Why is NCERT Solution for Class 9 Chapter 2: Polynomials Important?

The NCERT solution for Class 9 Chapter 2: Polynomials is important as it provides a structured approach to learning, ensuring that students develop a strong understanding of foundational concepts early in their academic journey. By mastering these basics, students can build confidence and readiness for tackling more difficult concepts in their further education.

Does the NCERT Solution for Class 9 Chapter 2: Polynomials help the students in preparing for the exam?

Yes, the NCERT solution for Class 9 Chapter 2: Polynomials is quite useful for students in preparing for their exams. The solutions are simple, clear, and concise allowing students to understand them better. They can solve the practice questions and exercises that allow them to get exam-ready in no time.

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You can get all the NCERT solutions for Class 9 Maths Chapter 2 from the official website of the Orchids International School. These solutions are tailored by subject matter experts and are very easy to understand.

Do I Need to Practice All Questions Provided in NCERT Solution for Class 9 Maths Chapter 2: Polynomials?

Yes, students must practice all the questions provided in the NCERT solution for Class 9 Maths Chapter 2: Polynomials as it will help them gain a comprehensive understanding of the concept, identify their weak areas, and strengthen their preparation.

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Students can utilize the NCERT solution for Class 9 Maths Chapter 2 effectively by practicing the solutions regularly. Solve the exercises and practice questions given in the solution.

NCERT Solutions For Class 9 Maths Chapter 2 Polynomials

Exercise 2.1

Question 1 :

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

Answer :

(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.

Question 2 :

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

(i)

(ii)

(iii)

(iv)

(v)

Answer :

(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 3 :

Write the coefficients of in each of the following :

(i)

(ii)

(iii)

(iv)

Answer :

(i)

The coefficient ofin the polynomialis 1.

(ii)

The coefficient ofin the polynomialis -1.

(iii)

The coefficient ofin the polynomialis.

(iv)

The coefficient ofin the polynomial is 0.

Question 4 :

Write the degree of each of the following polynomials :

(i)

(ii)

(iii)

(iv)3

Answer :

(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 :

Give one example each of a binomial of degree 35, and of a monomial of degree 100.

Answer :

The binomial of degree 35 can be

The binomial of degree 100 can be

Exercise 2.2

Question 1 :

Verify whether the following are zeroes of the polynomial, indicated against them.
(i) p(x) = 3x + 1,x = –1/3
(ii) p (x) = 5x – π, x = 4/5
(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 = – $\frac { 1 }{ \sqrt { 3 } }$ ,
(viii) p (x) = 2x + 1, x = 1/2

Answer :

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

NCERT Solutions for Class 9 Maths chapter 2-Polynomials

(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

NCERT Solutions for Class 9 Maths chapter 2-Polynomials

(vii) We have, p(x) = 3x2 – 1

NCERT Solutions for Class 9 Maths chapter 2-Polynomials

(viii) We have, p(x) = 2x + 1

Since, ≠ 0, so, x = is not a zero of 2x + 1.

Question 2 :

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

Answer :

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 3 :

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)

Answer :

(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

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.

Answer :

(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 = -5/2

Thus, zero of 2x + 5 is -5/2.

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

(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 ⇒ x= -d/c
Thus, zero of cx + d is -d/c.

Exercise 2.3

Question 1 :

Find the remainder when x³+3x²+3x+1 is divided by

(i) x+1

(ii) x-1/2

(iii) x

(iv)

(v) 5+2x

Answer :

(i) x+1

We need to find the zero of the polynomial.

x+1=0 => x = -1

While applying the remainder theorem, we need to put the zero of the polynomial x+1 in the polynomial x³+3x²+3x+1, to get

p(x) = x³+3x²+3x+1

p(0)

= -1+3-3+1

= 0

Therefore, we conclude that on dividing the polynomial x³+3x²+3x+1 by x+1, we will get the remainder as 0.

(ii)x=-1/2

We need to find the zero of the polynomial x-1/2.

x-1/2 = 0 => x = 1/2

While applying the remainder theorem, we need to put the zero of the polynomial x-1/2 in the polynomial x³+3x²+3x+1, to get

p(x) = x³+3x²+3x+1

= p(1/2) = (1/2)3 + 3(1/2)2 + 3(1/2) + 1

= 1/8 + 3(1/4) + 3/2 +1

= 1+6+12+8/8

= 27/8

Therefore, we conclude that on dividing the polynomial x³+3x²+3x+1 by x-1/2 , we will get the remainder as 27/8.

(iii)

We need to find the zero of the polynomial x.

x=0

While applying the remainder theorem, we need to put the zero of the polynomial x in the polynomial x³+3x²+3x+1, to get

p(x) = x³+3x²+3x+1

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

=0+0+0+1

Therefore, we conclude that on dividing the polynomial x³+3x²+3x+1 by 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 polynomial in the polynomial x³+3x²+3x+1, to get

p(x) = x³+3x²+3x+1

Therefore, we conclude that on dividing the polynomial x³+3x²+3x+1 by, we will get the remainder as .

(v) 5+2x

We need to find the zero of the polynomial 5+2x.

5+2x = 0 => x = - 5/2

While applying the remainder theorem, we need to put the zero of the polynomial 5+2x in the polynomial x³+3x²+3x+1, to get

p(x) = x³+3x²+3x+1

p(-5/2) = (-5/2)3 + 3(-5/2)2 + 3(-5/2) +1

= -125/8 + 3(25/4) - 15/2 + 1

= -125/8 + 75/4 - 15/2 + 1

= -125 + 150 - 60 + 8 / 8

= -27/4

Therefore, we conclude that on dividing the polynomial x³+3x²+3x+1 by 5+2x, we will get the remainder as -27/4.

Question 2 :

Find the remainder when x³ - ax² + 6x - a is divided by x-a.

Answer :

We need to find the zero of the polynomial x-a.

x-a = 0 => x = a

While applying the remainder theorem, we need to put the zero of the polynomial x-a in the polynomial x³ - ax² + 6x - a, to get

p(x) = x³ - ax² + 6x - a

p(a) = (a)³ - a(a)² + 6(a) - a

= a³ - a³ + 6a - a

= 5a

Therefore, we conclude that on dividing the polynomial x³ - ax² + 6x - a byx-a , we will get the remainder as 5a.

Question 3 :

Check whether 7+3x is a factor of 3x³+7x.

Answer :

We know that if the polynomial 7 + 3x is a factor of 3x³+7x, then on dividing the polynomial 3x³+7x by 7 + 3x, we must get the remainder as 0.

We need to find the zero of the polynomial 7 + 3x.

7 + 3x = 0 => x = - 7/3

While applying the remainder theorem, we need to put the zero of the polynomial 7 + 3x inthe polynomial 3x³+7x, to get

p(x) = 3x³+7x

= 3( - 7/3)³ + 7(- 7/3)

= 3( -343/27) - 49/3

= - 343/9 - 49/3

= - 343 - 147/9

= - 490/9

We conclude that on dividing the polynomial 3x³+7x by 7 + 3x, we will get the remainder as -490/9, which is not 0.

Therefore, we conclude that 7 + 3x is not a factor of 3x³+7x.

Exercise 2.4

Question 1 :

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

Answer :

(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 2 :

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

Answer :

(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)

Question 3 :

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

Answer :

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 4 :

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

Answer :

(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 5 :

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

Answer :

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 = 3/4

Exercise 2.5

Question 1 :

Factorize the following using appropriate identities :

(i)

(ii)

(iii)

Answer :

(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

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image061.png

Question 2 :

Expand each of the following, using suitable identities :

(i)

(ii)

(iii)

(iv)

(v)

(vi)

Answer :

(i)

We know that.

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

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image069.png

(ii)

We know that.

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

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image070.png

(iii)

We know that.

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

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image072.png

(iv)

We know that.

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

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image072.png

(v)

We know that.

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

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image073.png

(vi) NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image067.png

We know that.

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image076.jpg

Question 3 :

Write the following cubes in expanded form :

(i)

(ii)

(iii)

(iv)

Answer :

(i)

We know that.

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image092.png

Therefore, the expansion of the expressionis .

(ii)

We know that.

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image095.png

Therefore, the expansion of the expression NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image089.png is .

(iii) NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image089.png

We know that.

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image099.png

Therefore, the expansion of the expression NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image089.png is.

(iv) NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image090.png

We know that.

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image099.png

Therefore, the expansion of the expression NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image090.png is.

Question 4 :

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

Answer :

(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 5 :

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

Answer :

(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)

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/ A8

Question 6 :

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

Answer :

(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 7 :

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

Answer :

(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 8 :

Factorise 27x3 +y3 +z3 -9xyz.

Answer :

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 9 :

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

Answer :

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 10 :

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

Answer :

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 11 :

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

Answer :

(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 12 :

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

Answer :

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).

Question 13 :

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

Answer :

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).

Question 14 :

Evaluate the following products without multiplying directly :

(i)