### RS Aggarwal Solutions Class 9 Chapter 2 Polynomials

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## 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.

[embed]https://www.youtube.com/watch?v=ffLLmV4mZwU[/embed]

**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.

**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).

**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.

**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 _{} 1^{2} + a _{} 1 + b = 0

⇒ 1 – 10 + a + b = 0

⇒ a + b = 9 ….(i)

And f(2) = 2^{3} – 10 _{} 2^{2} + 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) = (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.

**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:**

**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)

**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

**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.

**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.

**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).

**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.

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**Complete RS Aggarwal Solutions Class 9**

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