Unit 4: Factorization - Solved Exercise 4.2 - Mathematics For Class IX (Science Group)

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Unit 4: Factorization
Solved Exercise 4.2

1. Factorize the following:
(i) a⁴ + a²x² + x⁴
Solution:
⇒ a⁴ + a²x² + x⁴
⇒ (a⁴ + x⁴) + a²x² (Rearrange the terms)
By adding & subtracting 2a²x², we get:
⇒ (a⁴ + 2a²x² + x⁴) - 2a²x² + a²x²
By using formula: [∵ a² + 2ab + b² = (a + b)²] a perfect square
⇒ {(a²)² + 2(a²)(x²) + (x²)²} - a²x²
⇒ (a² + x²)² - a²x²
Now By using formula [∵ a² - b² = (a - b)(a + b)] difference of two squares
⇒ (a² + x²)² - (ax)²
⇒ {(a² + x²) + (ax)}{(a² + x²) - (ax)}
⇒ (a² + x² + ax)(a² + x² - ax)
⇒ (a² + ax + x²)(a² - ax + x²) Ans

(ii) b⁴ + b² + 1
Solution:
⇒ b⁴ + b² + 1
⇒ (b⁴ + 1) + b² (Rearrange the terms)
By adding & subtracting 2b², we get:
⇒ (b⁴ + 2b² + 1) - 2b² + b²
By using formula: [∵ a² + 2ab + b² = (a + b)²] a perfect square
⇒ [(b²)² + 2(b²)(1) + (1)²] - b²
⇒ (b² + 1)² - b²
Now By using formula: [∵ a² - b² = (a - b)(a + b)] difference of two squares
⇒ (a² + 1)² - (b)²
⇒ {(a² + 1) + b}{(a² + 1) - b}
⇒ (a² + 1+ b)(a²& + 1 - b)
⇒ (a² + b + 1)(a² - b + 1) Ans

(iii) a⁸ + a⁴x⁴ + x⁸
Solution:
⇒ a⁸ + a⁴x⁴ + x⁸
⇒ (a⁸ + x⁸) + a⁴x⁴ (Rearrange the terms)
By adding & subtracting 2a⁴x⁴, we get:
⇒ (a⁸ + 2a⁴x⁴ + x⁸) - 2a⁴x⁴ + a⁴x⁴
By using formula: [∵ a² + 2ab + b² = (a + b)²] a perfect square
⇒ {(a⁴)² + 2(a⁴)(x⁴) + (x⁴)²} - a⁴x⁴
⇒ (a⁴ + x⁴)² - a⁴x⁴
Now By using formula [∵ a² - b² = (a - b)(a + b)] difference of two squares
⇒ (a⁴ + x⁴)² - (a²x²)²
⇒ {(a⁴ + x⁴) + (a²x²)}{(a⁴ + x⁴) - (a²x²)}
⇒ (a⁴ + x⁴ + a²x²)(a⁴ + x⁴ - a²x²)
⇒ {(a⁴ + x⁴) + a²x²}(a⁴ - a²x² + x⁴) (Rearrange the terms)
By adding & subtracting 2a²x² in first expression, we get:
⇒ {(a⁴ + 2a²x² + x⁴) - 2a²x² + a²x²}(a⁴ - a²x² + x⁴)
By using formula: [∵ a² + 2ab + b² = (a + b)²] a perfect square
⇒ [{(a²)² + 2(a²)(x²) + (x²)²} - a²x²](a⁴ - a²x² + x⁴)
⇒ {(a² + x²)² - a²x²}(a⁴ - a²x² + x⁴)
Now By using formula [∵ a² - b² = (a - b)(a + b)] difference of two squares
⇒ {(a² + x²)² - (ax)²}(a⁴ - a²x² + x⁴)
⇒ [{(a² + x²) + (ax)}{(a² + x²) - (ax)}](a⁴ - a²x² + x⁴)
⇒ {(a² + x² + ax)(a² + x² - ax)}(a⁴ - a²x² + x⁴)
⇒ (a² + ax + x²)(a² - ax + x²)(a⁴ - a²x² + x⁴) Ans

(iv) z⁸ + z⁴ + 1
Solution:
⇒ z⁸ + z⁴ + 1
⇒ (z⁸ + 1) + z⁴ (Rearrange the terms)
By adding & subtracting 2z⁴, we get:
⇒ (z⁸ + 2z⁴ + 1) - 2z⁴ + z⁴
By using formula: [∵ a² + 2ab + b² = (a + b)²] a perfect square
⇒ {(z⁴)² + 2(z⁴)(1) + (1)²} - z⁴
⇒ (z⁴ + 1)² - z⁴
Now By using formula [∵ a² - b² = (a - b)(a + b)] difference of two squares
⇒ (z⁴ + 1)² - (z²)²
⇒ {(z⁴ + 1) + z²}{(z⁴ + 1) - z²}
⇒ (z⁴ + 1 + z²)(z⁴ + 1 - z²)
⇒ {(z⁴ + 1) + z²}(z⁴ - z² + 1) (Rearrange the terms)
By adding & subtracting 2z²in first expression, we get:
⇒ {(z⁴ + 2z² + 1) - 2z² + z²}(z⁴ - z² + 1)
By using formula: [∵ a² + 2ab + b² = (a + b)²] a perfect square
⇒ [{(z²)² + 2(z²)(1) + (1)²} - z²](z⁴ - z² + 1)
⇒ {(z² + 1)² - z²}(z⁴ - z² + 1)
Now By using formula [∵ a² - b² = (a - b)(a + b)] difference of two squares
⇒ {(z² + 1)² - (z)²}(z⁴ - z² + 1)
⇒ [{(z² + 1) + (z)}{(z² + 1) - (z)}](z⁴ - z² + 1)
⇒ {(z² +1 + z)(a² + 1 - z)}(z⁴ - z² + 1)
⇒ (z² + z + 1)(z² - z + 1)(z⁴ - z² + 1) Ans

2. Factorize:
(i) 42x² - 8x - 2
Solution:
⇒ 42x² - 8x - 2
By taking 2 as common
⇒ 2(21x² - 4x - 1)
It can be factorize by breaking middle term

ROUGH WORK:

• Step 1: Take the product of coefficient of 1^st & last term i.e 21 x 1 = 21
• Step 2: Find prime factors of 21 = 3 x 7
• Step 3: As the last term has minus sign so middle term will obtained by subtracting two numbers from prime factors of 21.
  i.e 7 - 3 = 8
• Step 4: As mid term has minus sign so greater value has minus sign (-7x) & smaller value has plus sign (+3x).

⇒ 2(21x² - 4x - 1)
⇒ 2(21x² - 7x + 3x - 1
⇒ 2{7x(3x - 1) + 1(3x - 1)
⇒ 2(7x + 1)(3x - 1) Ans

(ii) 21z² - 4z - 1
Solution:
It can be factorize by breaking middle term

ROUGH WORK:

• Step 1: Take the product of coefficient of 1^st & last term i.e 21 x 1 = 21
• Step 2: Find prime factors of 21 = 3 x 7
• Step 3: As the last term has minus sign so middle term will obtained by difference of two numbers from prime factors of 21.
  i.e 7 - 3 = 4
• Step 4: As mid term has minus sign so greater value has minus sign (- 7z) & smaller value has plus sign (+ 3z).

⇒ 21z² - 4z - 1
⇒ 21z² - 7z + 3z - 1
⇒ 7z(3z - 1) + 1(3z - 1)
⇒ (7z + 1)(3z - 1) Ans

(iii) 9y² - 21yz⁴ - 8y²
Solution:
⇒ 9y² - 8y² - 21yz⁴
⇒ y² - 21yz⁴
By Taking y as common
⇒ y(y - 21z⁴) Ans

(iv) 24a² - 18a + 27
(Note: In book, this question is wrong. It can be solved either
24a² - 18a - 27 OR 24a² - 81a + 27)
Solution:
This can be possible to solve if we take it as
⇒ 24a² - 18a - 27 (change sign of last term to minus)
By Taking 3 as common
⇒ 3(8a² - 6a - 9)
The bracket expression can be factorized by breaking middle term

ROUGH WORK:

• Step 1: Take the product of coefficient of 1^st & last term i.e 8 x 9 = 72
• Step 2: Find prime factors of 72 = 2 x 2 x 2 x 3 x 3
• Step 3: As the last term has minus sign so middle term will obtained by subtracting two numbers. Now multiply Prime factors of 72 in a way to get two numbers whose difference is equal to 6
  i.e (2 x 2 x 3) - (2 x 3) = 12 - 6 = 6
• Step 4: As mid term has minus sign so greater value has minus sign (- 12a) & smaller value has plus sign (+ 6a).

⇒ 3(8a² - 12a + 6a - 9)
⇒ 3{4a(2a - 3) + 3(2a - 3)}
⇒ 3(4a + 3)(2a - 3) Ans

OR

Solution:
This can be possible to solve if we take it as
⇒ 24a² - 81a + 27 (change coefficient of mid term from 18 to 81)
By Taking 3 as common
⇒ 3(8a² - 27a + 9)
The bracket expression can be factorized by breaking middle term

ROUGH WORK:

• Step 1: Take the product of coefficient of 1^st & last term i.e 8 x 9 = 72
• Step 2: Find prime factors of 72 = 2 x 2 x 2 x 3 x 3
• Step 3: As the last term has plus sign so middle term will obtained by adding two numbers. Now multiply Prime factors of 72 in a way to get two numbers whose sum is equal to 27
  i.e (2 x 2 x 2 x 3) + (3) = 24 + 3 = 27
• Step 4: Both mid term should have same sign to add them and as it has minus sign so both terms value will have minus sign i.e (- 24a & -3a)

⇒ 3(8a² - 24 - 3a + 9)
⇒ 3{8a(a - 3) - 3(a - 3)}
⇒ 3(8a - 3)(a - 3) Ans

3. Factorize:
(i) x⁴ + 4y²
Solution:
⇒ x⁴ + 4y²
By adding & subtracting 4x²y, we get
⇒ (x⁴ + 4y²) + 4x²y - 4x²y
⇒ (x⁴ + 4x²y + 4y²) - 4x²y
By using formula: [∵ a² + 2ab + b² = (a + b)²] a perfect square
⇒ {(x²)² + 2(x²)(2y) + (2y)²} - 4x²y
⇒ (x² + 2y)² - 4x²y Ans

(ii) 36x⁴z⁴ + 9y⁴
Solution:
⇒ 36x⁴z⁴ + 9y⁴
By taking 9 as common, we get:
⇒ 9(4x⁴z⁴ + y⁴)
(Note: in book above value is given as answer but it can factorize further)
By adding & subtracting 4x²y²z², we get
⇒ 9(4x⁴z⁴ + y⁴) + 4x²y²z² - 4x²y²z²
⇒ 9(4x⁴z⁴ + 4x²y²z² + y⁴) - 4x²y²z²
By using formula: [∵ a² + 2ab + b² = (a + b)²] a perfect square
⇒ 9{(2x²z²)² + 2(2x²z²)(y²) + (y²)²} - 4x²y²z²
⇒ 9(2x²z² + y²)² - 4x²y²z²
Now By using formula [∵ a² - b² = (a - b)(a + b)] difference of two squares
⇒ {3(2x²z² + y²)}² - (2xyz)²
⇒ {3(2x²z² + y²) + 2xyz}{3(2x²z² + y²) - 2xyz}
By taking 3 as common, we get
⇒ 3{(2x²z² + 2xyz + y²)(2x²z² - 2xyz + y² )} Ans.

(iii) 4t⁴ + 625
Solution:
⇒ 4t⁴ + 625
By adding & subtracting 100t², we get
⇒ (4t⁴ + 625) + 100t² - 100t²
⇒ (4t⁴ + 625 + 100t²) - 100t²
⇒ (4t⁴ + 100t² + 625) - 100t²
By using formula: [∵ a² + 2ab + b² = (a + b)²] a perfect square
⇒ (2t²)² + 2(2t²)(25) + (25)²) - 100t²
⇒ (2t² + 25)² - 100t²
Now By using formula [∵ a² - b² = (a - b)(a + b)] difference of two squares
⇒ (2t² + 25)² - (10t)²
⇒ (2t² + 25 + 10t)(2t² + 25 - 10t)
⇒ (2t² + 10t + 25)(2t² - 10t + 25) Ans

(iv) 4t⁴ + 1
Solution:
⇒ 4t⁴ + 1
By adding & subtracting 4t², we get
⇒ (4t⁴ + 1) + 4t² - 4t²
⇒ (4t⁴ + 1 + 4t²) - 4t²
⇒ (4t⁴ + 4t² + 1) - 4t²
By using formula: [∵ a² + 2ab + b² = (a + b)²] a perfect square
⇒ (2t²)² + 2(2t²)(1) + (1)²) - 4t²
⇒ (2t² + 1)² - 4t²
Now By using formula [∵ a² - b² = (a - b)(a + b)] difference of two squares
⇒ (2t² + 1)² - (2t)²
⇒ (2t² + 1 + 2t)(2t² + 1- 2t)
⇒ (2t² + 2t + 2)(2t² - 2t + 1) Ans

4. Resolve into factors:
(i) x² + 3x - 10
Solution:
⇒ x² + 3x - 10
It can be factorized by breaking middle term

ROUGH WORK:

• Step 1: Take the product of coefficient of 1^st & last term i.e 1 x 10 = 10
• Step 2: Find prime factors of 10 = 2 x 5
• Step 3: As the last term has minus sign so middle term will obtained by subtracting two numbers from prime factors of 10
  i.e 5 - 2 = 3
• Step 4: As mid term has plus sign so greater value has plus sign (+5x) & smaller value has minus sign (-2x).

⇒ x² + 5x - 2x - 10
⇒ x(x + 5) - 2(x + 5)
⇒ (x + 5)(x - 2) Ans.

(ii) a²b² - 3ab - 10
Solution:
⇒ a²b² - 3ab - 10
It can be factorized by breaking middle term

ROUGH WORK:

• Step 1: Take the product of coefficient of 1^st & last term i.e 1 x 10 = 10
• Step 2: Find prime factors of 10 = 2 x 5
• Step 3: As the last term has minus sign so middle term will obtained by subtracting two numbers from prime factors of 10
  i.e 5 - 2 = 3
• Step 4: As mid term has minus sign so greater value has minus sign (-5ab) & smaller value has plus sign (+2ab).

⇒ a²b² - 5ab + 2ab - 10
⇒ ab(ab - 5) + 2(ab - 5)
⇒ (ab - 5)(ab + 2) Ans.

(iii) y² + 7y - 98
Solution:
⇒ y² + 7y - 98
It can be factorized by breaking middle term

ROUGH WORK:

• Step 1: Take the product of coefficient of 1^st & last term i.e 1 x 98 = 98
• Step 2: Find prime factors of 98 = 2 x 7 x 7
• Step 3: As the last term has minus sign so middle term will obtained by subtracting two numbers from prime factors of 98
  i.e (2 x 7) - 7 = 14 - 7 = 7
• Step 4: As mid term has plus sign so greater value has plus sign (+14y) & smaller value has minus sign (-7y).

⇒ y² + 14y - 7y - 98
⇒ y(y + 14) - 7(y + 14)
⇒ (Y + 14)(Y - 7) Ans.

(iv) x²y²z² + 2xyz - 24
Solution:
⇒ x²y²z² + 2xyz - 24
It can be factorized by breaking middle term

ROUGH WORK:

• Step 1: Take the product of coefficient of 1^st & last term i.e 1 x 24 = 24
• Step 2: Find prime factors of 24 = 2 x 2 x 2 x 3
• Step 3: As the last term has minus sign so middle term will obtained by subtracting two numbers from prime factors of 24
  i.e (2 x 3) - (2 x 2) = 6 - 4 = 2
• Step 4: As mid term has plus sign so greater value has plus sign (+6xyz) & smaller value has minus sign (-4xyz).

⇒ x²y²z² + 6xyz - 4xyz - 24
⇒ xyz(xyz + 6) - 4(xyz + 6)
⇒ (xyz + 6)(xyz - 4) Ans.

5. Resolve into factors:
(i) 121x⁴ + 11x² + 2
Solution:
The question can not be factorize because:
i) there is no common in all the three terms.
ii) Perfect square [a² ± 2ab + b² = (a + b)²] Or Difference of twi squares [a² - b² = (a - b)(a + b)] can not be followed by expression.
iii) The expression is not valid for breaking method as:
⇒ 121x⁴ + 11x² + 2

ROUGH WORK:

• Step 1: Taking the product of coefficient of 1^st & last term we get 121 x 2 = 242
• Step 2: Find prime factors of 242 = 2 x 11 x 11
• Step 3: As the last term has plus sign so middle term will obtained by sum of two numbers from prime factors of 242, which is impossible

Ans: Therefore the above expression can not be factorized.
Note: It can be factorized if the last term has minus sign as
⇒ 121x⁴ + 11x² - 2
Now it can be factorized by breaking middle term

ROUGH WORK:

• Step 1: Take the product of coefficient of 1^st & last term i.e 121 x 2 = 242
• Step 2: Find prime factors of 242 = 2 x 11 x 11
• Step 3: As the last term has minus sign so middle term will obtained by subtracting two numbers from prime factors of 242
  i.e (11 x 2) - (11) = 22 - 11 = 11
• Step 4: As mid term has plus sign so greater value has plus sign (+22x) & smaller value has minus sign (-11x).

⇒ 121x⁴ + 22x² - 11x² - 2
⇒ 11x²(11x² + 2) - 1(11x² + 2)
⇒ (11x² + 2)(11x² - 1) Ans

(ii) 42z⁴ + 50z² + 8
Solution:
⇒ 42z⁴ + 50z² + 8
By taking 2 as common, we get:
⇒ 2(21z⁴ + 25z² + 4)
It can be factorized by breaking middle term

ROUGH WORK:

• Step 1: Take the product of coefficient of 1^st & last term i.e 21 x 4 = 84
• Step 2: Find prime factors of 84 = 2 x 2 x 3 x 7
• Step 3: As the last term has plus sign so middle term will obtained by adding two numbers from prime factors of 84
  i.e (7 x 3) + (2 x 2) = 21 + 4 = 25
• Step 4: As mid term has plus sign so both values have plus sign (+21z²) & (+4z²).

⇒ 2(21z⁴ + 21z² + 4z² + 4)
⇒ 2{21z²(z² + 1) + 4(z² + 1)}
⇒ 2(21z² + 4)(z² + 1) Ans

(iii) 4x² + 12x + 5
Solution:
⇒ 4x² + 12x + 5
It can be factorized by breaking middle term

ROUGH WORK:

• Step 1: Take the product of coefficient of 1^st & last term i.e 4 x 5 = 20
• Step 2: Find prime factors of 20 = 2 x 2 x 5
• Step 3: As the last term has plus sign so middle term will obtained by addining two numbers from prime factors of 20
  i.e (2 x 5) + (2) = 10 + 2 = 12
• Step 4: As mid term has plus sign so both values have plus sign (+10x) & (+2x).

⇒ 4x² + 10x + 2x + 5
⇒ 2x(2x + 5) + 1(2x + 5)
⇒ (2x + 5)(2x + 1) Ans

(iv) 3x² - 38xy - 13y²
Solution:
⇒ 3x² - 38xy - 13y²
It can be factorized by breaking middle term

ROUGH WORK:

• Step 1: Take the product of coefficient of 1^st & last term i.e 3 x 13 = 39
• Step 2: Find prime factors of 39 = 1 x 39
• Step 3: As the last term has minus sign so middle term will obtained by subtracting two numbers from prime factors of 39
  i.e 39 - 1 = 38
• Step 4: As mid term has minus sign so greater value has minus sign (-39xy) & smaller value has plus sign (+xy).

⇒ 3x² - 39xy + xy - 13y²
⇒ 3x(x - 13y) + y(x - 13y)
⇒ (x - 13y)(3x + y) Ans

6. Resolve into factors:
(i) 81x⁴ + 36x²y² + 16y⁴
Solution:
⇒ (81x⁴ + 16y⁴) + 36x²y² (Rearrange the terms)
By adding & subtracting 2(9x²)(4y²) = 72x²y², we get:
⇒ (81x⁴ + 72x²y² + 16y⁴) - 72x²y² + 36x²y²
By using formula: [∵ a² + 2ab + b² = (a + b)²] a perfect square
⇒ {(9x²)² + 2(9x²)(4y²) + (4y²)² - 36x²y²
⇒ (9x² + 4y²)² - 36x²y²
Now By using formula [∵ a² - b² = (a - b)(a + b)] difference of two squares
⇒ (9x² + 4y²)² - (6xy)²
⇒ (9x² + 4y² - 6xy)(9x² + 4y² + 6xy)
⇒ (9x² - 6xy + 4y²)(9x² + 6xy + 4y²) Ans

(ii) x⁴ + x² + 25
Solution:
⇒ (x⁴ + 25) + x² (Rearrange the terms)
By adding & subtracting 2(x²)(5) = 10x², we get:
⇒ (x⁴ + 10x² + 25) - 10x²y² + x²
By using formula: [∵ a² + 2ab + b² = (a + b)²] a perfect square
⇒ {(x²)² + 2(x²)(5) + (5)² - 9x²y²
⇒ (x² + 5)² - 9x²
Now By using formula [∵ a² - b² = (a - b)(a + b)] difference of two squares
⇒ (x² + 5)² - (3x)²
⇒ (x² + 5 - 3x)(x² + 5 + 3x)
⇒ (x² - 3x + 5)(x² + 3x + 5) Ans

(iii) y⁴ - 7y² - 8
Solution:
⇒ y⁴ - 7y² - 8
It can be factorized by breaking middle term

ROUGH WORK:

• Step 1: Take the product of coefficient of 1^st & last term i.e 1 x 8 = 8
• Step 2: As the last term has minus sign so middle term will obtained by subtracting two numbers
  i.e 8 - 1 = 7
• Step 3: As mid term has minus sign so greater value has minus sign (-8y²) & smaller value has plus sign (+y²).

⇒ y⁴ - 8y² + y² - 8
⇒ y²(y² - 8) + 1(y² - 8)
⇒ (y² + 1)(y² - 8) Ans

(iv) 16a⁴ - 97a²b² + 81b⁴
Solution:
⇒ (16a⁴ + 81b⁴) - 97a²b² (Rearrange the terms)
By adding & subtracting 2(4a²)(9b²) = 72a²b², It can be solved by two methods:

METHOD 1:
⇒ (16a⁴ + 72a²b² + 81b⁴) - 72a²b² - 97a²b²
By using formula: [∵ a² + 2ab + b² = (a + b)²] a perfect square
⇒ {(4a²)² + 2(4a²)(9b²) + (9b²)² - 169a²b²
⇒ (4a² + 9b²)² - 169a²b²
Now By using formula [∵ a² - b² = (a - b)(a + b)] difference of two squares
⇒ (4a² + 9b²)² - (13ab)²
⇒ (4a² + 9b² - 13ab)(4a² + 9b² + 13ab)
⇒ (4a² - 13ab + 9b²)(4a² + 13ab + 9b²)
It can be further factorized by breaking middle term

ROUGH WORK:

• Step 1: As the last term has plus sign so middle term will obtained by adding two numbers
• Step 2: The sum of coefficient of 1^st & last term is equal to 13 i.e 9 + 4 = !#
• Step 3:In first bracket the mid term has minus sign so both values have minus sign i.e (-9ab) & (-4ab).
  While in second bracket the mid term has plus sign so both values have plus sign (+9ab) & (+4ab)

⇒ (4a² - 9ab - 4ab + 9b²)(4a² + 9ab + 4ab + 9b²)
⇒ {a(4a - 9b) - b(4a - 9b)}{a(4a + 9b) + b(4a + 9b)}
⇒ {(a - b)(4a - 9b)}{(a + b)(4a + 9b)}
⇒ (a + b)(a - b)(4a - 9b)(4a + 9b) Ans

OR
METHOD 2:
⇒ (16a⁴ - 72a²b² + 81b⁴) + 72a²b² - 97a²b²
By using formula: [∵ a² + 2ab + b² = (a + b)²] a perfect square
⇒ {(4a²)² - 2(4a²)(9b²) + (9b²)² - 25a²b²
⇒ (4a² - 9b²)² - 25a²b²
Now By using formula [∵ a² - b² = (a - b)(a + b)] difference of two squares
⇒ (4a² - 9b²)² - (5ab)²
⇒ (4a² - 9b² - 5ab)(4a² - 9b² + 5ab)
⇒ (4a² - 5ab - 9b²)(4a² + 5ab - 9b²)
It can be further factorized by breaking middle term

ROUGH WORK:

• Step 1: As the last term has minus sign so middle term will obtained by subtracting two numbers
• Step 2: The difference of coefficient of 1^st & last term is equal to 5 i.e 9 - 4 = 5
• Step 3:In first bracket the mid term has minus sign so greater value has minus sign (-9ab) & smaller value has plus sign (+4ab).
  While in second bracket the mid term has plus sign so greater value has plus sign (+9ab) & smaller value has minus sign (-4ab)

⇒ (4a² - 9ab + 4ab - 9b²)(4a² + 9ab - 4ab - 9b²)
⇒ {a(4a - 9b) + b(4a - 9b)}{a(4a + 9b) - b(4a + 9b)}
⇒ {(a + b)(4a - 9b)}{(a - b)(4a + 9b)}
⇒ (a + b)(a - b)(4a - 9b)(4a + 9b) Ans