NCERT Solutions for Class 9 Maths chapter-2 Polynomials

Class 9 Maths Chapter 2 – Polynomials is an important chapter that builds the base of algebra for students. In NCERT Class 9 Maths Chapter 2, students are introduced to the concept of polynomials and their role in mathematics. A polynomial is an algebraic expression made up of variables, constants, and powers of variables. This chapter clearly explains the meaning of polynomials and classifies them into monomials, binomials, and trinomials based on the number of terms. In Chapter 2 Maths Class 9, students learn key ideas such as the degree of a polynomial, coefficients, and terms. The chapter also covers different operations on polynomials, including addition, subtraction, and multiplication. Another important topic discussed is the zeroes (roots) of a polynomial and how to find them. These concepts are essential for solving problems not only in Class 9 Maths Chapter 2 but also in higher classes.

Understanding Maths Class 9 Chapter 2 helps students develop logical thinking and problem-solving skills. Polynomials are widely used in higher algebra, coordinate geometry, and even real-life applications, making this chapter very important from an exam point of view.To support students, NCERT Solutions for Polynomials Class 9 are prepared by the academic team at Physics Wallah. These Class 9 Maths Chapter 2 Solutions provide clear, step-by-step answers for all exercises given in the NCERT textbook. Before solving questions from Maths Ch 2 Class 9, students should carefully read the theory and revise all formulas.

Physics Wallah also offers detailed notes, short formula sheets, and additional practice questions for Class 9 Chapter 2 Maths. These resources help students strengthen their concepts and confidently prepare for exams using Polynomials Class 9 NCERT Solutions.

NCERT Solutions for Class 9 Maths Exercise 2.1

Question 1. Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.

(i)

(ii)

(iii)

(iv)

(v)

Solution:

(i)

We can observe that in the polynomial , we have x as the only variable and the powers of x in each term are a whole number.

Therefore, we conclude that is a polynomial in one variable.

(ii)

We can observe that in the polynomial , we have y as the only variable and the powers of y in each term are a whole number.

Therefore, we conclude that is a polynomial in one variable.

(iii)

We can observe that in the polynomial, we have t as the only variable and the powers of t in each term are not a whole number.

Therefore, we conclude thatis not a polynomial in one variable.

(iv)

We can observe that in the polynomial, we have y as the only variable and the powers of y in each term are not a whole number.

Therefore, we conclude thatis not a polynomial in one variable.

(v)

We can observe that in the polynomial, we have x, y and t as the variables and the powers of x, y and t in each term is a whole number.

Therefore, we conclude that is a polynomial but not a polynomial in one variable.

Question 2. Write the coefficients of in each of the following :

(i)

(ii)

(iii)

(iv)

Solution:

(i)

The coefficient ofin the polynomialis 1.

(ii)

The coefficient ofin the polynomialis.

(iii)

The coefficient ofin the polynomialis.

(iv)

The coefficient ofin the polynomial is 0.

Question 3. Give one example each of a binomial of degree 35, and of a monomial of degree 100.
Solution:
The binomial of degree 35 can be.

The binomial of degree 100 can be.

Check out: NCERT Solutions for Class 9 Maths chapter-1 Number Systems

Question 4. Write the degree of each of the following polynomials :
(i)

(ii)

(iii)

(iv)3

Solution:
(i)

We know that the degree of a polynomial is the highest power of the variable in the polynomial.

We can observe that in the polynomial, the highest power of the variable x is 3.

Therefore, we conclude that the degree of the polynomialis 3.

(ii)

We know that the degree of a polynomial is the highest power of the variable in the polynomial.

We can observe that in the polynomial, the highest power of the variable y is 2.

Therefore, we conclude that the degree of the polynomialis 2.

(iii)

We know that the degree of a polynomial is the highest power of the variable in the polynomial.

We observe that in the polynomial, the highest power of the variable t is 1.

Therefore, we conclude that the degree of the polynomialis 1.

(iv)3

We know that the degree of a polynomial is the highest power of the variable in the polynomial.

We can observe that in the polynomial 3, the highest power of the assumed variable x is 0.

Therefore, we conclude that the degree of the polynomial 3 is 0.

Question 5. Classify the following as linear, quadratic and cubic polynomials.
(i) x2+ x
(ii) x – x3
(iii) y + y2+4
(iv) 1 + x
(v) 3t
(vi) r2
(vii) 7x3

Solution :
(i) The degree of x2 + x is 2. So, it is a quadratic polynomial.
(ii) The degree of x – x3 is 3. So, it is a cubic polynomial.
(iii) The degree of y + y2 + 4 is 2. So, it is a quadratic polynomial.
(iv) The degree of 1 + x is 1. So, it is a linear polynomial.
(v) The degree of 3t is 1. So, it is a linear polynomial.
(vi) The degree of r2 is 2. So, it is a quadratic polynomial.
(vii) The degree of 7x3 is 3. So, it is a cubic polynomial.

NCERT Solutions for Class 9 Maths Exercise 2.2

Question 1. Find the value of the polynomial 5x – 4x2 + 3 at
(i) x = 0
(ii) x = – 1
(iii) x = 2

Solution:
Let p(x) = 5x – 4x2 + 3
(i) p(0) = 5(0) – 4(0)2 + 3 = 0 – 0 + 3 = 3
Thus, the value of 5x – 4x2 + 3 at x = 0 is 3.
(ii) p(-1) = 5(-1) – 4(-1)2 + 3
= – 5x – 4x2 + 3 = -9 + 3 = -6
Thus, the value of 5x – 4x2 + 3 at x = -1 is -6.
(iii) p(2) = 5(2) – 4(2)2 + 3 = 10 – 4(4) + 3
= 10 – 16 + 3 = -3
Thus, the value of 5x – 4x2 + 3 at x = 2 is – 3.

Question 2. Find p (0), p (1) and p (2) for each of the following polynomials.
(i) p(y) = y2 – y +1
(ii) p (t) = 2 +1 + 2t2 -t3
(iii) P (x) = x3
(iv) p (x) = (x-1) (x+1)

Solution:
(i) Given that p(y) = y2 – y + 1.
∴ P(0) = (0)2 – 0 + 1 = 0 – 0 + 1 = 1
p(1) = (1)2 – 1 + 1 = 1 – 1 + 1 = 1
p(2) = (2)2 – 2 + 1 = 4 – 2 + 1 = 3
(ii) Given that p(t) = 2 + t + 2t2 – t3
∴p(0) = 2 + 0 + 2(0)2 – (0)3
= 2 + 0 + 0 – 0=2
P(1) = 2 + 1 + 2(1)2 – (1)3
= 2 + 1 + 2 – 1 = 4
p( 2) = 2 + 2 + 2(2)2 – (2)3
= 2 + 2 + 8 – 8 = 4
(iii) Given that p(x) = x3
∴ p(0) = (0)3 = 0, p(1) = (1)3 = 1
p(2) = (2)3 = 8
(iv) Given that p(x) = (x – 1)(x + 1)
∴ p(0) = (0 – 1)(0 + 1) = (-1)(1) = -1
p(1) = (1 – 1)(1 +1) = (0)(2) = 0
P(2) = (2 – 1)(2 + 1) = (1)(3) = 3

Check out: Class 9th Books

Question 3. Verify whether the following are zeroes of the polynomial, indicated against them.
(i) p(x) = 3x + 1,x = –
(ii) p (x) = 5x – π, x =
(iii) p (x) = x2 – 1, x = x – 1
(iv) p (x) = (x + 1) (x – 2), x = – 1,2
(v) p (x) = x2, x = 0
(vi) p (x) = 1x + m, x = –
(vii) P (x) = 3x2 – 1, x = – ,
(viii) p (x) = 2x + 1, x =

Solution:
(i) We have , p(x) = 3x + 1

(ii) We have, p(x) = 5x – π
∴
(iii) We have, p(x) = x2 – 1
∴ p(1) = (1)2 – 1 = 1 – 1=0
Since, p(1) = 0, so x = 1 is a zero of x2 -1.
Also, p(-1) = (-1)2 -1 = 1 – 1 = 0
Since p(-1) = 0, so, x = -1, is also a zero of x2 – 1.

(iv) We have, p(x) = (x + 1)(x – 2)
∴ p(-1) = (-1 +1) (-1 – 2) = (0)(- 3) = 0
Since, p(-1) = 0, so, x = -1 is a zero of (x + 1)(x – 2).
Also, p( 2) = (2 + 1)(2 – 2) = (3)(0) = 0
Since, p(2) = 0, so, x = 2 is also a zero of (x + 1)(x – 2).

(v) We have, p(x) = x2
∴ p(o) = (0)2 = 0
Since, p(0) = 0, so, x = 0 is a zero of x2.

(vi) We have, p(x) = lx + m

(vii) We have, p(x) = 3x2 – 1

(viii) We have, p(x) = 2x + 1
∴
Since, ≠ 0, so, x = is not a zero of 2x + 1.

Question 4. Find the zero of the polynomial in each of the following cases
(i) p(x)=x+5
(ii) p (x) = x – 5
(iii) p (x) = 2x + 5
(iv) p (x) = 3x – 2
(v) p (x) = 3x
(vi) p (x)= ax, a≠0
(vii) p (x) = cx + d, c ≠ 0 where c and d are real numbers.

Solution:
(i) We have, p(x) = x + 5. Since, p(x) = 0
⇒ x + 5 = 0
⇒ x = -5.
Thus, zero of x + 5 is -5.

(ii) We have, p(x) = x – 5.
Since, p(x) = 0 ⇒ x – 5 = 0 ⇒ x = -5
Thus, zero of x – 5 is 5.

(iii) We have, p(x) = 2x + 5. Since, p(x) = 0
⇒ 2x + 5 =0
⇒ 2x = -5
⇒ x =
Thus, zero of 2x + 5 is .

(iv) We have, p(x) = 3x – 2. Since, p(x) = 0
⇒ 3x – 2 = 0
⇒ 3x = 2
⇒ x =
Thus, zero of 3x – 2 is

(v) We have, p(x) = 3x. Since, p(x) = 0
⇒ 3x = 0 ⇒ x = 0
Thus, zero of 3x is 0.

(vi) We have, p(x) = ax, a ≠ 0.
Since, p(x) = 0 => ax = 0 => x-0
Thus, zero of ax is 0.

(vii) We have, p(x) = cx + d. Since, p(x) = 0
⇒ cx + d = 0 ⇒ cx = -d ⇒
Thus, zero of cx + d is

NCERT Solutions for Class 9 Maths Exercise 2.3

Question 1. Find the remainder whenis divided by

(i)

(ii)

(iii) x

(iv)

(v)

Solution:
(i)

We need to find the zero of the polynomial.

While applying the remainder theorem, we need to put the zero of the polynomialin the polynomial, to get

=-1+3-3+1

=0

Therefore, we conclude that on dividing the polynomialby, we will get the remainder as 0.

(ii)

We need to find the zero of the polynomial.

While applying the remainder theorem, we need to put the zero of the polynomialin the polynomial, to get

Therefore, we conclude that on dividing the polynomialby, we will get the remainder as.

(iii)

We need to find the zero of the polynomial.

While applying the remainder theorem, we need to put the zero of the polynomialin the polynomial, to get

=0+0+0+1

=1

Therefore, we conclude that on dividing the polynomialby x, we will get the remainder as 1.

(iv)

We need to find the zero of the polynomial.

While applying the remainder theorem, we need to put the zero of the polynomialin the polynomial, to get

Therefore, we conclude that on dividing the polynomialby, we will get the remainder as.

(v)

We need to find the zero of the polynomial.

While applying the remainder theorem, we need to put the zero of the polynomial in the polynomial, to get

Therefore, we conclude that on dividing the polynomialby, we will get the remainder as.

 

Question 2. Find the remainder whenis divided by.

Solution:

We need to find the zero of the polynomial.

While applying the remainder theorem, we need to put the zero of the polynomialin the polynomial, to get

= 5a

Therefore, we conclude that on dividing the polynomialby, we will get the remainder as.

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Question 3. Check whether is a factor of.

Solution:
We know that if the polynomialis a factor of, then on dividing the polynomialby, we must get the remainder as 0.

We need to find the zero of the polynomial.

While applying the remainder theorem, we need to put the zero of the polynomialinthe polynomial, to get

We conclude that on dividing the polynomialby, we will get the remainder as, which is not 0.

Therefore, we conclude thatis not a factor of.

NCERT Solutions for Class 9 Maths Exercise 2.4

Question 1. Determine which of the following polynomials has (x +1) a factor.
(i) x3+x2+x +1
(ii) x4 + x3 + x2 + x + 1
(iii) x4 + 3x3 + 3x2 + x + 1
(iv) x3 – x2 – (2 +√2 )x + √2

Solution :
The zero of x + 1 is -1.
(i) Let p (x) = x3 + x2 + x + 1
∴ p (-1) = (-1)3 + (-1)2 + (-1) + 1 .
= -1 + 1 – 1 + 1
⇒ p (- 1) = 0
So, (x+ 1) is a factor of x3 + x2 + x + 1.

(ii) Let p (x) = x4 + x3 + x2 + x + 1
∴ P(-1) = (-1)4 + (-1)3 + (-1)2 + (-1)+1
= 1 – 1 + 1 – 1 + 1
⇒ P (-1) ≠ 1
So, (x + 1) is not a factor of x4 + x3 + x2 + x+ 1.

(iii) Let p (x) = x4 + 3x3 + 3x2 + x + 1 .
∴ p (-1)= (-1)4 + 3 (-1)3 + 3 (-1)2 + (- 1) + 1
= 1 – 3 + 3 – 1 + 1 = 1
⇒ p (-1) ≠ 0
So, (x + 1) is not a factor of x4 + 3x3 + 3x2 + x+ 1.

(iv) Let p (x) = x3 – x2 – (2 + √2) x + √2
∴ p (- 1) =(- 1)3- (-1)2 – (2 + √2)(-1) + √2
= -1 – 1 + 2 + √2 + √2
= 2√2
⇒ p (-1) ≠ 0
So, (x + 1) is not a factor of x3 – x2 – (2 + √2) x + √2.

Question 2. Use the Factor Theorem to determine whether g (x) is a factor of p (x) in each of the following cases
(i) p (x)= 2x3 + x2 – 2x – 1, g (x) = x + 1
(ii) p(x)= x3 + 3x2 + 3x + 1, g (x) = x + 2
(iii) p (x) = x3 – 4x2 + x + 6, g (x) = x – 3

Solution :

(i) We have, p (x)= 2x3 + x2 – 2x – 1 and g (x) = x + 1
∴ p(-1) = 2(-1)3 + (-1)2 – 2(-1) – 1
= 2(-1) + 1 + 2 – 1
= -2 + 1 + 2 -1 = 0
⇒ p(-1) = 0, so g(x) is a factor of p(x).

(ii) We have, p(x) x3 + 3x2 + 3x + 1 and g(x) = x + 2
∴ p(-2) = (-2)3 + 3(-2)2+ 3(-2) + 1
= -8 + 12 – 6 + 1
= -14 + 13
= -1
⇒ p(-2) ≠ 0, so g(x) is not a factor of p(x).

(iii) We have, = x3 – 4x2 + x + 6 and g (x) = x – 3
∴ p(3) = (3)3 – 4(3)2 + 3 + 6
= 27 – 4(9) + 3 + 6
= 27 – 36 + 3 + 6 = 0
⇒ p(3) = 0, so g(x) is a factor of p(x).

Question 3. Find the value of k, if x – 1 is a factor of p (x) in each of the following cases
(i) p (x) = x2 + x + k
(ii) p (x) = 2x2 + kx + √2
(iii) p (x) = kx2 – √2 x + 1
(iv) p (x) = kx2 – 3x + k

Solution :
For (x – 1) to be a factor of p(x), p(1) should be equal to 0.

(i) Here, p(x) = x2 + x + k
Since, p(1) = (1)2 +1 + k
⇒ p(1) = k + 2 = 0
⇒ k = -2.

(ii) Here, p (x) = 2x2 + kx + √2
Since, p(1) = 2(1)2 + k(1) + √2
= 2 + k + √2 =0
k = -2 – √2 = -(2 + √2)

(iii) Here, p (x) = kx2 – √2 x + 1
Since, p(1) = k(1)2 – (1) + 1
= k – √2 + 1 = 0
⇒ k = √2 -1

(iv) Here, p(x) = kx2 – 3x + k
p(1) = k(1)2 – 3(1) + k
= k – 3 + k
= 2k – 3 = 0
⇒ k =

Question 4. Factorise
(i) 12x2 – 7x +1
(ii) 2x2 + 7x + 3
(iii) 6x2 + 5x – 6
(iv) 3x2 – x – 4

Solution :
(i) We have,
12x2 – 7x + 1 = 12x2 – 4x- 3x + 1
= 4x (3x – 1 ) -1 (3x – 1)
= (3x -1) (4x -1)
Thus, 12x2 -7x + 3 = (2x – 1) (x + 3)

(ii) We have, 2x2 + 7x + 3 = 2x2 + x + 6x + 3
= x(2x + 1) + 3(2x + 1)
= (2x + 1)(x + 3)
Thus, 2×2 + 7x + 3 = (2x + 1)(x + 3)

(iii) We have, 6x2 + 5x – 6 = 6x2 + 9x – 4x – 6
= 3x(2x + 3) – 2(2x + 3)
= (2x + 3)(3x – 2)
Thus, 6x2 + 5x – 6 = (2x + 3)(3x – 2)

(iv) We have, 3x2 – x – 4 = 3x2 – 4x + 3x – 4
= x(3x – 4) + 1(3x – 4) = (3x – 4)(x + 1)
Thus, 3x2 – x – 4 = (3x – 4)(x + 1)

Question 5. Factorise
(i) x3 – 2x2 – x + 2
(ii) x3 – 3x2 – 9x – 5
(iii) x3 + 13x2 + 32x + 20
(iv) 2y3 + y2 – 2y – 1
Solution:

(i) We have, x3 – 2x2 – x + 2
Rearranging the terms, we have x3 – x – 2x2 + 2
= x(x2 – 1) – 2(x2 -1) = (x2 – 1)(x – 2)
= [(x)2 – (1)2](x – 2)
= (x – 1)(x + 1)(x – 2)
[∵ (a2 – b2) = (a + b)(a-b)]
Thus, x3 – 2x2 – x + 2 = (x – 1)(x + 1)(x – 2)

(ii) We have, x3 – 3x2 – 9x – 5
= x3 + x2 – 4x2 – 4x – 5x – 5 ,
= x2 (x + 1) – 4x(x + 1) – 5(x + 1)
= (x + 1)(x2 – 4x – 5)
= (x + 1)(x2 – 5x + x – 5)
= (x + 1)[x(x – 5) + 1(x – 5)]
= (x + 1)(x – 5)(x + 1)
Thus, x3 – 3x2 – 9x – 5 = (x + 1)(x – 5)(x +1)

(iii) We have, x3 + 13x2 + 32x + 20
= x3 + x2 + 12x2 + 12x + 20x + 20
= x2(x + 1) + 12x(x +1) + 20(x + 1)
= (x + 1)(x2 + 12x + 20)
= (x + 1)(x2 + 2x + 10x + 20)
= (x + 1)[x(x + 2) + 10(x + 2)]
= (x + 1)(x + 2)(x + 10)
Thus, x3 + 13x2 + 32x + 20
= (x + 1)(x + 2)(x + 10)

(iv) We have, 2y3 + y2 – 2y – 1
= 2y3 – 2y2 + 3y2 – 3y + y – 1
= 2y2(y – 1) + 3y(y – 1) + 1(y – 1)
= (y – 1)(2y2 + 3y + 1)
= (y – 1)(2y2 + 2y + y + 1)
= (y – 1)[2y(y + 1) + 1(y + 1)]
= (y – 1)(y + 1)(2y + 1)
Thus, 2y3 + y2 – 2y – 1
= (y – 1)(y + 1)(2y +1)

NCERT Solutions for Class 9 Maths Exercise 2.5

Question 1. Use suitable identities to find the following products :

(i)

(ii)

(iii)

(iv)

(v)

Solution:

(i)

We know that.

We need to apply the above identity to find the product

Therefore, we conclude that the productis.

(ii)

We know that.

We need to apply the above identity to find the product

=

Therefore, we conclude that the productis.

(iii)

We know that.

We need to apply the above identity to find the product

Therefore, we conclude that the productis.

(iv)

We know that.

We need to apply the above identity to find the product

Therefore, we conclude that the productis.

(v)

We know that.

We need to apply the above identity to find the product

Therefore, we conclude that the productis.

Question 2. Evaluate the following products without multiplying directly :

(i)

(ii)

(iii)

Solution:
(i)

We can observe that, we can apply the identity

=10000+1000+21

=11021

Therefore, we conclude that the value of the productis.

(ii)

We can observe that, we can apply the identity

=10000-900+20

=9120

Therefore, we conclude that the value of the productis .

(iii)

We can observe that, we can apply the identitywith respect to the expression, to get

=10000-16

=9984

Therefore, we conclude that the value of the productis.

Question 3. Factorize the following using appropriate identities :

(i)

(ii)

(iii)

Solution:
(i)

We can observe that, we can apply the identity

(ii)

We can observe that, we can apply the identity

(iii)

We can observe that, we can apply the identity

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Question 4. Expand each of the following, using suitable identities :

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Solution:

(i)

We know that.

We need to apply the above identity to expand the expression.

(ii)

We know that.

We need to apply the above identity to expand the expression.

(iii)

We know that.

We need to apply the above identity to expand the expression.

(iv)

We know that.

We need to apply the above identity to expand the expression.

(v)

We know that.

We need to apply the above identity to expand the expression.

(vi)

We know that.

Question 5. Factorize :

(i)

(ii)

Solution:
(i)

The expression can also be written as

We can observe that, we can apply the identitywith respect to the expression, to get

Therefore, we conclude that after factorizing the expression, we get.

(ii)

We need to factorize the expression.

The expression can also be written as

We can observe that, we can apply the identitywith respect to the expression, to get

Therefore, we conclude that after factorizing the expression, we get.

Question 6. Write the following cubes in expanded form :

(i)

(ii)

(iii)

(iv)

Solution :
(i)

We know that.

Therefore, the expansion of the expressionis .

(ii)

We know that.

Therefore, the expansion of the expressionis .

(iii)

We know that.

Therefore, the expansion of the expressionis.

(iv)

We know that.

Therefore, the expansion of the expressionis.

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Question 7. Evaluate the following using suitable identities
(i) (99)3
(ii) (102)3
(iii) (998)3

Solution:
(i) We have, 99 = (100 -1)
∴ 993 = (100 – 1)3
= (100)3 – 13 – 3(100)(1)(100 -1)
[Using (a – b)3 = a3 – b3 – 3ab (a – b)]
= 1000000 – 1 – 300(100 – 1)
= 1000000 -1 – 30000 + 300
= 1000300 – 30001 = 970299

(ii) We have, 102 =100 + 2
∴ 1023 = (100 + 2)3
= (100)3 + (2)3 + 3(100)(2)(100 + 2)
[Using (a + b)3 = a3 + b3 + 3ab (a + b)]
= 1000000 + 8 + 600(100 + 2)
= 1000000 + 8 + 60000 + 1200 = 1061208

(iii) We have, 998 = 1000 – 2
∴ (998)3 = (1000-2)3
= (1000)3– (2)3 – 3(1000)(2)(1000 – 2)
[Using (a – b)3 = a3 – b3 – 3ab (a – b)]
= 1000000000 – 8 – 6000(1000 – 2)
= 1000000000 – 8 – 6000000 +12000
= 994011992

Question 8. Factorise each of the following
(i) 8a3 +b3 + 12a2b+6ab2
(ii) 8a3 -b3-12a2b+6ab2
(iii) 27-125a3 -135a+225a2
(iv) 64a3 -27b3 -144a2b + 108ab2

Solution:
(i) 8a3 +b3 +12a2b+6ab2
= (2a)3 + (b)3 + 6ab(2a + b)
= (2a)3 + (b)3 + 3(2a)(b)(2a + b)
= (2 a + b)3
[Using a3 + b3 + 3 ab(a + b) = (a + b)3]
= (2a + b)(2a + b)(2a + b)

(ii) 8a3 – b3 – 12a2b + 6ab2
= (2a)3 – (b)3 – 3(2a)(b)(2a – b)
= (2a – b)3
[Using a3 + b3 + 3 ab(a + b) = (a + b)3]
= (2a – b) (2a – b) (2a – b)

(iii) 27 – 125a3 – 135a + 225a2
= (3)3 – (5a)3 – 3(3)(5a)(3 – 5a)
= (3 – 5a)3
[Using a3 + b3 + 3 ab(a + b) = (a + b)3]
= (3 – 5a) (3 – 5a) (3 – 5a)

(iv) 64a3 -27b3 -144a2b + 108ab2
= (4a)3 – (3b)3 – 3(4a)(3b)(4a – 3b)
= (4a – 3b)3
[Using a3 – b3 – 3 ab(a – b) = (a – b)3]
= (4a – 3b)(4a – 3b)(4a – 3b)

Question 9. Verify
(i) x3 + y3 = (x + y)-(x2 – xy + y2)
(ii) x3 – y3 = (x – y) (x2 + xy + y2)

Solution:
(i) ∵ (x + y)3 = x3 + y3 + 3xy(x + y)
⇒ (x + y)3 – 3(x + y)(xy) = x3 + y3
⇒ (x + y)[(x + y)2-3xy] = x3 + y3
⇒ (x + y)(x2 + y2 – xy) = x3 + y3
Hence, verified.

(ii) ∵ (x – y)3 = x3 – y3 – 3xy(x – y)
⇒ (x – y)3 + 3xy(x – y) = x3 – y3
⇒ (x – y)[(x – y)2 + 3xy)] = x3 – y3
⇒ (x – y)(x2 + y2 + xy) = x3 – y3
Hence, verified.

Question 10. Factorise each of the following
(i) 27y3 + 125z3
(ii) 64m3 – 343n3
[Hint See question 9]

Solution:
(i) We know that
x3 + y3 = (x + y)(x2 – xy + y2)
We have, 27y3 + 125z3 = (3y)3 + (5z)3
= (3y + 5z)[(3y)2 – (3y)(5z) + (5z)2]
= (3y + 5z)(9y2 – 15yz + 25z2)

(ii) We know that
x3 – y3 = (x – y)(x2 + xy + y2)
We have, 64m3 – 343n3 = (4m)3 – (7n)3
= (4m – 7n)[(4m)2 + (4m)(7n) + (7n)2]
= (4m – 7n)(16m2 + 28mn + 49n2)

Question 11. Factorise 27x3 +y3 +z3 -9xyz.

Solution:
We have,
27x3 + y3 + z3 – 9xyz = (3x)3 + (y)3 + (z)3 – 3(3x)(y)(z)
Using the identity,
x3 + y3 + z3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx)
We have, (3x)3 + (y)3 + (z)3 – 3(3x)(y)(z)
= (3x + y + z)[(3x)3 + y3 + z3 – (3x × y) – (y × 2) – (z × 3x)]
= (3x + y + z)(9x2 + y2 + z2 – 3xy – yz – 3zx)

Question 12. Verify that
x3 +y3 +z3 – 3xyz = (x + y+z)[(x-y)2 + (y – z)2 +(z – x)2]

Solution:
R.H.S
= (x + y + z)[(x – y)2+(y – z)2+(z – x)2]
= (x + y + 2)[(x2 + y2 – 2xy) + (y2 + z2 – 2yz) + (z2 + x2 – 2zx)]
= (x + y + 2)(x2 + y2 + y2 + z2 + z2 + x2 – 2xy – 2yz – 2zx)
= (x + y + z)[2(x2 + y2 + z2 – xy – yz – zx)]
= 2 x x (x + y + z)(x2 + y2 + z2 – xy – yz – zx)
= (x + y + z)(x2 + y2 + z2 – xy – yz – zx)
= x3 + y3 + z3 – 3xyz = L.H.S.
Hence, verified.

Question 13. If x + y + z = 0, show that x3 + y3 + z3 = 3 xyz.

Solution:
Since, x + y + z = 0
⇒ x + y = -z (x + y)3 = (-z)3
⇒ x3 + y3 + 3xy(x + y) = -z3
⇒ x3 + y3 + 3xy(-z) = -z3 [∵ x + y = -z]
⇒ x3 + y3 – 3xyz = -z3
⇒ x3 + y3 + z3 = 3xyz
Hence, if x + y + z = 0, then
x3 + y3 + z3 = 3xyz

Question 14. Without actually calculating the cubes, find the value of each of the following
(i) (- 12)3 + (7)3 + (5)3
(ii) (28)3 + (- 15)3 + (- 13)3

Solution:

(i) We have, (-12)3 + (7)3 + (5)3
Let x = -12, y = 7 and z = 5.
Then, x + y + z = -12 + 7 + 5 = 0
We know that if x + y + z = 0, then, x3 + y3 + z3 = 3xyz
∴ (-12)3 + (7)3 + (5)3 = 3[(-12)(7)(5)]
= 3[-420] = -1260

(ii) We have, (28)3 + (-15)3 + (-13)3
Let x = 28, y = -15 and z = -13.
Then, x + y + z = 28 – 15 – 13 = 0
We know that if x + y + z = 0, then x3 + y3 + z3 = 3xyz
∴ (28)3 + (-15)3 + (-13)3 = 3(28)(-15)(-13)
= 3(5460) = 16380

Question 15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given
(i) Area 25a2 – 35a + 12
(ii) Area 35y2 + 13y – 12

Solution:
Area of a rectangle = (Length) x (Breadth)
(i) 25a2 – 35a + 12 = 25a2 – 20a – 15a + 12 = 5a(5a – 4) – 3(5a – 4) = (5a – 4)(5a – 3)
Thus, the possible length and breadth are (5a – 3) and (5a – 4).

(ii) 35y2+ 13y -12 = 35y2 + 28y – 15y -12
= 7y(5y + 4) – 3(5y + 4) = (5 y + 4)(7y – 3)
Thus, the possible length and breadth are (7y – 3) and (5y + 4).

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Question 16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
(i) Volume 3x2 – 12x
(ii) Volume 12ky2 + 8ky – 20k

Solution:
Volume of a cuboid = (Length) x (Breadth) x (Height)
(i) We have, 3x2 – 12x = 3(x2 – 4x)
= 3 x (x – 4)
∴ The possible dimensions of the cuboid are 3, x and (x – 4).

(ii) We have, 12ky2 + 8ky – 20k
= 4[3ky2 + 2ky – 5k] = 4[k(3y2 + 2y – 5)]
= 4 x k x (3y2 + 2y – 5)
= 4k[3y2 – 3y + 5y – 5]
= 4k[3y(y – 1) + 5(y – 1)]
= 4k[(3y + 5) x (y – 1)]
= 4k x (3y + 5) x (y – 1)
Thus, the possible dimensions of the cuboid are 4k, (3y + 5) and (y -1).

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