Monday, 19 August 2024

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 2a2xin 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:
  • As the last term has minus sign so middle term will obtained by subtracting two numbers which are the factors of the product of the coefficient of the first & last term:
    i.e. 21 x 1 = 21
  • Factors of 21 = (1 x 21), (3 x 7)
  • Thus two terms are 7 & 3 as 7 - 3 = 4
  • For difference, as middle term has minus sign so the greater number has minus sign (-7x) & smaller number 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:
  • As the last term has minus sign so middle term will obtained by subtracting two numbers which are the factors of the product of the coefficient of the first & last term:
    i.e. 21 x 1 = 21
  • Factors of 21 = (1 x 21), (3 x 7)
  • Thus two terms are 7 & 3 as 7 - 3 = 4
  • For difference, as middle term has minus sign so the greater number has minus sign (-7z) & smaller number 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:
  • As the last term has minus sign so middle term will obtained by subtracting two numbers which are the factors of the product of the coefficient of the first & last term:
    i.e. 8 x 9 = 72
  • Factors of 72= (1 x 72), (2 x 36), (3 x 24), (4 x 18), (6 x 12), (8 x 9)
  • Thus two terms are 12 & 6 as 12 - 6 = 6
  • For difference, as middle term has minus sign so the greater number has minus sign (-12a) & small number 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:
  • As the last term has plus sign so middle term will obtained by adding two numbers which are the factors of the product of the coefficient of the first & last term:
    i.e. 8 x 9 = 72
  • Factors of 72= (1 x 72), (2 x 36), (3 x 24), (4 x 18), (6 x 12), (8 x 9)
  • Thus two terms are 24 & 3 as 24 + 3 = 27
  • For addition, as middle term has minus sign so both terms 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:
  • As the last term has minus sign so middle term will obtained by subtracting two numbers which are the factors of the product of the coefficient of the first & last term:
    i.e. 10 x 1 = 10
  • Factors of 10= (1 x 10), (2 x 5)
  • Thus two terms are 5 & 2 as 5 - 2 = 3
  • For difference, as middle term has plus sign so the greater number has plus sign (+5x) & smaller number 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:
  • As the last term has minus sign so middle term will obtained by subtracting two numbers which are the factors of the product of the coefficient of the first & last term:
    i.e. 10 x 1 = 10
  • Factors of 10= (1 x 10), (2 x 5)
  • Thus two terms are 5 & 2 as 5 - 2 = 3
  • For difference, as middle term has minus sign so the greater number has minus sign (-5x) & smaller number has plus sign (+2x).

⇒ 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:
  • As the last term has minus sign so middle term will obtained by subtracting two numbers which are the factors of the product of the coefficient of the first & last term:
    i.e. 98 x 1 = 98
  • Factors of 98= (1 x 98), (2 x 49), (7 x 14)
  • Thus two terms are 14 & 7 as 14 - 7 = 7
  • For difference, as middle term has plus sign so the greater number has plus sign (+14y) & smaller number 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:
  • As the last term has minus sign so middle term will obtained by subtracting two numbers which are the factors of the product of the coefficient of the first & last term:
    i.e. 24 x 1 = 24
  • Factors of 24= (1 x 24), (2 x 12), (3 x 8), (4 x 6)
  • Thus two terms are 6 & 2 as 6 - 4 = 2
  • For difference, as middle term has plus sign so the greater number has plus sign (+6xyz) & smaller number 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:
  • As the last term has plus sign so middle term will obtained by adding two numbers which are the factors of the product of the coefficient of the first & last term:
    i.e. 2 x 121 = 242
  • Factors of 242= (1 x 242), (2 x 121), (11 x 11)
  • 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:
  • As the last term has minus sign so middle term will obtained by subtracting two numbers which are the factors of the product of the coefficient of the first & last term:
    i.e. 2 x 121 = 242
  • Factors of 242= (1 x 242), (2 x 121), (11 x 22)
  • Thus two terms are 22 & 11 as 22 - 11 = 11
  • For difference, as middle term has plus sign so the greater number has plus sign (+22x) & smaller number 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:
  • As the last term has plus sign so middle term will obtained by adding two numbers which are the factors of the product of the coefficient of the first & last term:
    i.e. 21 x 4 = 84
  • Factors of 84= (1 x 84), (2 x 42), (3 x 28) (4 x 21) (6 x 14), (7 x 12)
  • Thus two terms are 21 & 4 as 21 + 4 = 25
  • For addition, as middle term has plus sign so both terms will have plus sign i.e  (+ 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:
  • As the last term has plus sign so middle term will obtained by adding two numbers which are the factors of the product of the coefficient of the first & last term:
    i.e. 4 x 5 = 20
  • Factors of 20= (1 x 20), (2 x 10), (4 x 5)
  • Thus two terms are 10 & 2 as 10 + 2 = 12
  • For addition, as middle term has plus sign so both terms will have plus sign i.e (+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:
  • As the last term has minus sign so middle term will obtained by subtracting two numbers which are the factors of the product of the coefficient of the first & last term:
    i.e. 3 x 13 = 39
  • Factors of 39= (1 x 39), (3 x 13)
  • Thus two terms are 39 & 1 as 39 - 1 = 38
  • For difference, as middle term has minus sign so the greater number has minus sign (-39xy) & smaller number 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:
  • As the last term has minus sign so middle term will obtained by subtracting two numbers which are the factors of the product of the coefficient of the first & last term:
    i.e. 1 x 8 = 8
  • Factors of 8= (1 x 8), (2 x 4)
  • Thus two terms are 8 & 1 as 8 - 1 = 7
  • For difference, as middle term has minus sign so the greater number has minus sign (-8y2) & smaller number 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:
  • As the last term has plus sign so middle term will obtained by adding two numbers which are the factors of the product of the coefficient of the first & last term:
    i.e. 9 x 4 = 36
  • Factors of 36= (1 x 36), (2 x 18), (3 x 12), (4 x 9), (6 x 6)
  • Thus two terms are 9 & 4 as 9 + 4 = 13
  • For addition, as middle term has minus sign in first bracket so both terms will have minus sign i.e (- 9qb & -4ab).
    * While middle term has plus sign in second bracket so both terme will have plus sign i.e. (+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:
  • As the last term has minus sign so middle term will obtained by subtracting two numbers which are the factors of the product of the coefficient of the first & last term:
    i.e. 9 x 4 = 36
  • Factors of 36 = (1 x 36), (2 x 18), (3 x 12) (4 x 9) (6 x 6)
  • Thus two terms are 4 & 9 as 9 - 4 = 5
  • For difference, as middle term has minus sign in first bracket so the greater number has minus sign (-9ab) & smaller number has plus sign (+4ab).
    * While middle term has plus sign in second bracket so the greater number has plus sign (+9ab) & smaller number 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


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