NCERT Solutions for Class 9 Maths chapter-2 Polynomials

Class 9 Maths chapter-2 Polynomials: Chapter 2 of the NCERT Class 9 Maths textbook, titled "Polynomials," introduces students to the fundamental concepts of polynomial algebra. This chapter looks into the definition and types of polynomials, including monomials, binomials, and trinomials, and explores their properties and operations. Students learn how to perform addition, subtraction, and multiplication of polynomials, and gain an understanding of polynomial division. Key concepts such as the degree of a polynomial, coefficients, and the roots of polynomials are covered, providing a foundation for solving polynomial equations. The chapter emphasizes the significance of polynomials in various mathematical contexts and real-life applications, laying the groundwork for more advanced topics in algebra and calculus. 

NCERT solutions for Class 9 Maths chapter 2 Polynomials is prepared by academic team of Physics Wallah. We have prepared NCERT solutions for all exercise of chapter 2. Given below is step by step solutions of all questions given in NCERT textbook for chapter 2. Read chapter 2 theory make sure you have gone through the theory part of chapter 2 from NCERT textbook and you have learned the formula of the given chapter. Physics Wallah prepared a detail notes and additional questions for class 9 maths with short notes of all maths formula of class 9 maths. Do read these contents before moving to solve the exercise of NCERT chapter 2.

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

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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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