Now, we know that

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


Solution
Given,
Now, we know that

So,

Hence the factorized form is: .

Question:27

Factorize the following :-

Answer:

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


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

Question:28

Expand the following

Answer:

(i)
Solution

Now, we know that

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

Hence the expanded form is
(ii)
Solution

Now, we know that

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

Hence the expanded form is
(iii)
Solution
Now, we know that

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

Hence the expanded form is .

Question:29

Factorize the following

Answer:

Solution
Given,
This can be written as:
…(i)
We know that,

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


Hence the factorized form is

Solution
Given,
This can be written as:
…(i)
We know that,

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


Hence the factorized form is

Solution
Given,
This can be written as:
…(i)
We know that,

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


Hence the factorized form is .

Question:30

If and , find .

Answer:

29
Solution
Given ab+bc+ac=26, a+b+c=9
As we know that
…..(1)
Put a+b+c=9 , ab+bc+ac=26 in (1)

Hence the answer is 29.

Question:31

Expand the following :

Answer:

Solution: Given
We know that

So we get:


Hence the expanded form is


Solution: Given
We know that

So we get:



Hence the expanded form is


Solution: Given
We know that

So we get:



Hence the expanded form is .

Question:32

Factorize the following :

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)
Given,
The given equation can be written as:
…(i)
Now we know that:

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

Hence the factorized form is

Question:33

Find the following product
(i)
(ii)

Answer:

(i)
Given:

(ii)
Given

Question:34

Factorise:
(i)
(ii)

Answer:

(i)
Given

we know that
so

(ii)
Given
This can be written as

we know that
so

Question:35

Find the following product

Answer:

Solution
Given:
This can be written as

We know that

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

Question:36

Factorize
(i)
(ii)

Answer:

(i)
Solution:
Given:
This can be written as:

We know that

Using this identity we get




(i)
Solution:
Given:
This can be written as:

We know that

Using this identity we get


=

Question:37

Without actually calculating the cubes, find the value of
(i)
(ii)

Answer:

(i)
Given
We know that

When a + b + c = 0, we get
Here,

So using the above identity, we get:

Hence the answer is
(ii)-0.018
Solution
Given:
We know that

When a + b + c = 0, we get
Here,

So,

Hence the answer is -0.018.

Question:38

Without finding the cubes, factorize

Answer:

solution :
given
We know that

When a + b + c = 0, we get
Here,

So using the above identity, we get:


Hence the answer is .

Question:39

Find the value of
(i) when
(ii) when

Answer:

(i) 0
Solution :-
Here
We know that

When a + b + c = 0, we get
Here,

So using the above identity, we get:

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

When a + b + c = 0, we get
Here,

So using the above identity, we get:

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

Hence the answer is 0.

Question:40

Give possible expressions for the length and breadth of the rectangle whose area is given by

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

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).

Similarly

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

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)

According to question p(-1)=19

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

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

Question:3

If both x – 2 and are factors of px² + 5x + r, show that p = r.

Answer:

Let
We know that if is factor of the polynomial f(x), then it always satisfies
(x-2) is a factor of f(x) then f(2) = 0
So,
…..(1)
Also is a factor of f(x) then

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

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:

Answer:

Solution
We know that

Question:6

Multiply: by

Answer:

Solution
We have,
This can be written as:

We know that

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

Hence the answer is .

Question:7

If a, b, c are all non-zero and a + b + c = 0, prove that

Answer:

Given,
L.H.S.
Taking LCM of denominators, we get
L.H.S
Now we know that if a + b + c = 0 then
Putting the value in above equation:
L.H.S = = 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

Now,

Putting the values, we get:

Hence proved.

Question:9

prove that:

Answer:

prove that:
Taking left hand side

Now using the identity

So,

Now using

So equation (i) becomes:

Now,

Hence proved.