RS Aggarwal Solutions Class 9 Chapter 2 Polynomials

RS Aggarwal Solutions Class 9 Chapter 2 Polynomials

NCERT Exemplar Class 9 Maths ¡s very important resource for students preparing for IX Board Examination. Here we have provided NCERT Exemplar Problems Solutions along with NCERT Exemplar Problems Class 9. Here you will find all the answers to the NCERT textbook questions of Chapter 2 – Polynomials.

RS Aggarwal Class 9 Solutions

RS Aggarwal Solutions Class 9 Chapter 2 Polynomials a1RS Aggarwal Solutions Class 9 Chapter 2 Polynomials a2

Exercise 2A

Question 1:
(i) It is a polynomial, Degree = 5.
(ii) It is polynomial, Degree = 3.
(iii) It is polynomial, Degree = 2.
(iv) It is not a polynomial.
(v) It is not a polynomial.
(vi) It is polynomial, Degree = 108.
(vii) It is not a polynomial.
(viii) It is a polynomial, Degree = 2.
(ix) It is not a polynomial.
(x) It is a polynomial, Degree = 0.
(xi) It is a polynomial, Degree = 0.
(xii) It is a polynomial, Degree = 2.

Question 2:
The degree of a polynomial in one variable is the highest power of the variable.

(i) Degree of 2x – \(\sqrt { 5 } \) is 1.
(ii) Degree of 3 – x + x2 – 6x3 is 3.
(iii) Degree of 9 is 0.
(iv) Degree of 8x4 – 36x + 5x7 is 7.
(v) Degree of x9 – x5 + 3x10 + 8 is 10.
(vi) Degree of 2 – 3x2 is 2.

Question 3:
(i) Coefficient of x3 in 2x + x2 – 5x3 + x4 is -5
(ii) Coefficient of x in
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Q3
(iii) Coefficient of x2 in RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Q3.1
(iv) Coefficient of x2 in 3x – 5 is 0.

Question 4:
(i) x27 – 36
(ii) y16
(iii) 5x3 – 8x + 7

Question 5:
(i) It is a quadratic polynomial.
(ii) It is a cubic polynomial.
(iii) It is a quadratic polynomial.
(iv) It is a linear polynomial.
(v) It is a linear polynomial.
(vi) It is a cubic polynomial.

Exercise 2B

Question 1:
p(x) = 5 – 4x + 2x2
(i) p(0) = 5 – 4(0) + 2(0)2 = 5

(ii) p(3) = 5 – 4(3) + 2(3)2
= 5 – 12 + 18
= 23 – 12 = 11

(iii) p(-2) = 5 – 4(-2) + 2(-2)2
= 5 + 8 + 8 = 21

Question 2:
p(y) = 4 + 3y – y2 + 5y3
(i) p(0) = 4 + 3(0) – 02 + 5(0)3
= 4 + 0 – 0 + 0 = 4

(ii) p(2) = 4 + 3(2) – 22 + 5(2)3
= 4 + 6 – 4 + 40
= 10 – 4 + 40 = 46

(iii) p(-1) = 4 + 3(-1) – (-1)2 + 5(-1)3
= 4 – 3 – 1 – 5 = -5

Question 3:
f(t) = 4t2 – 3t + 6
(i) f(0) = 4(0)2 – 3(0) + 6
= 0 – 0 + 6 = 6

(ii) f(4) = 4(4)2 – 3(4) + 6
= 64 – 12 + 6 = 58

(iii) f(-5) = 4(-5)2 – 3(-5) + 6
= 100 + 15 + 6 = 121

Question 4:
(i) p(x) = 0
⇒ x – 5 = 0
⇒ x = 5
⇒ 5 is the zero of the polynomial p(x).

(ii) q(x) = 0
⇒ x + 4 = 0
⇒ x = -4
⇒ -4 is the zero of the polynomial q(x).

(iii) p(t) = 0
⇒ 2t – 3 = 0
⇒ 2t =3
⇒ t =  \(\frac { 3 }{ 2 }  \)
⇒ t =  \(\frac { 3 }{ 2 }  \) is the zero of the polynomial p(t).

(iv) f(x) = 0
⇒ 3x + 1= 0
⇒ 3x = -1
⇒ x =  \(\frac { -1 }{ 3 }  \)
⇒ x =  \(\frac { -1 }{ 3 }  \) is the zero of the polynomial f(x).

(v) g(x) = 0
⇒ 5 – 4x = 0
⇒ -4x = -5
⇒ x =  \(\frac { 5 }{ 4 }  \)
⇒ x =   \(\frac { 5 }{ 4 }  \) is the zero of the polynomial g(x).

(vi) h(x) = 0
⇒ 6x – 1 = 0
⇒ 6x = 1
⇒ x =  \(\frac { 1 }{ 6 }  \)
⇒ x =  \(\frac { 1 }{ 6 }  \) is the zero of the polynomial h(x).

(vii) p(x) = 0
⇒ ax + b = 0
⇒ ax = -b
⇒ x =  \(\frac { -b }{ a }  \)
⇒ x =  \(\frac { -b }{ a }  \)is the zero of the polynomial p(x)

(viii) q(x) = 0
⇒ 4x = 0
⇒ x = 0
⇒ 0 is the zero of the polynomial q(x).

(ix) p(x) = 0
⇒ ax = 0
⇒ x = 0
⇒ 0 is the zero of the polynomial p(x).

Question 5:
(i) p(x) = x – 4
Then, p(4) = 4 – 4 = 0
⇒ 4 is a zero of the polynomial p(x).

(ii) p(x) = x – 3
Then, p(-3) = -3 – 3 = -6
⇒ -3 is not a zero of the polynomial p(x).

(iii) p(y) = 2y + 1
Then, RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Q5.2
⇒ \(\frac { -1 }{ 2 }  \) is a zero of the polynomial p(y).

(iv) p(x) = 2 – 5x
Then, RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Q5.1
⇒ \(\frac { 2 }{ 5 }  \) is a zero of the polynomial p(x).

(v) p(x) = (x – 1) (x – 2)
Then, p(1) = (1 – 1) (1 – 2) = 0 -1 = 0
⇒ 1 is a zero of the polynomial p(x).
Also, p(2) = (2 – 1)(2 – 2) = 1 0 = 0
⇒ 2 is a zero of the polynomial p(x).
Hence, 1 and 2 are the zeroes of the polynomial p(x).

(vi) p(x) = x2 – 3x.
Then, p(0) = 02 – 3(0) = 0
p(3) = (32) – 3(3) = 9 – 9 = 0
⇒ 0 and 3 are the zeroes of the polynomial p(x).

(vii) p(x) = x2 + x – 6
Then, p(2) = 22 + 2 – 6
= 4 + 2 – 6
= 6 – 6 = 0
⇒ 2 is a zero of the polynomial p(x).
Also, p(-3) = (-3)2 – 3 – 6
= 9 – 3 – 6 = 0
⇒ -3 is a zero of the polynomial p(x).
Hence, 2 and -3 are the zeroes of the polynomial p(x).

Exercise 2C

RS Aggarwal Solutions Class 9 Chapter 2 Polynomials c1

Question 1:
f(x) = x3 – 6x2 + 9x + 3
Now, x – 1 = 0  ⇒ x = 1
By the remainder theorem, we know that when f(x) is divided by (x – 1) the remainder is f(1).
Now, f(1) = 13 – 6 × 12 + 9 × 1 + 3
= 1 – 6 + 9 + 3
= 13 – 6 = 7
∴ The required remainder is 7.

Question 2:
f(x) = (2x3 – 5x2 + 9x – 8)
Now, x – 3 = 0  ⇒ x = 3
By the remainder theorem, we know that when f(x) is divided by (x – 3) the remainder is f(3).
Now, f(3) = 2 × 33 – 5 × 32 + 9 × 3 – 8
= 54 – 45 + 27 – 8
= 81 – 53 = 28
∴ The required remainder is 28.

Question 3:
f(x) = (3x4 – 6x2 – 8x + 2)
Now, x – 2 = 0  ⇒ x = 2
By the remainder theorem, we know that when f(x) is divided by (x – 2) the remainder is f(2).
Now, f(2) = 3 × 24 – 6 × 22 – 8 × 2 + 2
= 48 – 24 – 16 + 2
= 50 – 40 = 10
∴ The required remainder is 10.

Question 4:
f(x) = x3 – 7x2 + 6x + 4
Now, x – 6 = 0  ⇒ x = 6
By the remainder theorem, we know that when f(x) is divide by (x – 6) the remainder is f(6)
Now, f(6) = 63 – 7 × 62 + 6 × 6 + 4
= 216 – 252 + 36 + 4
= 256 – 252 = 4
∴ The required remainder is 4.

Question 5:
f(x) = (x3 – 6x2 + 13x + 60)
Now, x + 2 = 0  ⇒ x = -2
By the remainder the theorem, we know that when f(x) is divide by (x + 2) the remainder is f(-2).
Now, f(-2) = (-2)3 – 6(-2)2 + 13(-2) + 60
= -8 – 24 – 26 + 60
= -58 + 60 = 2
∴ The required remainder is 2.

Question 6:
f(x) = (2x4 + 6x3 + 2x2 + x – 8)
Now, x + 3 = 0  ⇒ x = -3
By the remainder the theorem, we know that when f(x) is divide by (x + 3) the remainder is f(-3).
f(-3) = 2(-3)4 + 6(-3)3 + 2(-3)2 – 3 – 8
= 162 – 162 + 18 – 3 – 8
= 18 – 11 = 7
∴ The required remainder is 7.

Question 7:
f(x) = (4x3 – 12x2 + 11x – 5)
Now, 2x – 1 = 0  ⇒ x =  \(\frac { 1 }{ 2 }  \)
By the remainder theorem, we know that when f(x) is divided by (2x – 1) the remainder is  \(f\left( \frac { 1 }{ 2 }  \right)     \)
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2c 7.1
∴ The required remainder is -2.

Question 8:
f(x) = (81x4 + 54x3 – 9x2 – 3x + 2)
Now, 3x + 2 = 0  ⇒ x =  \(\frac { -2 }{ 3 }  \)
By the remainder theorem, we know that when f(x) is divided by (3x+ 2) the remainder is  \(f\left( \frac { -2 }{ 3 }  \right)     \)
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2c 8.1
∴ The required remainder is 0.

Question 9:
f(x) = (x3 – ax2 + 2x – a)
Now, x – a = 0 x  ⇒ = a
By the remainder theorem, we know that when f(x) is divided by (x – a) the remainder is f(a)
Now, f(a) = a3 – a a2 + 2 a – a
= a3 – a3 + 2a – a
= a
∴ The required remainder is a.

Question 10:
Let f(x) = ax3 + 3x2 – 3
and g(x) = 2x3 – 5x + a
∴ f(4) = a × 43 + 3 × 42 – 3
= 64a + 48 – 3
= 64a + 45
g(4) = 2 × 43 – 5 × 4 + a
= 128 – 20 + a
= 108 + a
It is given that:
f(4) = g(4)
⇒ 64a + 45 = 108 + a
⇒ 64a – a = 108 – 45
⇒ 63a = 63
⇒ a =  \(\frac { 63 }{ 63 }  \)  = 1
∴ The value of a is 1.

Question 11:
Let f(x) = (x4 – 2x3 + 3x2 – ax + b)
∴ From the given information,
f(1) = 14 – 2(1)3 + 3(1)2 – a (1 ) + b = 5
⇒ 1 – 2 + 3 – a + b = 5
⇒ 2 – a + b = 5 ….(i)
And,
f(-1) = (-1)4 – 2(-1)3 + 3(-1)2 – a(-1) + b = 19
⇒ 1 + 2 + 3 + a + b = 19
⇒ 6 + a + b = 19 ….(ii)
Adding (i) and (ii), we get
⇒ 8 + 2b = 24
⇒ 2b = 24 – 8 = 16
⇒ b =  \(\frac { 16 }{ 2 }  \)
Substituting the value of b = 8 in (i), we get
2 – a + 8 = 5
⇒ -a + 10 = 5
⇒ -a = -10 + 5
⇒ -a = -5
⇒ a = 5
∴ a = 5 and b = 8
f(x) = x4 – 2x3 + 3x2 – ax + b
= x4 – 2x3 + 3x2 – 5x + 8
∴ f(2) = (2)4 – 2(2)3 + 3(2)2 – 5(2) + 8
= 16 – 16 + 12 – 10 + 8
= 20 – 10 = 10
∴ The required remainder is 10.

Exercise 2D

RS Aggarwal Solutions Class 9 Chapter 2 Polynomials d1

Question 1:
f(x) = (x3 – 8)
By the Factor Theorem, (x – 2) will be a factor of f(x) if f(2) = 0.
Here, f(2) = (2)3 – 8
= 8 – 8 = 0
∴ (x – 2) is a factor of (x3 – 8).

Question 2:
f(x) = (2x3 + 7x2 – 24x – 45)
By the Factor Theorem, (x – 3) will be a factor of f(x) if f(3) = 0.
Here, f(3) = 2 × 33 + 7 × 32 – 24 × 3 – 45
= 54 + 63 – 72 – 45
= 117 – 117 = 0
∴ (x – 3) is a factor of (2x3 + 7x2 – 24x – 45).

Question 3:
f(x) = (2x4 + 9x3 + 6x2 – 11x – 6)
By the Factor Theorem, (x – 1) will be a factor of f(x) if f(1) = 0.
Here, f(1) = 2 × 14 + 9 × 13 + 6 × 12 – 11 × 1 – 6
= 2 + 9 + 6 – 11 – 6
= 17 – 17 = 0
∴ (x – 1) is factor of (2x4 + 9x3 + 6x2 – 11x – 6).

Question 4:
f(x) = (x4 – x2 – 12)
By the Factor Theorem, (x + 2) will be a factor of f(x) if f(-2) = 0.
Here, f(-2) = (-2)4 – (-2)2 – 12
= 16 – 4 – 12
= 16 – 16 = 0
∴ (x + 2) is a factor of (x4 – x2 – 12).

Question 5:
f(x) = 2x3 + 9x2 – 11x – 30
By the Factor Theorem, (x + 5) will be a factor of f(x) if f(-5) = 0.
Here, f(-5) = 2(-5)3 + 9(-5)2 – 11(-5) – 30
= -250 + 225 + 55 – 30
= -280 + 280 = 0
∴ (x + 5) is a factor of (2x3 + 9x2 – 11x – 30).

Question 6:
f(x) = (2x4 + x3 – 8x2 – x + 6)
By the Factor Theorem, (x – a) will be a factor of f(x) if f(a) = 0.
Here, 2x – 3 = 0  ⇒ x =  \(\frac { 3 }{ 2 }  \)
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2d 6.1
∴ (2x – 3) is a factor of (2x4 + x3 – 8x2 – x + 6).

Question 7:
f(x) = (7x2 – \(4\sqrt { 2 }         \) x – 6 = 0)
By the Factor Theorem, (x – a) will be a factor of f(x) if f(a) = 0.
Here, RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2d 7.1
= 14 – 8 – 6
= 14 – 14 = 0
∴ (x – \(\sqrt { 2 }    \)) is a factor of (7 – \(4\sqrt { 2 }         \) x – 6 = 0).

Question 8:

f(x) =   (\(4\sqrt { 2 }         \)x2  + 5x +\(\sqrt { 2 }    \) = 0)
By the Factor Theorem, (x – a) will be a factor of f(x) if f(a) = 0.
Here, RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2d 8.1
∴ (x + \(\sqrt { 2 }    \)) is a factor of (\(4\sqrt { 2 }         \)x2  + 5x +\(\sqrt { 2 }    \) = 0).

Question 9:
f(x) = (2x3 + 9x2 + x + k)
x – 1 = 0  ⇒ x = 1
∴ f(1) = 2 × 13 + 9 × 12 + 1 + k
= 2 + 9 + 1 + k
= 12 + k
Given that (x – 1) is a factor of f(x).
By the Factor Theorem, (x – a) will be a factor of f(x) if f(a) = 0 and therefore f(1) = 0.
⇒ f(1) = 12 + k = 0
⇒ k = -12.

Question 10:
f(x) = (2x3 – 3x2 – 18x + a)
x – 4 = 0  ⇒ x = 4
∴ f(4) = 2(4)3 – 3(4)2 – 18 × 4 + a
= 128 – 48 – 72 + a
= 128 – 120 + a
= 8 + a
Given that (x – 4) is a factor of f(x).
By the Factor Theorem, (x – a) will be a factor of f(x) if f(a) = 0 and therefore f(4) = 0.
⇒ f(4) = 8 + a = 0
⇒ a = -8

Question 11:
f(x) = x4 – x3 – 11x2 – x + a
x + 3 = 0  ⇒ x = -3
∴ f(-3) = (-3)4 – (-3)3 -11 (-3)2 – (-3) + a
= 81 + 27 – 11 × 9 + 3 + a
= 81 + 27 – 99 + 3 + a
= 111 – 99 + a
= 12 + a
Given that f(x) is divisible by (x + 3), that is (x+3) is a factor of f(x).
By the Factor Theorem, (x – a) will be a factor of f(x) if f(a) = 0 and therefore f(-3) = 0.
⇒ f(-3) = 12 + a =0
⇒ a = -12.

Question 12:
f(x) = (2x3 + ax2 + 11x + a + 3)
2x – 1 = 0  ⇒ x =  \(\frac { 1 }{ 2 }  \)
Given that f(x) is exactly divisible by (2x – 1), that is (2x – 1) is a factor of f(x).
By the Factor Theorem, (x – a) will be a factor of f(x) if f(a) = 0
and therefore \(f\left( \frac { 1 }{ 2 }  \right)     \) ≠ 0.
Therefore, we have
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2d 12.1
∴ The value of a = -7.

Question 13:
Let f(x) = (x3 – 10x2 + ax + b), then by factor theorem
(x – 1) and (x – 2) will be factors of f(x) if f(1) = 0 and f(2) = 0.
f(1) = 13 – 10 12 + a 1 + b = 0
⇒ 1 – 10 + a + b = 0
⇒ a + b = 9 ….(i)
And f(2) = 23 – 10 22 + a 2 + b = 0
⇒ 8 – 40 + 2a + b = 0
⇒ 2a + b = 32 ….(ii)
Subtracting (i) from (ii), we get
a = 23
Substituting the value of a = 23 in (i), we get
⇒ 23 + b = 9
⇒ b = 9 – 23
⇒ b = -14
∴ a = 23 and b = -14.

Question 14:
Let f(x) = (x4 + ax3 – 7x2 – 8x + b)
Now, x + 2 = 0 x = -2 and x + 3 = 0 x = -3
By factor theorem, (x + 2) and (x + 3) will be factors of f(x) if f(-2) = 0 and f(-3) = 0
∴ f(-2) = (-2)4 + a (-2)3 – 7 (-2)2 – 8 (-2) + b = 0
⇒ 16 – 8a – 28 + 16 + b = 0
⇒ -8a + b = -4
⇒ 8a – b = 4 ….(i)
And, f(-3) = (-3)4 + a (-3)3 – 7 (-3)2 – 8 (-3) + b = 0
⇒ 81 – 27a – 63 + 24 + b = 0
⇒ -27a + b = -42
⇒ 27a – b = 42 ….(ii)
Subtracting (i) from (ii), we get,
19a = 38
So, a = 2
Substituting the value of a = 2 in (i), we get
8(2) – b = 4
⇒ 16 – b = 4
⇒ -b = -16 + 4
⇒ -b = -12
⇒ b = 12
∴ a = 2 and b = 12.

Question 15:
Let f(x) = x3 – 3x2 – 13x + 15
Now, x2 + 2x – 3 = x2 + 3x – x – 3
= x (x + 3) – 1 (x + 3)
= (x + 3) (x – 1)
Thus, f(x) will be exactly divisible by x2 + 2x – 3 = (x + 3) (x – 1) if (x + 3) and (x – 1) are both factors of f(x), so by factor theorem, we should have f(-3) = 0 and f(1) = 0.
Now, f(-3) = (-3)3 – 3 (-3)2 – 13 (-3) + 15
= -27 – 3 × 9 + 39 + 15
= -27 – 27 + 39 + 15
= -54 + 54 = 0
And, f(1) = 13 – 3 × 12 – 13 × 1 + 15
= 1 – 3 – 13 + 15
= 16 – 16 = 0
∴ f(-3) = 0 and f(1) = 0
So, x2 + 2x – 3 divides f(x) exactly.

Question 16:
Let f(x) = (x3 + ax2 + bx + 6)
Now, by remainder theorem, f(x) when divided by (x – 3) will leave a remainder as f(3).
So, f(3) = 33 + a × 32 + b × 3 + 6 = 3
⇒ 27 + 9a + 3b + 6 = 3
⇒ 9a + 3b + 33 = 3
⇒ 9a + 3b = 3 – 33
⇒ 9a + 3b = -30
⇒ 3a + b = -10 ….(i)
Given that (x – 2) is a factor of f(x).
By the Factor Theorem, (x – a) will be a factor of f(x) if f(a) = 0 and therefore f(2) = 0.
f(2) =  23 + a × 22 + b × 2 + 6 = 0
⇒ 8 + 4a+ 2b + 6 = 0
⇒ 4a + 2b = -14
⇒ 2a + b = -7 ….(ii)
Subtracting (ii) from (i), we get,
⇒ a = -3
Substituting the value of a = -3 in (i), we get,
⇒ 3(-3) + b = -10
⇒ -9 + b = -10
⇒ b = -10 + 9
⇒ b = -1
∴ a = -3 and b = -1.

Exercise 2E

RS Aggarwal Solutions Class 9 Chapter 2 Polynomials e1

Question 1:
9x2 + 12xy = 3x (3x + 4y)

Question 2:
18x2y – 24xyz = 6xy (3x – 4z)

Question 3:
27a3b3 – 45a4b2 = 9a3b2 (3b – 5a)

Question 4:
2a (x + y) – 3b (x + y) = (x + y) (2a – 3b)

Question 5:
2x (p2 + q2) + 4y (p2 + q2)
= (2x + 4y) (p2 + q2)
= 2(x+ 2y) (p2 + q2)

Question 6:
x (a – 5) + y (5 – a)
= x (a – 5) + y (-1) (a – 5)
= (x – y) (a – 5)

Question 7:
4 (a + b) – 6 (a + b)2
= (a + b) [4 – 6 (a + b)]
= 2 (a + b) (2 – 3a – 3b)
= 2 (a + b) (2 – 3a – 3b)

Question 8:
8 (3a – 2b)2 – 10 (3a – 2b)
= (3a – 2b) [8(3a – 2b) – 10]
= (3a – 2b) 2[4 (3a – 2b) – 5]
= 2 (3a – 2b) (12 a – 8b – 5)

Question 9:
x (x + y)3 – 3x2y (x + y)
= x (x + y) [(x + y)2 – 3xy]
= x (x + y) (x2 + y2 + 2xy – 3xy)
= x (x + y) (x2 + y2 – xy)

Question 10:
x3 + 2x2 + 5x + 10
= x2 (x + 2) + 5 (x + 2)
= (x2 + 5) (x + 2)

Question 11:
x2 + xy – 2xz – 2yz
= x (x + y) – 2z (x + y)
= (x+ y) (x – 2z)

Question 12:
a3b – a2b + 5ab – 5b
= a2b (a – 1) + 5b (a – 1)
= (a – 1) (a2b + 5b)
= (a – 1) b (a2 + 5)
= b (a – 1) (a2 + 5)

Question 13:
8 – 4a – 2a3 + a4
= 4(2 – a) – a3 (2 – a)
= (2 – a) (4 – a3)

Question 14:
x3 – 2x2y + 3xy2 – 6y3
= x2 (x – 2y) + 3y2 (x – 2y)
= (x – 2y) (x2 + 3y2)

Question 15:
px + pq – 5q – 5x
= p(x + q) – 5 (q + x)
= (x + q) (p – 5)

Question 16:
x2 – xy + y – x
= x (x – y) – 1 (x – y)
= (x – y) (x – 1)

Question 17:
(3a – 1)2 – 6a + 2
= (3a – 1)2 – 2 (3a – 1)
= (3a – 1) [(3a – 1) – 2]
= (3a – 1) (3a – 3)
= 3(3a – 1) (a – 1)

Question 18:
(2x – 3)2 – 8x + 12
= (2x – 3)2 – 4 (2x – 3)
= (2x – 3) (2x – 3 – 4)
= (2x – 3) (2x – 7)

Question 19:
a3 + a – 3a2 – 3
= a(a2 + 1) – 3 (a2 + 1)
= (a – 3) (a2 + 1)

Question 20:
3ax – 6ay – 8by + 4bx
= 3a (x – 2y) + 4b (x – 2y)
= (x – 2y) (3a + 4b)

Question 21:
abx2 + a2x + b2x +ab
= ax (bx + a) + b (bx + a)
= (bx + a) (ax + b)

Question 22:
x3 – x2 + ax + x – a – 1
= x3 – x2 + ax – a + x – 1
= x2 (x – 1) + a (x – 1) + 1 (x – 1)
= (x – 1) (x2 + a + 1)

Question 23:
2x + 4y – 8xy – 1
= 2x – 1 – 8xy + 4y
= (2x – 1) – 4y (2x – 1)
= (2x – 1) (1 – 4y)

Question 24:
ab (x2 + y2) – xy (a2 + b2)
= abx2 + aby2 – a2xy – b2xy
= abx2 – a2xy + aby2 – b2xy
= ax (bx – ay) + by(ay – bx)
= (bx – ay) (ax – by)

Question 25:
a2 + ab (b + 1) + b3
= a2 + ab2 + ab + b3
= a2 + ab + ab2 + b3
= a (a + b) + b2 (a + b)
= (a + b) (a + b2)

Question 26:
a3 + ab (1 – 2a) – 2b2
= a3 + ab – 2a2b – 2b2
= a (a2 + b) – 2b (a2 + b)
= (a2 + b) (a – 2b)

Question 27:
2a2 + bc – 2ab – ac
= 2a2 – 2ab – ac + bc
= 2a (a – b) – c (a – b)
= (a – b) (2a – c)

Question 28:
(ax + by)2 + (bx – ay)2
= a2x2 + b2y2 + 2abxy + b2x2 + a2y2 – 2abxy
= a2x2 + b2y2 + b2x2 + a2y2
= a2x2 + b2x2 + b2y2 + a2y2
= x2 (a2 + b2) + y2(a2 + b2)
= (a2 + b2) (x2 + y2)

Question 29:
a (a + b – c) – bc
= a2 + ab – ac – bc
= a(a + b) – c (a + b)
= (a – c) (a + b)

Question 30:
a(a – 2b – c) + 2bc
= a2 – 2ab – ac + 2bc
= a (a – 2b) – c (a – 2b)
= (a – 2b) (a – c)

Question 31:
a2x2 + (ax2 + 1)x + a
= a2x2 + ax3 + x + a
= ax2 (a + x) + 1 (x + a)
= (ax2 + 1) (a + x)

Question 32:
ab (x2 + 1) + x (a2 + b2)
= abx2 + ab + a2x + b2x
= abx2 + a2x + ab + b2x
= ax (bx + a) + b (bx + a)
= (bx + a) (ax + b)

Question 33:
x2 – (a + b) x + ab
= x2 – ax – bx + ab
= x (x – a) – b(x – a)
= (x – a) (x – b)

Question 34:
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2e 34.1

Exercise 2F

Question 1:
25x2 – 64y2
= (5x)2 – (8y)2
= (5x + 8y) (5x – 8y)
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 1.1

Question 2:
100 – 9x2
= (10)2 – (3x)2
= (10 + 3x) (10 – 3x)
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 2.1

Question 3:
5x2 – 7y2
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 3.1

Question 4:
(3x + 5y)2 – 4z2
= (3x + 5y)2 – (2z)2
= (3x + 5y + 2z) (3x + 5y – 2z)
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 4.1

Question 5:
150 – 6x2
= 6 (25 – x2)
= 6 (52 – x2)
= 6 (5 + x) (5 – x)
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 5.1

Question 6:
20x2 – 45
= 5(4x2 – 9)
= 5 [(2x)2 – (3)2]
= 5 (2x + 3) (2x – 3)
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 6.1

Question 7:
3x3 – 48x
= 3x (x2 – 16)
= 3x [(x)2 – (4)2]
= 3x (x + 4) (x – 4)
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 7.1

Question 8:
2 – 50x2
= 2 (1 – 25x2)
= 2 [(1)2 – (5x)2]
= 2 (1 + 5x) (1 – 5x)
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 8.1

Question 9:
27a2 – 48b2
= 3 (9a2 – 16b2)
= 3 [(3a)2 – (4b)2]
= 3(3a + 4b) (3a – 4b)
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 9.1

Question 10:
x – 64x3
= x (1 – 64x2)
= x[(1)2 – (8x)2]
= x (1 + 8x) (1 – 8x)
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 10.1

Question 11:
8ab2 – 18a3
= 2a (4b2 – 9a2)
= 2a [(2b)2 – (3a)2]
= 2a (2b + 3a) (2b – 3a)
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 24.2

Question 12:
3a3b – 243ab3
= 3ab (a2 – 81 b2)
= 3ab [(a)2 – (9b)2]
= 3ab (a + 9b) (a – 9b)
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 24.2

Question 13:
(a + b)3 – a – b
= (a + b)3 – (a + b)
= (a + b) [(a + b)2 – 12]
= (a + b) (a + b + 1) (a + b – 1)
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 11.1

Question 14:
108a2 – 3(b – c)2
= 3 [(36a2 – (b -c)2]
= 3 [(6a)2 – (b – c)2]
= 3 (6a + b – c) (6a – b + c)
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 12.1

Question 15:
x3 – 5x2 – x + 5
= x2 (x – 5) – 1 (x – 5)
= (x – 5) (x2 – 1)
= (x – 5) (x + 1) (x – 1)
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 13.1

Question 16:
a2 + 2ab + b2 – 9c2
= (a + b)2 – (3c)2
= (a + b + 3c) (a + b – 3c)
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 14.1

Question 17:
9 – a2 + 2ab – b2
= 9 – (a2 – 2ab + b2)
= 32 – (a – b)2
= (3 + a – b) (3 – a + b)
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 15.1

Question 18:
a2 – 4ac + 4c2 – b2
= a2 – 4ac + 4c2 – b2
= a2 – 2 a 2c + (2c)2 – b2
= (a – 2c)2 – b2
= (a – 2c + b) (a – 2c – b)
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 16.1

Question 19:
9a2 + 3a – 8b – 64b2
= 9a2 – 64b2 + 3a – 8b
= (3a)2 – (8b)2 + (3a – 8b)
= (3a + 8b) (3a – 8b) + (3a – 8b)
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 17.1
= (3a – 8b) (3a + 8b + 1)

Question 20:
x2 – y2 + 6y – 9
= x2 – (y2 – 6y + 9)
= x2 – (y2 – 2 y 3 + 32)
= x2 – (y – 3)2
= [x + (y – 3)] [x – (y – 3)]
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 18.1
= (x + y – 3) (x – y + 3)

Question 21:
4x2 – 9y2 – 2x – 3y
= (2x)2 – (3y)2 – (2x + 3y)
= (2x + 3y) (2x – 3y) – (2x + 3y)
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 19.1
= (2x + 3y) (2x – 3y – 1)

Question 22:
x4 – 1
= (x2 )2 – 12
= (x2 + 1) (x2 – 1) RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 21.1
= (x2 + 1) (x + 1) (x – 1)
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 20.1

Question 23:
a – b – a2 + b2
= (a – b) – (a2 – b2)
= (a – b) – (a – b) (a + b)
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 22.1
= (a – b) (1 – a – b)

Question 24:
x4 – 625
= (x2)2 – (25)2
= (x2 + 25) (x2 – 25)
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 23.1
= (x2 + 25) (x2 – 52)
= (x2 + 25) (x + 5) (x – 5)
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 24.1

Exercise 2G

Question 1:
x2 + 11x + 30
= x2 + 6x + 5x + 30
= x (x + 6) + 5 (x + 6)
= (x + 6) (x + 5).

Question 2:
x2 + 18x + 32
= x2 + 16x + 2x + 32
= x (x + 16) + 2 (x + 16)
= (x + 16) (x + 2).

Question 3:
x2 + 7x – 18
= x2 + 9x – 2x – 18
= x (x + 9) – 2 (x + 9)
= (x + 9) (x – 2).

Question 4:
x2 + 5x – 6
= x2 + 6x – x – 6
= x (x + 6) – 1 (x+ 6)
= (x + 6) (x – 1).

Question 5:
y2 – 4y + 3
= y2 – 3y – y + 3
= y (y – 3) – 1 (y – 3)
= (y – 3) (y – 1).

Question 6:
x2 – 21x + 108
= x2 – 12x – 9x + 108
= x (x – 12) – 9 (x – 12)
= (x – 12) (x – 9).

Question 7:
x2 – 11x – 80
= x2 – 16x + 5x – 80
= x (x – 16) + 5 (x – 16)
= (x – 16) (x + 5).

Question 8:
x2 – x – 156
= x2 – 13x + 12x – 156
= x (x – 13) + 12 (x – 13)
= (x – 13) (x + 12).

Question 9:
z2 – 32z – 105
= z2 – 35z + 3z – 105
= z (z – 35) + 3 (z – 35)
= (z – 35) (z + 3)

Question 10:
40 + 3x – x2
= 40 + 8x – 5x – x2
= 8 (5 + x) -x (5 + x)
= (5 + x) (8 – x).

Question 11:
6 – x – x2
= 6 + 2x – 3x – x2
= 2(3 + x) – x (3 + x)
= (3 + x) (2 – x).

Question 12:
7x2 + 49x + 84
= 7(x2 + 7x + 12)
= 7 [x2 + 4x + 3x + 12]
= 7 [x (x + 4) + 3 (x + 4)]
= 7 (x + 4) (x + 3).

Question 13:
m2 + 17mn – 84n2
= m2 + 21mn – 4mn – 84n2
= m (m + 21n) – 4n (m + 21n)
= (m + 21n) (m – 4n).

Question 14:
5x2 + 16x + 3
= 5x2 + 15x + x + 3
= 5x (x + 3) + 1 (x + 3)
= (5x + 1) (x + 3).

Question 15:
6x2 + 17x + 12
= 6x2 + 9x + 8x + 12
= 3x (2x + 3) + 4(2x + 3)
= (2x + 3) (3x + 4).

Question 16:
9x2 + 18x + 8
= 9x2 + 12x + 6x + 8
= 3x (3x+ 4) +2 (3x + 4)
= (3x + 4) (3x + 2).

Question 17:
14x2 + 9x + 1
= 14x2 + 7x + 2x + 1
= 7x (2x + 1) + (2x + 1)
= (7x + 1) (2x + 1).

Question 18:
2x2 + 3x – 90
= 2x2 – 12x + 15x – 90
= 2x (x – 6) + 15 (x – 6)
= (x – 6) (2x + 15).

Question 19:
2x2 + 11x – 21
= 2x2 + 14x – 3x – 21
= 2x (x + 7) – 3 (x + 7)
= (x + 7) (2x – 3).

Question 20:
3x2 – 14x + 8
= 3x2 – 12x – 2x +8
= 3x (x – 4) – 2(x – 4)
= (x – 4) (3x – 2).

Question 21:
18x2 + 3x – 10
= 18x2 – 12x + 15x – 10
= 6x (3x – 2) + 5 (3x – 2)
= (6x + 5) (3x – 2).

Question 22:
15x2 + 2x – 8
= 15x2 – 10x + 12x – 8
= 5x (3x – 2) + 4 (3x – 2)
= (3x – 2) (5x + 4).

Question 23:
6x2 + 11x – 10
= 6x2 + 15x – 4x – 10
= 3x (2x + 5) – 2(2x+ 5)
= (2x + 5) (3x – 2).

Question 24:
30x2 + 7x – 15
= 30x2 – 18x + 25x – 15
= 6x (5x – 3) + 5 (5x- 3)
= (5x – 3) (6x + 5).

Question 25:
24x2 – 41x + 12
= 24x2 – 32x – 9x + 12
= 8x (3x – 4) – 3 (3x – 4)
= (3x – 4) (8x – 3).

Question 26:
2x2 – 7x – 15
= 2x2 – 10x + 3x – 15
= 2x (x – 5) + 3 (x – 5)
= (x – 5) (2x + 3).

Question 27:
6x2 – 5x – 21
= 6x2 + 9x – 14x – 21
= 3x (2x + 3) – 7 (2x + 3)
= (3x – 7) (2x + 3).

Question 28:
10x2 – 9x – 7
= 10x2 + 5x – 14x – 7
= 5x (2x + 1) – 7 (2x+ 1)
= (2x + 1) (5x – 7).

Question 29:
5x2 – 16x – 21
= 5x2 + 5x – 21x – 21
= 5x (x + 1) -21 (x + 1)
= (x + 1) (5x – 21).

Question 30:
2x2 – x – 21
= 2x2 + 6x – 7x – 21
= 2x (x + 3) – 7 (x + 3)
= (x + 3) (2x – 7).

Question 31:
15x2 – x – 28
= 15x2 + 20x – 21x – 28
= 5x (3x + 4) – 7 (3x + 4)
= (3x + 4) (5x – 7).

Question 32:
8a2 – 27ab + 9b2
= 8a2 – 24ab – 3ab + 9b2
= 8a (a – 3b) – 3b (a – 3b)
= (a – 3b) (8a – 3b).

Question 33:
5x2 + 33xy – 14y2
= 5x2 + 35xy – 2xy – 14y2
= 5x (x + 7y) – 2y (x + 7y)
= (x + 7y) (5x – 2y).

Question 34:
3x3 – x2 – 10x
= x (3x2 – x – 10)
= x [3x2 – 6x + 5x – 10]
= x [3x (x – 2) + 5 (x – 2)]
= x (x – 2) (3x + 5).

Question 35:
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 35.1

Question 36:
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 36.1

Question 37:
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 37.1

Question 38:
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 38.1

Question 39:
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 39.1

Question 40:
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 40.1

Question 41:
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 41.1

Question 42:
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 42.1

Question 43:
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 43.1

Question 44:
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 44.1

Question 45:
Let x + y = z
Then, 2 (x + y)2 – 9 (x + y) – 5
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 45.1
Now, replacing z by (x + y), we get
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 45.2

Question 46:
Let 2a – b = c
Then, 9 (2a – b)2 – 4 (2a – b) -13
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 46.1
Now, replacing c by (2a – b) , we get
9 (2a – b)2 – 4 (2a – b) – 13
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 46.2

Question 47:
Let x – 2y = z
Then, 7 (x – 2y)2 – 25 (x – 2y) + 12
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 47.1
Now replace z by (x – 2y), we get
7 (x – 2y)2 – 25 (x – 2y) + 12
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 47.2

Question 48:
Let x2 = y
Then, 4x4 + 7x2 – 2
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 48.1
Now replacing y by x2, we get
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2f 48.2

Exercise 2H

RS Aggarwal Solutions Class 9 Chapter 2 Polynomials h1

Question 1:
We know:
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
(i) (a + 2b + 5c)2
= (a)2 + (2b)2 + (5c)2 + 2(a) (2b) + 2 (2b) (5c) + 2(5c) (a)
= a2 + 4b2 + 25c2 + 4ab + 20bc + 10ac
(ii) (2a – b + c)2
= (2a)2 + (-b)2 + (c)2 + 2 (2a) (-b) + 2(-b) (c) + 2 (c) (2a)
= 4a2 + b2 + c2 – 4ab – 2bc + 4ac.
(iii) (a – 2b – 3c)2
= (a)2 + (-2b)2 + (-3c)2 + 2(a) (-2b) + 2(-2b) (-3c) + 2 (-3c) (a)
= a2 + 4b2 + 9c2 – 4ab + 12bc – 6ac.

Question 2:
We know:
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
(i) (2a – 5b – 7c)2
= (2a)2 + (-5b)2 + (-7c)2 + 2 (2a) (-5b) + 2 (-5b) (-7c) + 2 (-7c) (2a)
= 4a2 + 25b2 + 49c2 – 20ab + 70bc – 28ac.
(ii) (-3a + 4b – 5c)2
= (-3a)2 + (4b)2 + (-5c)2 + 2 (-3a) (4b) + 2 (4b) (-5c) + 2 (-5c) (-3a)
= 9a2 + 16b2 + 25c2 – 24ab – 40bc + 30ac.
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2h 2.1

Question 3:
4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz
= (2x)2 + (3y)2 + (-4z)2 + 2 (2x) (3y) + 2(3y) (-4z) + 2 (-4z) (2x)
= (2x + 3y – 4z)2

Question 4:
9x2 + 16y2 + 4z2 – 24xy + 16yz – 12xz
= (-3x)2 + (4y)2 + (2z)2 + 2 (-3x) (4y) + 2 (4y) (2z) + 2 (2z) (-3x)
= (-3x + 4y + 2z)2.

Question 5:
25x2 + 4y2 + 9z2 – 20xy – 12yz + 30xz
= (5x)2 + (-2y)2 + (3z)2 + 2(5x) (-2y) + 2(-2y) (3z) + 2(3z) (5x)
= (5x – 2y + 3z)2

Question 6:
(i) (99)2
= (100 – 1)2
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2h 6.1
= (100)2 – 2(100) (1) + (1)2
= 10000 – 200 + 1
= 9801.
(ii) (998)2
= (1000 – 2)2
= (1000)2 – 2 (1000) (2) + (2)2
= 1000000 – 4000 + 4
= 996004.

Exercise 2I

RS Aggarwal Solutions Class 9 Chapter 2 Polynomials i1

Question 1:
(i) (3x + 2)3
= (3x)3 + (2)3 + 3 × 3x × 2 (3x + 2)
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2i 1.1
= 27x3 + 8 + 18x (3x + 2)
= 27x3 + 8 + 54x2 + 36x.
(ii) (3a – 2b)3
= (3a)3 – (2b)3 – 3 × 3a × 2b (3a – 2b)
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Q1.1
= 27a3 – 8b3 – 18ab (3a – 2b)
= 27 a3 – 8b3 – 54a2b + 36ab2.
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2i 1.2

Question 2:
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2i 2.1

Question 3:
(i) (95)3
= (100 – 5)3
= (100)3 – (5)3 – 3 × 100 × 5 (100 – 5)
= 1000000 – 125 – (1500 95)
= 857375.
(ii) (999)3
= (1000 – 1)3
= (1000)3 – (1)3 – 3 × 1000 × 1 (1000 – 1)
= 1000000000 – 1 – 3000 (1000 – 1)
= 1000000000 – 1 – (3000 999)
= 997002999.

Exercise 2J

RS Aggarwal Solutions Class 9 Chapter 2 Polynomials j1

Question 1:
x3 + 27
= x3 + 33
= (x + 3) (x2 – 3x + 9)
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2j 1.1

Question 2:
8x3 + 27y3
= (2x)3 + (3y)3
= (2x+ 3y) [(2x)2 – (2x) (3y) + (3y)2]
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2j 1.1
= (2x + 3y) (4x2 – 6xy + 9y2).

Question 3:
343 + 125 b3
= (7)3 + (5b)3
= (7 + 5b) [(7)2 – (7) (5b) + (5b)2]
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2j 2.1
= (7 + 5b) (49 – 35b + 25b2)

Question 4:
1 + 64x3
= (1)3 + (4x)3
= (1 + 4x) [(1)2 – 1 (4x) + (4x)2]
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2j 3.1
= (1 + 4x) (1 – 4x + 16x2).

Question 5:
125a3 + \(\frac { 1 }{ 8 } \)
We know that
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2j 4.1
Let us rewrite
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2j 5.1

Question 6:
216x3 + \(\frac { 1 }{ 125 } \)
We know that
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Q6.1
Let us rewrite
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2j 6.1

Question 7:
16x 4 + 54x
= 2x (8x 3 + 27)
= 2x [(2x)3 + (3)3]
= 2x (2x + 3) [(2x)2 – 2x(3) + 32]
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2j 7.1
=2x(2x+3)(4x2 -6x +9)

Question 8:
7a3 + 56b3
= 7(a3 + 8b3)
= 7 [(a)3 + (2b)3]
= 7 (a + 2b) [a2 – a 2b + (2b)2]
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2j 8.1
= 7 (a + 2b) (a2 – 2ab + 4b2).

Question 9:
x5 + x2
= x2(x3 + 1)
= x2 (x + 1) [(x)2 – x (1) + (1)2]
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2j 9.1
= x2 (x + 1) (x2 – x + 1).

Question 10:
a3 + 0.008
= (a)3 + (0.2)3
= (a + 0.2) [(a)2 – a(0.2) + (0.2)2]
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2j 10.1
= (a + 0.2) (a2 – 0.2a + 0.04).

Question 11:
x6 + y6
= (x2)3 + (y2)3
= (x2 + y2) [(x2)2 – x2 (y2)+ (y2)2]
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2j 11.1
= (x2 + y2) (x4 – x2y2 + y4).

Question 12:
2a3 + 16b3 – 5a – 10b
= 2 (a3 + 8b3) – 5 (a + 2b)
= 2 [(a)3 + (2b)3] – 5 (a + 2b)
= 2 (a + 2b) [(a)2 – a (2b) + (2b)2 ] – 5 (a + 2b)
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2j 12.1
= (a + 2b) [2(a2 – 2ab + 4b2) – 5]

Question 13:
x3 – 512
= (x)3 – (8)3
= (x – 8) [(x)2 + x (8) + (8)2]
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2j 13.1
= (x – 8) (x2 + 8x + 64).

Question 14:
64x3 – 343
= (4x)3 – (7)3
= (4x – 7) [(4x)2 + 4x (7) + (7)2]
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2j 14.1
= (4x – 7) (16x2 + 28x + 49).

Question 15:
1 – 27x3
= (1)3 – (3x)3
= (1 – 3x) [(1)2 + 1 (3x) + (3x)2]
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2j 15.1
= (1 – 3x) (1 + 3x + 9x2).

Question 16:
1 – 27x3
= (1)3 – (3x)3
= (1 – 3x) [(1)2 + 1 (3x) + (3x)2]
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2j 15.1
= (1 – 3x) (1 + 3x + 9x2).

Question 17:
We know that
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2j 21.1
Let us rewrite
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2j 17.1

Question 18:
a3 – 0.064
= (a)3 – (0.4)3
= (a – 0.4) [(a)2 + a (0.4) + (0.4)2]
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2j 18.1
= (a – 0.4) (a2 + 0.4 a + 0.16).

Question 19:
(a + b)3 – 8
= (a + b)3 – (2)3
= (a + b – 2) [(a + b)2 + (a + b) 2 + (2)2]
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2j 19.1
= (a + b – 2) [a2 + b2 + 2ab + 2 (a + b) + 4].

Question 20:
x6 – 729
= (x2)3 – (9)3
= (x2 – 9) [(x2)2 + x2 9 + (9)2]
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2j 16.1
= (x2 – 9) (x4 + 9x2 + 81)
= (x + 3) (x – 3) [(x2 + 9)2 – (3x)2]
= (x + 3) (x – 3) (x2 + 3x + 9) (x2 – 3x + 9).

Question 21:
We know that,
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2j 21.1
Therefore,
(a + b)3 – (a – b)3
= [a + b – (a – b)] [ (a + b)2 + (a + b) (a – b) + (a – b)2]
= (a + b – a + b) [ a2 + b2 + 2ab + a2 – b2 + a2 + b2 – 2ab]
= 2b (3a2 + b2).

Question 22:
x – 8xy3
= x (1 – 8y3)
= x [(1)3 – (2y)3]
= x (1 – 2y) [(1)2 + 1 (2y) + (2y)2]
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2j 22.1
= x (1 – 2y) (1 + 2y + 4y2).

Question 23:
32x4 – 500x
= 4x (8x3 – 125)
= 4x [(2x)3 – (5)3]
= 4x [(2x – 5) [(2x)2 + 2x (5) + (5)2]
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2j 23.1
= 4x (2x – 5) (4x2 + 10x + 25).

Question 24:
3a7b – 81a4b4
= 3a4b (a3 – 27b3)
= 3a4b [(a)3 – (3b)3]
= 3a4b (a – 3b) [(a)2 + a (3b) + (3b)2]
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2j 24.1
= 3a4b (a – 3b) (a2 + 3ab + 9b2).

Question 25:
We know that
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2j 21.1
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2j 25.1

Question 26:
8a3 – b3 – 4ax + 2bx
= 8a3 – b3 – 2x (2a – b)
= (2a)3 – (b)3 – 2x (2a – b)
= (2a – b) [(2a)2 + 2a (b) + (b)2] – 2x (2a – b)
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2j 26.1
= (2a – b) (4a2 + 2ab + b2) – 2x (2a – b)
= (2a – b) (4a2 + 2ab + b2 – 2x).

Question 27:
8a3 – b3 – 4ax + 2bx
= 8a3 – b3 – 2x (2a – b)
= (2a)3 – (b)3 – 2x (2a – b)
= (2a – b) [(2a)2 + 2a (b) + (b)2] – 2x (2a – b)
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2j 26.1
= (2a – b) (4a2 + 2ab + b2) – 2x (2a – b)
= (2a – b) (4a2 + 2ab + b2 – 2x).

Exercise 2K

RS Aggarwal Solutions Class 9 Chapter 2 Polynomials k1

Question 1:
125a3 + b3 + 64c3 – 60abc
= (5a)3 + (b)3 + (4c)3 – 3 (5a) (b) (4c)
= (5a + b + 4c) [(5a)2 + b2 + (4c)2 – (5a) (b) – (b) (4c) – (5a) (4c)]
[∵ a3 + b3 + c3 – 3abc = (a+ b + c) (a2 + b2 + c2 – ab – bc – ca)]
= (5a + b + 4c) (25a2 + b2 + 16c2 – 5ab – 4bc – 20ac).

Question 2:
a3 + 8b3 + 64c3 – 24abc
= (a)3 + (2b)3 + (4c)3 – 3 a 2b 4c
= (a + 2b + 4c) [a2 + 4b2 + 16c2 – 2ab – 8bc – 4ca).

Question 3:
1 + b3 + 8c3 – 6bc
= 1 + (b)3 + (2c)3 – 3 (b) (2c)
= (1 + b + 2c) [1 + b2 + (2c)2 – b – b 2c – 2c]
= (1 + b + 2c) (1 + b2 + 4c2 – b – 2bc – 2c).

Question 4:
216 + 27b3 + 8c3 – 108bc
= (6)3 + (3b)3 + (2c)2 – 3 6 3b 2c
= (6 + 3b + 2c) [(6)2 + (3b)2 + (2c)2 – 6 3b – 3b 2c – 2c 6]
= (6 + 3b + 2c) (36 + 9b2 + 4c2 – 18b – 6bc – 12c).

Question 5:
27a3 – b3 + 8c3 + 18abc
= (3a)3 + (-b)3 + (2c)3 + 3(3a) (-b) (2c)
= [3a + (-b) + 2c] [(3a)2 + (-b)2 + (2c)2 – 3a (-b) – (-b) (2c) – (2c) (3a)]
= (3a – b + 2c) (9a2 + b2 + 4c2 + 3ab + 2bc – 6ca).

Question 6:
8a3 + 125b3 – 64c3 + 120abc
= (2a)3 + (5b)3 + (-4c)3 – 3 (2a) (5b) (-4c)
= (2a + 5b – 4c) [(2a)2 + (5b)2 + (-4c)2 – (2a) (5b) – (5b) (-4c) – (-4c) (2a)]
= (2a + 5b – 4c) (4a2 + 25b2 + 16c2 – 10ab + 20bc + 8ca).

Question 7:
8 – 27b3 – 343c3 – 126bc
= (2)3 + (-3b)3 + (-7c)3 – 3(2) (-3b) (-7c)
= (2 – 3b – 7c) [(2)2 + (-3b)2 + (-7c)2 – (2) (-3b) – (-3b) (-7c) – (-7c) (2)]
= (2 – 3b – 7c) (4 + 9b2 + 49c2 + 6b – 21bc + 14c).

Question 8:
125 – 8x3 – 27y3 – 90xy
= (5)3 + (-2x)3 + (-3y)3 – 3 (5) (-2x) (-3y)
= (5 – 2x – 3y) [(5)2 + (-2x)2 + (-3y)2 – (5) (-2x) – (-2x) (-3y) – (-3y) (5)]
= (5 – 2x – 3y) (25 + 4x2 + 9y2 + 10x – 6xy + 15y).

Question 9:
RS Aggarwal Solutions Class 9 Chapter 2 Polynomials 2k 9.1

Question 10:
x3 + y3 – 12xy + 64
= x3 + y3 + 64 – 12xy
= (x)3 + (y)3 + (4)3 – 3 (x) (y) (4)
= (x + y + 4) [(x)2 + (y)2 + (4)2 – x × y – y × 4 – 4 × x ]
= (x + y + 4) (x2 + y2 + 16 – xy – 4y – 4x).

Question 11:
Putting (a – b) = x, (b – c) = y and (c – a) = z, we get,
(a – b)3 + (b – c)3 + (c – a)3
= x3 + y3 + z3, where (x + y + z) = (a – b) + (b – c) + (c – a) = 0
= 3xyz [∵ (x + y + z) = 0 ⇒ (x3 + y3 + z3) = 3xyz]
= 3(a – b) (b – c) (c – a).

Question 12:
We have:
(3a – 2b) + (2b – 5c) + (5c – 3a) = 0
So, (3a – 2b)3 + (2b – 5c)3 + (5c – 3a)3
= 3(3a – 2b) (2b – 5c) (5c – 3a).

Question 13:
a3 (b – c)3 + b3 (c – a)3 + c3 (a – b)3
= [a (b – c)]3 + [b (c – a)]3 + [c (a – b)]3
Now, since, a (b – c) + b (c -a) + c (a – b)
= ab – ac + bc – ba + ca – bc = 0
So, a3 (b – c)3 + b3 (c – a)3 + c3 (a – b)3
= 3a (b – c) b (c – a) c (a – b)
= 3abc (a – b) (b – c) (c – a).

Question 14:
(5a – 7b)3 + (9c – 5a)3 + (7b – 9c)3
Since, (5a – 7b) + (9c – 5a) + (7b – 9c)
= 5a – 7b + 9c – 5a + 7b – 9c = 0
So, (5a – 7b)3 + (9c – 5a)3 + (7b – 9c)3
= 3(5a – 7b) (9c – 5a) (7b – 9c).

Question 15:
(x + y – z) (x2 + y2 + z2 – xy + yz + zx)
= [x + y + (-z)] [(x)2 + (y)2 + (-z)2 – (x) (y) – (y) (-z) – (-z) (x)]
= x3 + y3 – z3 + 3xyz.

Question 16:
(x – 2y + 3) (x2 + 4y2 + 2xy – 3x + 6y + 9)
= [x + (-2y) + 3] [(x)2 + (-2y)2 + (3) – (x) (-2y) – (-2y) (3) – (3) (x)]
= (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
= a3 + b3 + c3 – 3abc
Where, x = a, (-2y) = b and 3 = c
(x – 2y + 3) (x2 + 4y2 + 2xy – 3x + 6y + 9)
= (x)3 + (-2y)3 + (3)2 – 3 (x) (-2y) (3)
= x3 – 8y3 + 27 + 18xy.

Question 17:
(x – 2y – z) (x2 + 4y2 + z2 + 2xy + zx – 2yz)
= [x + (-2y) + (-z)] [(x)2 + (-2y)2 + (-z)2 – (x) (-2y) – (-2y) (-z) – (-z) (x)]
= (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
= a3 + b3 + c3 – 3abc
Where x = a, (-2y) = b and (-z) = c
(x – 2y – z) (x2 + 4y2 + z2 + 2xy + zx – 2yz)
= (x)3 + (-2y)3 + (-z)3 – 3 (x) (-2y) (-z)
= x3 – 8y3 – z3 – 6xyz.

Question 18:
Given, x + y + 4 = 0
We have (x3 + y3 – 12xy + 64)
= (x)3 + (y)3 + (4)3 – 3 (x) (y) (4)
= 0.
Since, we know a + b + c = 0 ⇒ (a3 + b3 + c3) = 3abc

Question 19:
Given x = 2y + 6
Or, x – 2y – 6 = 0
We have, (x3 – 8y3 – 36xy – 216)
= (x3 – 8y3 – 216 – 36xy)
= (x)3 + (-2y)3 + (-6)3 – 3 (x) (-2y) (-6)
= (x – 2y – 6) [(x)2 + (-2y)2 + (-6)2 – (x) (-2y) – (-2y) (-6) – (-6) (x)]
= (x – 2y – 6) (x2 + 4y2 + 36 + 2xy – 12y + 6x)
= 0 (x2 + 4y2 + 36 + 2xy – 12y + 6x)
= 0.

NCERT SolutionsMathsScienceRD Sharma

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