Polynomials RS Aggarwal Solution for Class 9th Maths

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RS Aggarwal Class 9 Solutions Chapter 2 Polynomials

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-class-9-solutions-polynomials-2a-q3-1
(iii) Coefficient of x2 in rs-aggarwal-class-9-solutions-polynomials-2a-q3-2
(iv) Coefficient of x2 in 3x – 5 is 0.

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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-class-9-solutions-polynomials-2b-q5-1
\frac { -1 }{ 2 }   is a zero of the polynomial p(y).

(iv) p(x) = 2 – 5x
Then, rs-aggarwal-class-9-solutions-polynomials-2b-q5-2
\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

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-class-9-solutions-polynomials-2c-q7
∴ 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-class-9-solutions-polynomials-2c-q8
∴ 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

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-class-9-solutions-polynomials-2d-q6
∴ (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-class-9-solutions-polynomials-2d-q7
= 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-class-9-solutions-polynomials-2d-q8
∴ (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-class-9-solutions-polynomials-2d-q12
∴ 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

algebraic-identities

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-class-9-solutions-polynomials-2e-q34

Exercise 2F

Question 1:
25x2 – 64y2
= (5x)2 – (8y)2
= (5x + 8y) (5x – 8y)
rs-aggarwal-class-9-solutions-polynomials-2f-q1

Question 2:
100 – 9x2
= (10)2 – (3x)2
= (10 + 3x) (10 – 3x)
rs-aggarwal-class-9-solutions-polynomials-2f-q1

Question 3:
5x2 – 7y2
rs-aggarwal-class-9-solutions-polynomials-2f-q3

Question 4:
(3x + 5y)2 – 4z2
= (3x + 5y)2 – (2z)2
= (3x + 5y + 2z) (3x + 5y – 2z)
rs-aggarwal-class-9-solutions-polynomials-2f-q1

Question 5:
150 – 6x2
= 6 (25 – x2)
= 6 (52 – x2)
= 6 (5 + x) (5 – x)
rs-aggarwal-class-9-solutions-polynomials-2f-q1

Question 6:
20x2 – 45
= 5(4x2 – 9)
= 5 [(2x)2 – (3)2]
= 5 (2x + 3) (2x – 3)
rs-aggarwal-class-9-solutions-polynomials-2f-q1

Question 7:
3x3 – 48x
= 3x (x2 – 16)
= 3x [(x)2 – (4)2]
= 3x (x + 4) (x – 4)
rs-aggarwal-class-9-solutions-polynomials-2f-q1

Question 8:
2 – 50x2
= 2 (1 – 25x2)
= 2 [(1)2 – (5x)2]
= 2 (1 + 5x) (1 – 5x)
rs-aggarwal-class-9-solutions-polynomials-2f-q1

Question 9:
27a2 – 48b2
= 3 (9a2 – 16b2)
= 3 [(3a)2 – (4b)2]
= 3(3a + 4b) (3a – 4b)
rs-aggarwal-class-9-solutions-polynomials-2f-q1

Question 10:
x – 64x3
= x (1 – 64x2)
= x[(1)2 – (8x)2]
= x (1 + 8x) (1 – 8x)
rs-aggarwal-class-9-solutions-polynomials-2f-q1

Question 11:
8ab2 – 18a3
= 2a (4b2 – 9a2)
= 2a [(2b)2 – (3a)2]
= 2a (2b + 3a) (2b – 3a)
rs-aggarwal-class-9-solutions-polynomials-2f-q1

Question 12:
3a3b – 243ab3
= 3ab (a2 – 81 b2)
= 3ab [(a)2 – (9b)2]
= 3ab (a + 9b) (a – 9b)
rs-aggarwal-class-9-solutions-polynomials-2f-q1

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-class-9-solutions-polynomials-2f-q1

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-class-9-solutions-polynomials-2f-q1

Question 15:
x3 – 5x2 – x + 5
= x2 (x – 5) – 1 (x – 5)
= (x – 5) (x2 – 1)
= (x – 5) (x + 1) (x – 1)
rs-aggarwal-class-9-solutions-polynomials-2f-q1

Question 16:
a2 + 2ab + b2 – 9c2
= (a + b)2 – (3c)2
= (a + b + 3c) (a + b – 3c)
rs-aggarwal-class-9-solutions-polynomials-2f-q1

Question 17:
9 – a2 + 2ab – b2
= 9 – (a2 – 2ab + b2)
= 32 – (a – b)2
= (3 + a – b) (3 – a + b)
rs-aggarwal-class-9-solutions-polynomials-2f-q1

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-class-9-solutions-polynomials-2f-q1

Question 19:
9a2 + 3a – 8b – 64b2
= 9a2 – 64b2 + 3a – 8b
= (3a)2 – (8b)2 + (3a – 8b)
= (3a + 8b) (3a – 8b) + (3a – 8b)
rs-aggarwal-class-9-solutions-polynomials-2f-q1
= (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-class-9-solutions-polynomials-2f-q1
= (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-class-9-solutions-polynomials-2f-q1
= (2x + 3y) (2x – 3y – 1)

Question 22:
x4 – 1
= (x2 )2 – 12
= (x2 + 1) (x2 – 1) rs-aggarwal-class-9-solutions-polynomials-2f-q1
= (x2 + 1) (x + 1) (x – 1)
rs-aggarwal-class-9-solutions-polynomials-2f-q1

Question 23:
a – b – a2 + b2
= (a – b) – (a2 – b2)
= (a – b) – (a – b) (a + b)
rs-aggarwal-class-9-solutions-polynomials-2f-q1
= (a – b) (1 – a – b)

Question 24:
x4 – 625
= (x2)2 – (25)2
= (x2 + 25) (x2 – 25)
rs-aggarwal-class-9-solutions-polynomials-2f-q1
= (x2 + 25) (x2 – 52)
= (x2 + 25) (x + 5) (x – 5)
rs-aggarwal-class-9-solutions-polynomials-2f-q1

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-class-9-solutions-polynomials-2g-q35

Question 36:
rs-aggarwal-class-9-solutions-polynomials-2g-q36

Question 37:
rs-aggarwal-class-9-solutions-polynomials-2g-q37

Question 38:
rs-aggarwal-class-9-solutions-polynomials-2g-q38

Question 39:
rs-aggarwal-class-9-solutions-polynomials-2g-q39

Question 40:
rs-aggarwal-class-9-solutions-polynomials-2g-q40

Question 41:
rs-aggarwal-class-9-solutions-polynomials-2g-q41

Question 42:
rs-aggarwal-class-9-solutions-polynomials-2g-q42

Question 43:
rs-aggarwal-class-9-solutions-polynomials-2g-q43

Question 44:
rs-aggarwal-class-9-solutions-polynomials-2g-q44

Question 45:
Let x + y = z
Then, 2 (x + y)2 – 9 (x + y) – 5
rs-aggarwal-class-9-solutions-polynomials-2g-q45-1
Now, replacing z by (x + y), we get
rs-aggarwal-class-9-solutions-polynomials-2g-q45-2

Question 46:
Let 2a – b = c
Then, 9 (2a – b)2 – 4 (2a – b) -13
rs-aggarwal-class-9-solutions-polynomials-2g-q46-1
Now, replacing c by (2a – b) , we get
9 (2a – b)2 – 4 (2a – b) – 13
rs-aggarwal-class-9-solutions-polynomials-2g-q46-2

Question 47:
Let x – 2y = z
Then, 7 (x – 2y)2 – 25 (x – 2y) + 12
rs-aggarwal-class-9-solutions-polynomials-2g-q47-1
Now replace z by (x – 2y), we get
7 (x – 2y)2 – 25 (x – 2y) + 12
rs-aggarwal-class-9-solutions-polynomials-2g-q47-2

Question 48:
Let x2 = y
Then, 4x4 + 7x2 – 2
rs-aggarwal-class-9-solutions-polynomials-2g-q48-1
Now replacing y by x2, we get
rs-aggarwal-class-9-solutions-polynomials-2g-q48-2

Exercise 2H

algebraic-identities

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-class-9-solutions-polynomials-2h-q2

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-class-9-solutions-polynomials-2h-q6
= (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

algebraic-identities

Question 1:
(i) (3x + 2)3
= (3x)3 + (2)3 + 3 × 3x × 2 (3x + 2)
rs-aggarwal-class-9-solutions-polynomials-2i-q1-1
= 27x3 + 8 + 18x (3x + 2)
= 27x3 + 8 + 54x2 + 36x.
(ii) (3a – 2b)3
= (3a)3 – (2b)3 – 3 × 3a × 2b (3a – 2b)
rs-aggarwal-class-9-solutions-polynomials-2i-q1-2
= 27a3 – 8b3 – 18ab (3a – 2b)
= 27 a3 – 8b3 – 54a2b + 36ab2.
rs-aggarwal-class-9-solutions-polynomials-2i-q1-3

Question 2:
rs-aggarwal-class-9-solutions-polynomials-2i-q2

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

algebraic-identities

Question 1:
x3 + 27
= x3 + 33
= (x + 3) (x2 – 3x + 9)
rs-aggarwal-class-9-solutions-polynomials-2j-q1

Question 2:
8x3 + 27y3
= (2x)3 + (3y)3
= (2x+ 3y) [(2x)2 – (2x) (3y) + (3y)2]
rs-aggarwal-class-9-solutions-polynomials-2j-q1
= (2x + 3y) (4x2 – 6xy + 9y2).

Question 3:
343 + 125 b3
= (7)3 + (5b)3
= (7 + 5b) [(7)2 – (7) (5b) + (5b)2]
rs-aggarwal-class-9-solutions-polynomials-2j-q1
= (7 + 5b) (49 – 35b + 25b2)

Question 4:
1 + 64x3
= (1)3 + (4x)3
= (1 + 4x) [(1)2 – 1 (4x) + (4x)2]
rs-aggarwal-class-9-solutions-polynomials-2j-q1
= (1 + 4x) (1 – 4x + 16x2).

Question 5:
125a3 + \frac { 1 }{ 8 }
We know that
rs-aggarwal-class-9-solutions-polynomials-2j-q5-1
Let us rewrite
rs-aggarwal-class-9-solutions-polynomials-2j-q5

Question 6:
216x3 + \frac { 1 }{ 125 }
We know that
rs-aggarwal-class-9-solutions-polynomials-2j-q5-1
Let us rewrite
rs-aggarwal-class-9-solutions-polynomials-2j-q6

Question 7:
16x 4 + 54x
= 2x (8x 3 + 27)
= 2x [(2x)3 + (3)3]
= 2x (2x + 3) [(2x)2 – 2x(3) + 32]
rs-aggarwal-class-9-solutions-polynomials-2j-q1
=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-class-9-solutions-polynomials-2j-q1
= 7 (a + 2b) (a2 – 2ab + 4b2).

Question 9:
x5 + x2
= x2(x3 + 1)
= x2 (x + 1) [(x)2 – x (1) + (1)2]
rs-aggarwal-class-9-solutions-polynomials-2j-q1
= 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-class-9-solutions-polynomials-2j-q1
= (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-class-9-solutions-polynomials-2j-q1
= (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-class-9-solutions-polynomials-2j-q1
= (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-class-9-solutions-polynomials-2j-q2
= (x – 8) (x2 + 8x + 64).

Question 14:
64x3 – 343
= (4x)3 – (7)3
= (4x – 7) [(4x)2 + 4x (7) + (7)2]
rs-aggarwal-class-9-solutions-polynomials-2j-q2
= (4x – 7) (16x2 + 28x + 49).

Question 15:
1 – 27x3
= (1)3 – (3x)3
= (1 – 3x) [(1)2 + 1 (3x) + (3x)2]
rs-aggarwal-class-9-solutions-polynomials-2j-q2
= (1 – 3x) (1 + 3x + 9x2).

Question 16:
1 – 27x3
= (1)3 – (3x)3
= (1 – 3x) [(1)2 + 1 (3x) + (3x)2]
rs-aggarwal-class-9-solutions-polynomials-2j-q2
= (1 – 3x) (1 + 3x + 9x2).

Question 17:
We know that
rs-aggarwal-class-9-solutions-polynomials-2j-q17
Let us rewrite
rs-aggarwal-class-9-solutions-polynomials-2j-q17-1

Question 18:
a3 – 0.064
= (a)3 – (0.4)3
= (a – 0.4) [(a)2 + a (0.4) + (0.4)2]
rs-aggarwal-class-9-solutions-polynomials-2j-q2
= (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-class-9-solutions-polynomials-2j-q2
= (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-class-9-solutions-polynomials-2j-q2
= (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-class-9-solutions-polynomials-2j-q17
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-class-9-solutions-polynomials-2j-q2
= 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-class-9-solutions-polynomials-2j-q2
= 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-class-9-solutions-polynomials-2j-q2
= 3a4b (a – 3b) (a2 + 3ab + 9b2).

Question 25:
We know that
rs-aggarwal-class-9-solutions-polynomials-2j-q17
rs-aggarwal-class-9-solutions-polynomials-2j-q25

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-class-9-solutions-polynomials-2j-q2
= (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-class-9-solutions-polynomials-2j-q2
= (2a – b) (4a2 + 2ab + b2) – 2x (2a – b)
= (2a – b) (4a2 + 2ab + b2 – 2x).

Exercise 2K

algebraic-identities

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-class-9-solutions-polynomials-2k-q9

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.

 

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