In this chapter, we provide RS Aggarwal Solutions for Class 9 Chapter 2 Polynomials Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 9 Chapter 2 Polynomials Maths pdf, free RS Aggarwal Solutions Class 9 Chapter 2 Polynomials Maths book pdf download. Now you will get step by step solution to each question.

## RS Aggarwal Solutions Class 9 Chapter 2 Polynomials

**RS Aggarwal Solutions Class 9 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 – [latex]sqrt { 5 } [/latex] is 1.

(ii) Degree of 3 – x + x^{2} – 6x^{3} is 3.

(iii) Degree of 9 is 0.

(iv) Degree of 8x^{4} – 36x + 5x^{7} is 7.

(v) Degree of x^{9} – x^{5} + 3x^{10} + 8 is 10.

(vi) Degree of 2 – 3x^{2} is 2.

**Question 3:**

(i) Coefficient of x^{3} in 2x + x^{2} – 5x^{3} + x^{4} is -5

(ii) Coefficient of x in

(iii) Coefficient of x^{2} in

(iv) Coefficient of x^{2} in 3x – 5 is 0.

**Question 4:**

(i) x^{27} – 36

(ii) y^{16}

(iii) 5x^{3} – 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.

**RS Aggarwal Solutions Class 9 Chapter 2 Polynomials** **Exercise 2B**

**Question 1:**

p(x) = 5 – 4x + 2x^{2}

(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 – y^{2} + 5y^{3}

(i) p(0) = 4 + 3(0) – 0^{2} + 5(0)^{3}

= 4 + 0 – 0 + 0 = 4

(ii) p(2) = 4 + 3(2) – 2^{2} + 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) = 4t^{2} – 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 = [latex s=2]frac { 3 }{ 2 } [/latex]

⇒ t = [latex s=2]frac { 3 }{ 2 } [/latex] is the zero of the polynomial p(t).

(iv) f(x) = 0

⇒ 3x + 1= 0

⇒ 3x = -1

⇒ x = [latex s=2]frac { -1 }{ 3 } [/latex]

⇒ x = [latex s=2]frac { -1 }{ 3 } [/latex] is the zero of the polynomial f(x).

(v) g(x) = 0

⇒ 5 – 4x = 0

⇒ -4x = -5

⇒ x = [latex s=2]frac { 5 }{ 4 } [/latex]

⇒ x = [latex s=2]frac { 5 }{ 4 } [/latex] is the zero of the polynomial g(x).

(vi) h(x) = 0

⇒ 6x – 1 = 0

⇒ 6x = 1

⇒ x = [latex s=2]frac { 1 }{ 6 } [/latex]

⇒ x = [latex s=2]frac { 1 }{ 6 } [/latex] is the zero of the polynomial h(x).

(vii) p(x) = 0

⇒ ax + b = 0

⇒ ax = -b

⇒ x = [latex s=2]frac { -b }{ a } [/latex]

⇒ x = [latex s=2]frac { -b }{ a } [/latex]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,

⇒ [latex s=2]frac { -1 }{ 2 } [/latex] is a zero of the polynomial p(y).

(iv) p(x) = 2 – 5x

Then,

⇒ [latex s=2]frac { 2 }{ 5 } [/latex] 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) = x^{2} – 3x.

Then, p(0) = 0^{2} – 3(0) = 0

p(3) = (3^{2}) – 3(3) = 9 – 9 = 0

⇒ 0 and 3 are the zeroes of the polynomial p(x).

(vii) p(x) = x^{2} + x – 6

Then, p(2) = 2^{2} + 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).

**RS Aggarwal Solutions Class 9 Chapter 2 Polynomials** **Exercise 2C**

**Question 1:**

f(x) = x^{3} – 6x^{2} + 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) = 1^{3} – 6 × 1^{2} + 9 × 1 + 3

= 1 – 6 + 9 + 3

= 13 – 6 = 7

∴ The required remainder is 7.

**Question 2:**

f(x) = (2x^{3} – 5x^{2} + 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 × 3^{3} – 5 × 3^{2} + 9 × 3 – 8

= 54 – 45 + 27 – 8

= 81 – 53 = 28

∴ The required remainder is 28.

**Question 3:**

f(x) = (3x^{4} – 6x^{2} – 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 × 2^{4} – 6 × 2^{2} – 8 × 2 + 2

= 48 – 24 – 16 + 2

= 50 – 40 = 10

∴ The required remainder is 10.

**Question 4:**

f(x) = x^{3} – 7x^{2} + 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) = 6^{3} – 7 × 6^{2} + 6 × 6 + 4

= 216 – 252 + 36 + 4

= 256 – 252 = 4

∴ The required remainder is 4.

**Question 5:**

f(x) = (x^{3} – 6x^{2} + 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) = (2x^{4} + 6x^{3} + 2x^{2} + 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) = (4x^{3} – 12x^{2} + 11x – 5)

Now, 2x – 1 = 0 ⇒ x = [latex s=2]frac { 1 }{ 2 } [/latex]

By the remainder theorem, we know that when f(x) is divided by (2x – 1) the remainder is [latex s=2]fleft( frac { 1 }{ 2 } right) [/latex]

∴ The required remainder is -2.

**Question 8:**

f(x) = (81x^{4} + 54x^{3} – 9x^{2} – 3x + 2)

Now, 3x + 2 = 0 ⇒ x = [latex s=2]frac { -2 }{ 3 } [/latex]

By the remainder theorem, we know that when f(x) is divided by (3x+ 2) the remainder is [latex s=2]fleft( frac { -2 }{ 3 } right) [/latex]

∴ The required remainder is 0.

**Question 9:**

f(x) = (x^{3} – ax^{2} + 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) = a^{3} – a a^{2} + 2 a – a

= a^{3} – a^{3} + 2a – a

= a

∴ The required remainder is a.

**Question 10:**

Let f(x) = ax^{3} + 3x^{2} – 3

and g(x) = 2x^{3} – 5x + a

∴ f(4) = a × 4^{3} + 3 × 4^{2} – 3

= 64a + 48 – 3

= 64a + 45

g(4) = 2 × 4^{3} – 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 = [latex]frac { 63 }{ 63 } [/latex] = 1

∴ The value of a is 1.

**Question 11:**

Let f(x) = (x^{4} – 2x^{3} + 3x^{2} – ax + b)

∴ From the given information,

f(1) = 1^{4} – 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 = [latex]frac { 16 }{ 2 } [/latex]

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) = x^{4} – 2x^{3} + 3x^{2} – ax + b

= x^{4} – 2x^{3} + 3x^{2} – 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.

**RS Aggarwal Solutions Class 9 Chapter 2 Polynomials** **Exercise 2D**

**Question 1:**

f(x) = (x^{3} – 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 (x^{3} – 8).

**Question 2:**

f(x) = (2x^{3} + 7x^{2} – 24x – 45)

By the Factor Theorem, (x – 3) will be a factor of f(x) if f(3) = 0.

Here, f(3) = 2 × 3^{3} + 7 × 3^{2} – 24 × 3 – 45

= 54 + 63 – 72 – 45

= 117 – 117 = 0

∴ (x – 3) is a factor of (2x^{3} + 7x^{2} – 24x – 45).

**Question 3:**

f(x) = (2x^{4} + 9x^{3} + 6x^{2} – 11x – 6)

By the Factor Theorem, (x – 1) will be a factor of f(x) if f(1) = 0.

Here, f(1) = 2 × 1^{4} + 9 × 1^{3} + 6 × 1^{2} – 11 × 1 – 6

= 2 + 9 + 6 – 11 – 6

= 17 – 17 = 0

∴ (x – 1) is factor of (2x^{4} + 9x^{3} + 6x^{2} – 11x – 6).

**Question 4:**

f(x) = (x^{4} – x^{2} – 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 (x^{4} – x^{2} – 12).

**Question 5:**

f(x) = 2x^{3} + 9x^{2} – 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 (2x^{3} + 9x^{2} – 11x – 30).

**Question 6:**

f(x) = (2x^{4} + x^{3} – 8x^{2} – x + 6)

By the Factor Theorem, (x – a) will be a factor of f(x) if f(a) = 0.

Here, 2x – 3 = 0 ⇒ x = [latex]frac { 3 }{ 2 } [/latex]

∴ (2x – 3) is a factor of (2x^{4} + x^{3} – 8x^{2} – x + 6).

**Question 7:**

f(x) = (7x^{2} – [latex]4sqrt { 2 } [/latex] x – 6 = 0)

By the Factor Theorem, (x – a) will be a factor of f(x) if f(a) = 0.

Here,

= 14 – 8 – 6

= 14 – 14 = 0

∴ (x – [latex]sqrt { 2 } [/latex]) is a factor of (7 – [latex]4sqrt { 2 } [/latex] x – 6 = 0).

**Question 8:**

f(x) = ([latex]4sqrt { 2 } [/latex]x^{2} + 5x +[latex]sqrt { 2 } [/latex] = 0)

By the Factor Theorem, (x – a) will be a factor of f(x) if f(a) = 0.

Here,

∴ (x + [latex]sqrt { 2 } [/latex]) is a factor of ([latex]4sqrt { 2 } [/latex]x^{2} + 5x +[latex]sqrt { 2 } [/latex] = 0).

**Question 9:**

f(x) = (2x^{3} + 9x^{2} + x + k)

x – 1 = 0 ⇒ x = 1

∴ f(1) = 2 × 1^{3} + 9 × 1^{2} + 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) = (2x^{3} – 3x^{2} – 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) = x^{4} – x^{3} – 11x^{2} – 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) = (2x^{3} + ax^{2} + 11x + a + 3)

2x – 1 = 0 ⇒ x = [latex s=2]frac { 1 }{ 2 } [/latex]

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 [latex s=2]fleft( frac { 1 }{ 2 } right) [/latex] ≠ 0.

Therefore, we have

∴ The value of a = -7.

**Question 13:**

Let f(x) = (x^{3} – 10x^{2} + 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) = 1^{3} – 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 geta = 23Substituting 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) = (x^{4} + ax^{3} – 7x^{2} – 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) = x^{3} – 3x^{2} – 13x + 15

Now, x^{2} + 2x – 3 = x^{2} + 3x – x – 3

= x (x + 3) – 1 (x + 3)

= (x + 3) (x – 1)

Thus, f(x) will be exactly divisible by x^{2} + 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) = 1^{3} – 3 × 1^{2} – 13 × 1 + 15

= 1 – 3 – 13 + 15

= 16 – 16 = 0

∴ f(-3) = 0 and f(1) = 0

So, x^{2} + 2x – 3 divides f(x) exactly.

**Question 16:**

Let f(x) = (x^{3} + ax^{2} + bx + 6)

Now, by remainder theorem, f(x) when divided by (x – 3) will leave a remainder as f(3).

So, f(3) = 3^{3} + a × 3^{2} + 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) = 2^{3} + a × 2^{2} + 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.

**RS Aggarwal Solutions Class 9 Chapter 2 Polynomials** **Exercise 2E**

**Question 1:**

9x^{2} + 12xy = 3x (3x + 4y)

**Question 2:**

18x^{2}y – 24xyz = 6xy (3x – 4z)

**Question 3:**

27a^{3}b^{3} – 45a^{4}b^{2} = 9a^{3}b^{2} (3b – 5a)

**Question 4:**

2a (x + y) – 3b (x + y) = (x + y) (2a – 3b)

**Question 5:**

2x (p^{2} + q^{2}) + 4y (p^{2} + q^{2})

= (2x + 4y) (p^{2} + q^{2})

= 2(x+ 2y) (p^{2} + q^{2})

**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} – 3x^{2}y (x + y)

= x (x + y) [(x + y)^{2} – 3xy]

= x (x + y) (x^{2} + y^{2} + 2xy – 3xy)

= x (x + y) (x^{2} + y^{2} – xy)

**Question 10:**

x^{3} + 2x^{2} + 5x + 10

= x^{2} (x + 2) + 5 (x + 2)

= (x^{2} + 5) (x + 2)

**Question 11:**

x^{2} + xy – 2xz – 2yz

= x (x + y) – 2z (x + y)

= (x+ y) (x – 2z)

**Question 12:**

a^{3}b – a^{2}b + 5ab – 5b

= a^{2}b (a – 1) + 5b (a – 1)

= (a – 1) (a^{2}b + 5b)

= (a – 1) b (a^{2} + 5)

= b (a – 1) (a^{2} + 5)

**Question 13:**

8 – 4a – 2a^{3} + a^{4}

= 4(2 – a) – a^{3} (2 – a)

= (2 – a) (4 – a^{3})

**Question 14:**

x^{3} – 2x^{2}y + 3xy^{2} – 6y^{3}

= x^{2} (x – 2y) + 3y^{2} (x – 2y)

= (x – 2y) (x^{2} + 3y^{2})

**Question 15:**

px + pq – 5q – 5x

= p(x + q) – 5 (q + x)

= (x + q) (p – 5)

**Question 16:**

x^{2} – 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:**

a^{3} + a – 3a^{2} – 3

= a(a^{2} + 1) – 3 (a^{2} + 1)

= (a – 3) (a^{2} + 1)

**Question 20:**

3ax – 6ay – 8by + 4bx

= 3a (x – 2y) + 4b (x – 2y)

= (x – 2y) (3a + 4b)

**Question 21:**

abx^{2} + a^{2}x + b^{2}x +ab

= ax (bx + a) + b (bx + a)

= (bx + a) (ax + b)

**Question 22:**

x^{3} – x^{2} + ax + x – a – 1

= x^{3} – x^{2} + ax – a + x – 1

= x^{2} (x – 1) + a (x – 1) + 1 (x – 1)

= (x – 1) (x^{2} + a + 1)

**Question 23:**

2x + 4y – 8xy – 1

= 2x – 1 – 8xy + 4y

= (2x – 1) – 4y (2x – 1)

= (2x – 1) (1 – 4y)

**Question 24:**

ab (x^{2} + y^{2}) – xy (a^{2} + b^{2})

= abx^{2} + aby^{2} – a^{2}xy – b^{2}xy

= abx^{2} – a^{2}xy + aby^{2} – b^{2}xy

= ax (bx – ay) + by(ay – bx)

= (bx – ay) (ax – by)

**Question 25:**

a^{2} + ab (b + 1) + b^{3}

= a^{2} + ab^{2} + ab + b^{3}

= a^{2} + ab + ab^{2} + b^{3}

= a (a + b) + b^{2} (a + b)

= (a + b) (a + b^{2})

**Question 26:**

a^{3} + ab (1 – 2a) – 2b^{2}

= a^{3} + ab – 2a^{2}b – 2b^{2}

= a (a^{2} + b) – 2b (a^{2} + b)

= (a^{2} + b) (a – 2b)

**Question 27:**

2a^{2} + bc – 2ab – ac

= 2a^{2} – 2ab – ac + bc

= 2a (a – b) – c (a – b)

= (a – b) (2a – c)

**Question 28:**

(ax + by)^{2} + (bx – ay)^{2}

= a^{2}x^{2} + b^{2}y^{2} + 2abxy + b^{2}x^{2} + a^{2}y^{2} – 2abxy

= a^{2}x^{2} + b^{2}y^{2} + b^{2}x^{2} + a^{2}y^{2}

= a^{2}x^{2} + b^{2}x^{2} + b^{2}y^{2} + a^{2}y^{2}

= x^{2} (a^{2} + b^{2}) + y^{2}(a^{2} + b^{2})

= (a^{2} + b^{2}) (x^{2} + y^{2})

**Question 29:**

a (a + b – c) – bc

= a^{2} + ab – ac – bc

= a(a + b) – c (a + b)

= (a – c) (a + b)

**Question 30:**

a(a – 2b – c) + 2bc

= a^{2} – 2ab – ac + 2bc

= a (a – 2b) – c (a – 2b)

= (a – 2b) (a – c)

**Question 31:**

a^{2}x^{2} + (ax^{2} + 1)x + a

= a^{2}x^{2} + ax^{3} + x + a

= ax^{2} (a + x) + 1 (x + a)

= (ax^{2} + 1) (a + x)

**Question 32:**

ab (x^{2} + 1) + x (a^{2} + b^{2})

= abx^{2} + ab + a^{2}x + b^{2}x

= abx^{2} + a^{2}x + ab + b^{2}x

= ax (bx + a) + b (bx + a)

= (bx + a) (ax + b)

**Question 33:**

x^{2} – (a + b) x + ab

= x^{2} – ax – bx + ab

= x (x – a) – b(x – a)

= (x – a) (x – b)

**Question 34:**

**RS Aggarwal Solutions Class 9 Chapter 2 Polynomials** **Exercise 2F**

**Question 1:**

25x^{2} – 64y^{2}

= (5x)^{2} – (8y)^{2}

= (5x + 8y) (5x – 8y)

**Question 2:**

100 – 9x^{2}

= (10)^{2} – (3x)^{2}

= (10 + 3x) (10 – 3x)

**Question 3:**

5x^{2} – 7y^{2}

**Question 4:**

(3x + 5y)^{2} – 4z^{2}

= (3x + 5y)^{2} – (2z)^{2}

= (3x + 5y + 2z) (3x + 5y – 2z)

**Question 5:**

150 – 6x^{2}

= 6 (25 – x^{2})

= 6 (5^{2} – x^{2})

= 6 (5 + x) (5 – x)

**Question 6:**

20x^{2} – 45

= 5(4x^{2} – 9)

= 5 [(2x)^{2} – (3)^{2}]

= 5 (2x + 3) (2x – 3)

**Question 7:**

3x^{3} – 48x

= 3x (x^{2} – 16)

= 3x [(x)^{2} – (4)^{2}]

= 3x (x + 4) (x – 4)

**Question 8:**

2 – 50x^{2}

= 2 (1 – 25x^{2})

= 2 [(1)^{2} – (5x)^{2}]

= 2 (1 + 5x) (1 – 5x)

**Question 9:**

27a^{2 }– 48b^{2}

= 3 (9a^{2} – 16b^{2})

= 3 [(3a)^{2} – (4b)^{2}]

= 3(3a + 4b) (3a – 4b)

**Question 10:**

x – 64x^{3}

= x (1 – 64x^{2})

= x[(1)^{2} – (8x)^{2}]

= x (1 + 8x) (1 – 8x)

**Question 11:**

8ab^{2} – 18a^{3}

= 2a (4b^{2} – 9a^{2})

= 2a [(2b)^{2} – (3a)^{2}]

= 2a (2b + 3a) (2b – 3a)

**Question 12:**

3a^{3}b – 243ab^{3}

= 3ab (a^{2} – 81 b^{2})

= 3ab [(a)^{2} – (9b)^{2}]

= 3ab (a + 9b) (a – 9b)

**Question 13:**

(a + b)^{3} – a – b

= (a + b)^{3} – (a + b)

= (a + b) [(a + b)^{2} – 1^{2}]

= (a + b) (a + b + 1) (a + b – 1)

**Question 14:**

108a^{2} – 3(b – c)^{2}

= 3 [(36a^{2} – (b -c)^{2}]

= 3 [(6a)^{2} – (b – c)^{2}]

= 3 (6a + b – c) (6a – b + c)

**Question 15:**

x^{3} – 5x^{2} – x + 5

= x^{2} (x – 5) – 1 (x – 5)

= (x – 5) (x^{2} – 1)

= (x – 5) (x + 1) (x – 1)

**Question 16:**

a^{2} + 2ab + b^{2} – 9c^{2}

= (a + b)^{2} – (3c)^{2}

= (a + b + 3c) (a + b – 3c)

**Question 17:**

9 – a^{2} + 2ab – b^{2}

= 9 – (a^{2} – 2ab + b^{2})

= 3^{2} – (a – b)^{2}

= (3 + a – b) (3 – a + b)

**Question 18:**

a^{2} – 4ac + 4c^{2} – b^{2}

= a^{2} – 4ac + 4c^{2} – b^{2}

= a^{2} – 2 a 2c + (2c)^{2} – b^{2}

= (a – 2c)^{2} – b^{2}

= (a – 2c + b) (a – 2c – b)

**Question 19:**

9a^{2} + 3a – 8b – 64b^{2}

= 9a^{2} – 64b^{2} + 3a – 8b

= (3a)^{2} – (8b)^{2} + (3a – 8b)

= (3a + 8b) (3a – 8b) + (3a – 8b)

= (3a – 8b) (3a + 8b + 1)

**Question 20:**

x^{2} – y^{2} + 6y – 9

= x^{2} – (y^{2} – 6y + 9)

= x^{2} – (y^{2} – 2 y 3 + 3^{2})

= x^{2} – (y – 3)^{2}

= [x + (y – 3)] [x – (y – 3)]

= (x + y – 3) (x – y + 3)

**Question 21:**

4x^{2} – 9y^{2} – 2x – 3y

= (2x)^{2} – (3y)^{2} – (2x + 3y)

= (2x + 3y) (2x – 3y) – (2x + 3y)

= (2x + 3y) (2x – 3y – 1)

**Question 22:**

x^{4} – 1

= (x^{2 })^{2} – 1^{2}

= (x^{2} + 1) (x^{2} – 1)

= (x^{2} + 1) (x + 1) (x – 1)

**Question 23:**

a – b – a^{2} + b^{2}

= (a – b) – (a^{2} – b^{2})

= (a – b) – (a – b) (a + b)

= (a – b) (1 – a – b)

**Question 24:**

x^{4} – 625

= (x^{2})^{2} – (25)^{2}

= (x^{2} + 25) (x^{2} – 25)

= (x^{2} + 25) (x^{2} – 5^{2})

= (x^{2} + 25) (x + 5) (x – 5)

**RS Aggarwal Solutions Class 9 Chapter 2 Polynomials** **Exercise 2G**

**Question 1:**

x^{2} + 11x + 30

= x^{2} + 6x + 5x + 30

= x (x + 6) + 5 (x + 6)

= (x + 6) (x + 5).

**Question 2:**

x^{2} + 18x + 32

= x^{2} + 16x + 2x + 32

= x (x + 16) + 2 (x + 16)

= (x + 16) (x + 2).

**Question 3:**

x^{2} + 7x – 18

= x^{2} + 9x – 2x – 18

= x (x + 9) – 2 (x + 9)

= (x + 9) (x – 2).

**Question 4:**

x^{2} + 5x – 6

= x^{2} + 6x – x – 6

= x (x + 6) – 1 (x+ 6)

= (x + 6) (x – 1).

**Question 5:**

y^{2} – 4y + 3

= y^{2} – 3y – y + 3

= y (y – 3) – 1 (y – 3)

= (y – 3) (y – 1).

**Question 6:**

x^{2} – 21x + 108

= x^{2} – 12x – 9x + 108

= x (x – 12) – 9 (x – 12)

= (x – 12) (x – 9).

**Question 7:**

x^{2} – 11x – 80

= x^{2} – 16x + 5x – 80

= x (x – 16) + 5 (x – 16)

= (x – 16) (x + 5).

**Question 8:**

x^{2} – x – 156

= x^{2} – 13x + 12x – 156

= x (x – 13) + 12 (x – 13)

= (x – 13) (x + 12).

**Question 9:**

z^{2} – 32z – 105

= z^{2} – 35z + 3z – 105

= z (z – 35) + 3 (z – 35)

= (z – 35) (z + 3)

**Question 10:**

40 + 3x – x^{2}

= 40 + 8x – 5x – x^{2}

= 8 (5 + x) -x (5 + x)

= (5 + x) (8 – x).

**Question 11:**

6 – x – x^{2}

= 6 + 2x – 3x – x^{2}

= 2(3 + x) – x (3 + x)

= (3 + x) (2 – x).

**Question 12:**

7x^{2} + 49x + 84

= 7(x^{2} + 7x + 12)

= 7 [x^{2} + 4x + 3x + 12]

= 7 [x (x + 4) + 3 (x + 4)]

= 7 (x + 4) (x + 3).

**Question 13:**

m^{2} + 17mn – 84n^{2}

= m^{2} + 21mn – 4mn – 84n^{2}

= m (m + 21n) – 4n (m + 21n)

= (m + 21n) (m – 4n).

**Question 14:**

5x^{2} + 16x + 3

= 5x^{2} + 15x + x + 3

= 5x (x + 3) + 1 (x + 3)

= (5x + 1) (x + 3).

**Question 15:**

6x^{2} + 17x + 12

= 6x^{2} + 9x + 8x + 12

= 3x (2x + 3) + 4(2x + 3)

= (2x + 3) (3x + 4).

**Question 16:**

9x^{2} + 18x + 8

= 9x^{2} + 12x + 6x + 8

= 3x (3x+ 4) +2 (3x + 4)

= (3x + 4) (3x + 2).

**Question 17:**

14x^{2} + 9x + 1

= 14x^{2} + 7x + 2x + 1

= 7x (2x + 1) + (2x + 1)

= (7x + 1) (2x + 1).

**Question 18:**

2x^{2} + 3x – 90

= 2x^{2} – 12x + 15x – 90

= 2x (x – 6) + 15 (x – 6)

= (x – 6) (2x + 15).

**Question 19:**

2x^{2} + 11x – 21

= 2x^{2} + 14x – 3x – 21

= 2x (x + 7) – 3 (x + 7)

= (x + 7) (2x – 3).

**Question 20:**

3x^{2} – 14x + 8

= 3x^{2} – 12x – 2x +8

= 3x (x – 4) – 2(x – 4)

= (x – 4) (3x – 2).

**Question 21:**

18x^{2} + 3x – 10

= 18x^{2} – 12x + 15x – 10

= 6x (3x – 2) + 5 (3x – 2)

= (6x + 5) (3x – 2).

**Question 22:**

15x^{2} + 2x – 8

= 15x^{2} – 10x + 12x – 8

= 5x (3x – 2) + 4 (3x – 2)

= (3x – 2) (5x + 4).

**Question 23:**

6x^{2} + 11x – 10

= 6x^{2} + 15x – 4x – 10

= 3x (2x + 5) – 2(2x+ 5)

= (2x + 5) (3x – 2).

**Question 24:**

30x^{2} + 7x – 15

= 30x^{2} – 18x + 25x – 15

= 6x (5x – 3) + 5 (5x- 3)

= (5x – 3) (6x + 5).

**Question 25:**

24x^{2} – 41x + 12

= 24x^{2} – 32x – 9x + 12

= 8x (3x – 4) – 3 (3x – 4)

= (3x – 4) (8x – 3).

**Question 26:**

2x^{2} – 7x – 15

= 2x^{2} – 10x + 3x – 15

= 2x (x – 5) + 3 (x – 5)

= (x – 5) (2x + 3).

**Question 27:**

6x^{2} – 5x – 21

= 6x^{2} + 9x – 14x – 21

= 3x (2x + 3) – 7 (2x + 3)

= (3x – 7) (2x + 3).

**Question 28:**

10x^{2} – 9x – 7

= 10x^{2} + 5x – 14x – 7

= 5x (2x + 1) – 7 (2x+ 1)

= (2x + 1) (5x – 7).

**Question 29:**

5x^{2} – 16x – 21

= 5x^{2} + 5x – 21x – 21

= 5x (x + 1) -21 (x + 1)

= (x + 1) (5x – 21).

**Question 30:**

2x^{2} – x – 21

= 2x^{2} + 6x – 7x – 21

= 2x (x + 3) – 7 (x + 3)

= (x + 3) (2x – 7).

**Question 31:**

15x^{2} – x – 28

= 15x^{2} + 20x – 21x – 28

= 5x (3x + 4) – 7 (3x + 4)

= (3x + 4) (5x – 7).

**Question 32:**

8a^{2} – 27ab + 9b^{2}

= 8a^{2} – 24ab – 3ab + 9b^{2}

= 8a (a – 3b) – 3b (a – 3b)

= (a – 3b) (8a – 3b).

**Question 33:**

5x^{2} + 33xy – 14y^{2}

= 5x^{2} + 35xy – 2xy – 14y^{2}

= 5x (x + 7y) – 2y (x + 7y)

= (x + 7y) (5x – 2y).

**Question 34:**

3x^{3} – x^{2} – 10x

= x (3x^{2} – x – 10)

= x [3x^{2} – 6x + 5x – 10]

= x [3x (x – 2) + 5 (x – 2)]

= x (x – 2) (3x + 5).

**Question 35:**

**Question 36:**

**Question 37:**

**Question 38:**

**Question 39:**

**Question 40:**

**Question 41:**

**Question 42:**

**Question 43:**

**Question 44:**

**Question 45:**

Let x + y = z

Then, 2 (x + y)^{2} – 9 (x + y) – 5

Now, replacing z by (x + y), we get

**Question 46:**

Let 2a – b = c

Then, 9 (2a – b)^{2} – 4 (2a – b) -13

Now, replacing c by (2a – b) , we get

9 (2a – b)^{2} – 4 (2a – b) – 13

**Question 47:**

Let x – 2y = z

Then, 7 (x – 2y)^{2} – 25 (x – 2y) + 12

Now replace z by (x – 2y), we get

7 (x – 2y)^{2} – 25 (x – 2y) + 12

**Question 48:**

Let x^{2} = y

Then, 4x^{4} + 7x^{2} – 2

Now replacing y by x^{2}, we get

**RS Aggarwal Solutions Class 9 Chapter 2 Polynomials** **Exercise 2H**

**Question 1:**

We know:

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca

(i) (a + 2b + 5c)^{2}

= (a)^{2} + (2b)^{2} + (5c)^{2} + 2(a) (2b) + 2 (2b) (5c) + 2(5c) (a)

= a^{2} + 4b^{2} + 25c^{2} + 4ab + 20bc + 10ac

(ii) (2a – b + c)^{2}

= (2a)^{2} + (-b)^{2} + (c)^{2} + 2 (2a) (-b) + 2(-b) (c) + 2 (c) (2a)

= 4a^{2} + b^{2} + c^{2} – 4ab – 2bc + 4ac.

(iii) (a – 2b – 3c)^{2}

= (a)^{2} + (-2b)^{2} + (-3c)^{2} + 2(a) (-2b) + 2(-2b) (-3c) + 2 (-3c) (a)

= a^{2} + 4b^{2} + 9c^{2} – 4ab + 12bc – 6ac.

**Question 2:**

We know:

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca

(i) (2a – 5b – 7c)^{2}

= (2a)^{2} + (-5b)^{2} + (-7c)^{2} + 2 (2a) (-5b) + 2 (-5b) (-7c) + 2 (-7c) (2a)

= 4a^{2} + 25b^{2} + 49c^{2} – 20ab + 70bc – 28ac.

(ii) (-3a + 4b – 5c)^{2}

= (-3a)^{2} + (4b)^{2} + (-5c)^{2} + 2 (-3a) (4b) + 2 (4b) (-5c) + 2 (-5c) (-3a)

= 9a^{2} + 16b^{2} + 25c^{2} – 24ab – 40bc + 30ac.

**Question 3:**

4x^{2} + 9y^{2} + 16z^{2} + 12xy – 24yz – 16xz

= (2x)^{2} + (3y)^{2} + (-4z)^{2} + 2 (2x) (3y) + 2(3y) (-4z) + 2 (-4z) (2x)

= (2x + 3y – 4z)^{2}

**Question 4:**

9x^{2} + 16y^{2} + 4z^{2} – 24xy + 16yz – 12xz

= (-3x)^{2} + (4y)^{2} + (2z)^{2} + 2 (-3x) (4y) + 2 (4y) (2z) + 2 (2z) (-3x)

= (-3x + 4y + 2z)^{2}.

**Question 5:**

25x^{2} + 4y^{2} + 9z^{2} – 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}

= (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.

**RS Aggarwal Solutions Class 9 Chapter 2 Polynomials** **Exercise 2I**

**Question 1:**

(i) (3x + 2)^{3}

= (3x)^{3} + (2)^{3} + 3 × 3x × 2 (3x + 2)

= 27x^{3} + 8 + 18x (3x + 2)

= 27x^{3} + 8 + 54x^{2} + 36x.

(ii) (3a – 2b)^{3}

= (3a)^{3} – (2b)^{3} – 3 × 3a × 2b (3a – 2b)

= 27a^{3} – 8b^{3} – 18ab (3a – 2b)

= 27 a^{3} – 8b^{3} – 54a^{2}b + 36ab^{2}.

**Question 2:**

**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.

**RS Aggarwal Solutions Class 9 Chapter 2 Polynomials** **Exercise 2J**

**Question 1:**

x^{3} + 27

= x^{3} + 3^{3}

= (x + 3) (x^{2} – 3x + 9)

**Question 2:**

8x^{3} + 27y^{3}

= (2x)^{3} + (3y)^{3}

= (2x+ 3y) [(2x)^{2} – (2x) (3y) + (3y)^{2}]

= (2x + 3y) (4x^{2} – 6xy + 9y^{2}).

**Question 3:**

343 + 125 b^{3}

= (7)^{3} + (5b)^{3}

= (7 + 5b) [(7)^{2} – (7) (5b) + (5b)^{2}]

= (7 + 5b) (49 – 35b + 25b^{2})

**Question 4:**

1 + 64x^{3}

= (1)^{3} + (4x)^{3}

= (1 + 4x) [(1)^{2} – 1 (4x) + (4x)^{2}]

= (1 + 4x) (1 – 4x + 16x^{2}).

**Question 5:**

125a^{3} + [latex]frac { 1 }{ 8 } [/latex]

We know that

Let us rewrite

**Question 6:**

216x^{3} + [latex]frac { 1 }{ 125 } [/latex]

We know that

Let us rewrite

**Question 7:**

16x^{ 4} + 54x

= 2x (8x^{ 3} + 27)

= 2x [(2x)^{3} + (3)^{3}]

= 2x (2x + 3) [(2x)^{2} – 2x(3) + 3^{2}]

=2x(2x+3)(4x^{2} -6x +9)

**Question 8:**

7a^{3} + 56b^{3}

= 7(a^{3} + 8b^{3})

= 7 [(a)^{3} + (2b)^{3}]

= 7 (a + 2b) [a^{2} – a 2b + (2b)^{2}]

= 7 (a + 2b) (a^{2} – 2ab + 4b^{2}).

**Question 9:**

x^{5} + x^{2}

= x^{2}(x^{3} + 1)

= x^{2} (x + 1) [(x)^{2} – x (1) + (1)^{2}]

= x^{2} (x + 1) (x^{2} – x + 1).

**Question 10:**

a^{3} + 0.008

= (a)^{3} + (0.2)^{3}

= (a + 0.2) [(a)^{2} – a(0.2) + (0.2)^{2}]

= (a + 0.2) (a^{2} – 0.2a + 0.04).

**Question 11:**

x^{6} + y^{6}

= (x^{2})^{3} + (y^{2})^{3}

= (x^{2} + y^{2}) [(x^{2})^{2} – x^{2} (y^{2})+ (y^{2})^{2}]

= (x^{2} + y^{2}) (x^{4} – x^{2}y^{2} + y^{4}).

**Question 12:**

2a^{3} + 16b^{3} – 5a – 10b

= 2 (a^{3} + 8b^{3}) – 5 (a + 2b)

= 2 [(a)^{3} + (2b)^{3}] – 5 (a + 2b)

= 2 (a + 2b) [(a)^{2} – a (2b) + (2b)^{2} ] – 5 (a + 2b)

= (a + 2b) [2(a^{2} – 2ab + 4b^{2}) – 5]

**Question 13:**

x^{3} – 512

= (x)^{3} – (8)^{3}

= (x – 8) [(x)^{2} + x (8) + (8)^{2}]

= (x – 8) (x^{2} + 8x + 64).

**Question 14:**

64x^{3} – 343

= (4x)^{3} – (7)^{3}

= (4x – 7) [(4x)^{2} + 4x (7) + (7)^{2}]

= (4x – 7) (16x^{2} + 28x + 49).

**Question 15:**

1 – 27x^{3}

= (1)^{3} – (3x)^{3}

= (1 – 3x) [(1)^{2} + 1 (3x) + (3x)^{2}]

= (1 – 3x) (1 + 3x + 9x^{2}).

**Question 16:**

1 – 27x^{3}

= (1)^{3} – (3x)^{3}

= (1 – 3x) [(1)^{2} + 1 (3x) + (3x)^{2}]

= (1 – 3x) (1 + 3x + 9x^{2}).

**Question 17:**

We know that

Let us rewrite

**Question 18:**

a^{3} – 0.064

= (a)^{3} – (0.4)^{3}

= (a – 0.4) [(a)^{2} + a (0.4) + (0.4)^{2}]

= (a – 0.4) (a^{2} + 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}]

= (a + b – 2) [a^{2} + b^{2} + 2ab + 2 (a + b) + 4].

**Question 20:**

x^{6} – 729

= (x^{2})^{3} – (9)^{3}

= (x^{2} – 9) [(x^{2})^{2} + x^{2} 9 + (9)^{2}]

= (x^{2} – 9) (x^{4} + 9x^{2} + 81)

= (x + 3) (x – 3) [(x^{2} + 9)^{2} – (3x)^{2}]

= (x + 3) (x – 3) (x^{2} + 3x + 9) (x^{2} – 3x + 9).

**Question 21:**

We know that,

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) [ a^{2} + b^{2} + 2ab + a^{2} – b^{2} + a^{2} + b^{2} – 2ab]

= 2b (3a^{2} + b^{2}).

**Question 22:**

x – 8xy^{3}

= x (1 – 8y^{3})

= x [(1)^{3} – (2y)^{3}]

= x (1 – 2y) [(1)^{2} + 1 (2y) + (2y)^{2}]

= x (1 – 2y) (1 + 2y + 4y^{2}).

**Question 23:**

32x^{4} – 500x

= 4x (8x^{3} – 125)

= 4x [(2x)^{3} – (5)^{3}]

= 4x [(2x – 5) [(2x)^{2} + 2x (5) + (5)^{2}]

= 4x (2x – 5) (4x^{2} + 10x + 25).

**Question 24:**

3a^{7}b – 81a^{4}b^{4}

= 3a^{4}b (a^{3} – 27b^{3})

= 3a^{4}b [(a)^{3} – (3b)^{3}]

= 3a^{4}b (a – 3b) [(a)^{2} + a (3b) + (3b)^{2}]

= 3a^{4}b (a – 3b) (a^{2} + 3ab + 9b^{2}).

**Question 25:**

We know that

**Question 26:**

8a^{3} – b^{3} – 4ax + 2bx

= 8a^{3} – b^{3} – 2x (2a – b)

= (2a)^{3} – (b)^{3} – 2x (2a – b)

= (2a – b) [(2a)^{2} + 2a (b) + (b)^{2}] – 2x (2a – b)

= (2a – b) (4a^{2} + 2ab + b^{2}) – 2x (2a – b)

= (2a – b) (4a^{2} + 2ab + b^{2} – 2x).

**Question 27:**

8a^{3} – b^{3} – 4ax + 2bx

= 8a^{3} – b^{3} – 2x (2a – b)

= (2a)^{3} – (b)^{3} – 2x (2a – b)

= (2a – b) [(2a)^{2} + 2a (b) + (b)^{2}] – 2x (2a – b)

= (2a – b) (4a^{2} + 2ab + b^{2}) – 2x (2a – b)

= (2a – b) (4a^{2} + 2ab + b^{2} – 2x).

**RS Aggarwal Solutions Class 9 Chapter 2 Polynomials** **Exercise 2K**

**Question 1:**

125a^{3} + b^{3} + 64c^{3} – 60abc

= (5a)^{3} + (b)^{3} + (4c)^{3} – 3 (5a) (b) (4c)

= (5a + b + 4c) [(5a)^{2} + b^{2} + (4c)^{2} – (5a) (b) – (b) (4c) – (5a) (4c)]

[∵ a^{3} + b^{3} + c^{3} – 3abc = (a+ b + c) (a^{2} + b^{2} + c^{2} – ab – bc – ca)]

= (5a + b + 4c) (25a^{2} + b^{2} + 16c^{2} – 5ab – 4bc – 20ac).

**Question 2:**

a^{3} + 8b^{3} + 64c^{3} – 24abc

= (a)^{3} + (2b)^{3} + (4c)^{3} – 3 a 2b 4c

= (a + 2b + 4c) [a^{2} + 4b^{2} + 16c^{2 }– 2ab – 8bc – 4ca).

**Question 3:**

1 + b^{3} + 8c^{3} – 6bc

= 1 + (b)^{3} + (2c)^{3} – 3 (b) (2c)

= (1 + b + 2c) [1 + b^{2} + (2c)^{2} – b – b 2c – 2c]

= (1 + b + 2c) (1 + b^{2} + 4c^{2} – b – 2bc – 2c).

**Question 4:**

216 + 27b^{3} + 8c^{3} – 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 + 9b^{2} + 4c^{2} – 18b – 6bc – 12c).

**Question 5:**

27a^{3} – b^{3} + 8c^{3} + 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) (9a^{2} + b^{2} + 4c^{2} + 3ab + 2bc – 6ca).

**Question 6:**

8a^{3} + 125b^{3} – 64c^{3} + 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) (4a^{2} + 25b^{2} + 16c^{2} – 10ab + 20bc + 8ca).

**Question 7:**

8 – 27b^{3} – 343c^{3} – 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 + 9b^{2} + 49c^{2} + 6b – 21bc + 14c).

**Question 8:**

125 – 8x^{3} – 27y^{3} – 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 + 4x^{2} + 9y^{2} + 10x – 6xy + 15y).

**Question 9:**

**Question 10:**

x^{3} + y^{3} – 12xy + 64

= x^{3} + y^{3} + 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) (x^{2} + y^{2} + 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}

= x^{3} + y^{3} + z^{3}, where (x + y + z) = (a – b) + (b – c) + (c – a) = 0

= 3xyz [∵ (x + y + z) = 0 ⇒ (x^{3} + y^{3} + z^{3}) = 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:**

a^{3} (b – c)^{3} + b^{3} (c – a)^{3} + c^{3} (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, a^{3} (b – c)^{3} + b^{3} (c – a)^{3} + c^{3} (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) (x^{2} + y^{2} + z^{2} – xy + yz + zx)

= [x + y + (-z)] [(x)^{2} + (y)^{2} + (-z)^{2} – (x) (y) – (y) (-z) – (-z) (x)]

= x^{3} + y^{3} – z^{3} + 3xyz.

**Question 16:**

(x – 2y + 3) (x^{2} + 4y^{2} + 2xy – 3x + 6y + 9)

= [x + (-2y) + 3] [(x)^{2} + (-2y)^{2} + (3) – (x) (-2y) – (-2y) (3) – (3) (x)]

= (a + b + c) (a^{2} + b^{2} + c^{2} – ab – bc – ca)

= a^{3} + b^{3} + c^{3} – 3abc

Where, x = a, (-2y) = b and 3 = c

(x – 2y + 3) (x^{2} + 4y^{2} + 2xy – 3x + 6y + 9)

= (x)^{3} + (-2y)^{3} + (3)^{2} – 3 (x) (-2y) (3)

= x^{3} – 8y^{3} + 27 + 18xy.

**Question 17:**

(x – 2y – z) (x^{2} + 4y^{2} + z^{2} + 2xy + zx – 2yz)

= [x + (-2y) + (-z)] [(x)^{2} + (-2y)^{2} + (-z)^{2} – (x) (-2y) – (-2y) (-z) – (-z) (x)]

= (a + b + c) (a^{2} + b^{2} + c^{2} – ab – bc – ca)

= a^{3} + b^{3} + c^{3} – 3abc

Where x = a, (-2y) = b and (-z) = c

(x – 2y – z) (x^{2} + 4y^{2} + z^{2} + 2xy + zx – 2yz)

= (x)^{3} + (-2y)^{3} + (-z)^{3} – 3 (x) (-2y) (-z)

= x^{3} – 8y^{3} – z^{3} – 6xyz.

**Question 18:**

Given, x + y + 4 = 0

We have (x^{3} + y^{3} – 12xy + 64)

= (x)^{3} + (y)^{3} + (4)^{3} – 3 (x) (y) (4)

= 0.

Since, we know a + b + c = 0 ⇒ (a^{3} + b^{3} + c^{3}) = 3abc

**Question 19:**

Given x = 2y + 6

Or, x – 2y – 6 = 0

We have, (x^{3} – 8y^{3} – 36xy – 216)

= (x^{3} – 8y^{3} – 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) (x^{2} + 4y^{2} + 36 + 2xy – 12y + 6x)

= 0 (x^{2} + 4y^{2} + 36 + 2xy – 12y + 6x)

= 0.

**All Chapter RS Aggarwal Solutions For Class 9 Maths**

—————————————————————————–**All Subject NCERT Exemplar Problems Solutions For Class 9**

**All Subject NCERT Solutions For Class 9**

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rsaggarwalsolutions.in to your friends.