Unit 4: Factorization - Solved Exercise 4.3 - Mathematics For Class IX (Science Group)

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Unit 4: Factorization
Solved Exercise 4.3

1. Factorize the following:
(i) (x² - 4x - 5)(x² - 4x - 12) - 144
Solution:
⇒ (x² - 4x - 5)(x² - 4x - 12) - 144
Let x² - 4x = t, then we have
⇒ (t - 5)(t - 12) - 144
⇒ {t(t - 12) - 5(t - 12)} - 144
⇒ (t² - 12t - 5t + 60) - 144
⇒ t² - 17t + 60 - 144
⇒ t² - 17t - 84
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. 84 x 1 = 84
• Factors of 84= (1 x 84), (2 x 42), (3 x 28) (4 x 21) (6 x 18), (7 x 14)
• Thus two terms are 21 & 4 as 21 - 4 = 17
• For difference, as middle term has minus sign so the greater number has minus sign (-21t) & smaller number has plus sign (+4t).

⇒ t² - 21t + 4t - 84
⇒ t(t - 21) + 4(t - 21)
⇒ (t - 21)(t + 4)
[∵ t = x² - 4x]
⇒ (x² - 4x - 21)(x² - 4x + 4)
Again it can be factorize further:
* 1^st Bracket: By breaking middle term &
* 2^nd Bracket: By using formula: [∵ a² + 2ab + b² = (a + b)²] a perfect square

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).

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

(ii) (x² + 5x + 6)(x² + 5x + 4) - 3
Solution:
⇒ (x² + 5x + 6)(x² + 5x + 4) - 3
Let x² + 5x = t, then we have
⇒ (t + 6)(t + 4) - 3
⇒ {t(t + 4) + 6(t + 4)} - 3
⇒ (t² + 4t + 6t + 24) - 3
⇒ t² + 10t + 24 - 3
⇒ t² + 10t + 21
It can be factorize 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 1 = 21
• Factors of 21 = (1 x 21), (3 x 7)
• Thus two terms are 7 & 3 as 7 + 3 = 10
• For addition, as middle term has plus sign so both terms will have plus sign i.e (+7t & +3t).

⇒ t² + 7t + 3t + 21
⇒ t(t + 7) + 3(t + 7)
⇒ (t + 7)(t + 3)
[∵ t = x² + 5x]
⇒ (x² + 5x + 7)(x² + 5x + 3) Ans.

(iii) (x² - 2x + 3)(x² - 2x + 4) - 42
Solution:
⇒ (x² - 2x + 3)(x² - 2x + 4) - 42
Let x² - 2x = t, then we have
⇒ (t + 3)(t + 4) - 42
⇒ {t(t + 4) + 3(t + 4)} - 42
⇒ (t² + 4t + 3t + 12) - 42
⇒ t² + 7t + 12 - 42
⇒ t² + 7t - 30
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. 30 x 1 = 30
• Factors of 30 = (1 x 30), (2 x 15), (3 x 10), (5 x 6)
• Thus two terms are 10 & 3 as 10 - 3 = 7
• For difference, as middle term has plus sign so the greater number has plus sign (+10t) & smaller number has minus sign (-3t).

⇒ t² + 10t - 3t - 30
⇒ t(t + 10) - 3(t + 10)
⇒ (t + 10)(t - 3)
[∵ t = x² - 2x]
⇒ (x² - 2x + 10)(x² - 2x - 3)
We can not factorize 1^st Bracket but 2^nd Bracket can be factorize by breaking middle term as:

ROUGH WORK WORK For 2^nd Bracket::

• 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 1 = 3
• Factors of 3 = (1 x 3)
• Thus two terms are 3 & 1 as 3 - 1 = 2
• For difference, as middle term has minus sign so the greater number has minus sign (-3x) & smaller number has plus sign (+x).

⇒ (x² - 2x + 10){(x² - 3x + x - 3)}
⇒ (x² - 2x + 10){x(x - 3) + 1(x - 3)}
⇒ (x² - 2x + 10)(x - 3)(x + 1) Ans.

(iv) (x² - 8x + 4)(x² - 8x - 4) + 15
Solution:
⇒ (x² - 8x + 4)(x² - 8x - 4) + 15
Let x² - 8x = t, then we have
⇒ (t + 4)(t - 4) + 15
[∵ (a + b)(a - b) = a² - b²]
⇒ {(t)² - (4)²} + 15
⇒ t² - 16 + 15
⇒ t² - 1
It can be factorize by the difference of two squares
[∵ a² - b² = (a + b)(a - b)]
⇒ (t)² - (1)²
⇒ (t - 1)(t + 1)
[∵ t = x² - 8x]
⇒ (x² - 8x - 1)(x² - 8x + 1) Ans.

(v) (x² + 9x - 1)(x² + 9x + 5) - 7
Solution:
⇒ (x² + 9x - 1)(x² + 9x + 5) - 7
Let x² + 9x = t, then we have
⇒ (t - 1)(t + 5) - 7
⇒ {t(t + 5) - 1(t + 5)} - 7
⇒ t² + 5t - t - 5 - 7
⇒ t² + 4t - 12
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.12 x 1 = 12
• Factors of 12 = (1 x 12), (2 x 6), (3 x 4)
• Thus two terms are 6 & 2 as 6 - 2 = 4
• For difference, as middle term has plus sign so the greater number has plus sign (+6t) & smaller number has minus sign (-2t).

⇒ t² + 6t - 2t - 12
⇒ t(t + 6) - 2(t + 6)
⇒ (t + 6)(t - 2)
[∵ t = x² + 9x]
⇒ (x² + 9x + 6)(x² + 9x - 2) Ans

(vi) (x² - 5x + 4)(x² - 5x + 6) - 120
Solution:
⇒ (x² - 5x + 4)(x² - 5x + 6) - 120
Let x² - 5x = t, then we have
⇒ (t + 4)(t + 6) - 120
⇒ {t(t + 6) + 4(t + 6)} - 120
⇒ (t² + 6t + 4t + 24) - 120
⇒ t² + 10t + 24 - 120
⇒ t² + 10t - 96
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. 96 x 1 = 96
• Factors of 96 = (1 x 96), (2 x 48), (3 x 32), (4 x 24), (6 x 16), (8 x 12)
• Thus two terms are 16 & 6 as 16 - 6 = 10
• For difference, as middle term has plus sign so the greater number has plus sign (+16t) & smaller number has minus sign (-6t).

⇒ t² + 16t - 6t - 96
⇒ t(t + 16) - 6(t + 16)
⇒ (t + 16)(t - 6)
[∵ t = x² - 5x]
⇒ (x² - 5x + 16)(x² - 5x - 6) Ans

2. Factorize:
(i) (x + 1)(x + 2)(x + 3)(x + 4) - 48
Solution:
⇒ (x + 1)(x + 2)(x + 3)(x + 4) - 48
Here 1 + 4 = 2 + 3 = 5
Therefore, by arranging the factors,we get:
⇒ (x + 1)(x + 4)(x + 2)(x + 3) - 48
⇒ {x (x + 4) + 1(x + 4)}{x(x + 3) + 2(x + 3)} - 48
⇒ {(x² + 4x + x + 4)}{(x² + 3x + 2x + 6)} - 48
⇒ (x² + 5x + 4)(x² + 5x + 6) - 48
Let x² + 5x = t
⇒ (t + 4)(t + 6) - 48
⇒ {t(t + 6) + 4(t + 6)} - 48
⇒ (t² + 6t + 4t + 24) - 48
⇒ t² + 10t + 24 - 48
⇒ t² + 10t - 24
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.24 x 1 = 23
• Factors of 24 = (1 x 24), (2 x 12), (3 x 8), (4 x 6)
• Thus two terms are 12 & 2 as 12 - 2 = 10
• For difference, as middle term has plus sign so the greater number has plus sign (+12t) & smaller number has minus sign (-2t).

⇒ t² + 12t - 2t - 24
⇒ t(t + 12) - 2(t + 12)
⇒ (t + 12)(t - 2)
[∵ t = x² + 5x]
⇒ (x² + 5x + 12)(x² + 5x - 2) Ans

(ii) (x + 2)(x + 3)(x + 4)(x + 5) + 24
Solution:
⇒ (x + 2)(x + 3)(x + 4)(x + 5) + 24
Here 2 + 5 = 3 + 4 = 7
Therefore, by arranging the factors,we get:
⇒ (x + 2)(x + 5)(x + 3)(x + 4) + 24
⇒ {x (x + 5) + 2(x + 5)}{x(x + 4) + 3(x + 4)} + 24
⇒ {(x² + 5x + 2x + 10)}{(x² + 4x + 3x + 12)} + 24
⇒ (x² + 7x + 10)(x² + 7x + 12) + 24
Let x² + 7x = t
⇒ (t + 10)(t + 12) + 24
⇒ {t(t + 12) + 10(t + 12)} + 24
⇒ (t² + 12t + 10t + 120) + 24
⇒ t² + 22t + 120 + 24
⇒ t² + 22t + 148
It can not be solved & factorize further
It is possible only if the last term is - 24

Suppose if (ii) (x + 2)(x + 3)(x + 4)(x + 5) - 24
Solution:
⇒ (x + 2)(x + 3)(x + 4)(x + 5) - 24
Here 2 + 5 = 3 + 4 = 7
Therefore, by arranging the factors,we get:
⇒ (x + 2)(x + 5)(x + 3)(x + 4) - 24
⇒ {x (x + 5) + 2(x + 5)}{x(x + 4) + 3(x + 4)} - 24
⇒ {(x² + 5x + 2x + 10)}{(x² + 4x + 3x + 12)} - 24
⇒ (x² + 7x + 10)(x² + 7x + 12) - 24
Let x² + 7x = t
⇒ (t + 10)(t + 12) - 24
⇒ {t(t + 12) + 10(t + 12)} - 24
⇒ (t² + 12t + 10t + 120) - 24
⇒ t² + 22t + 120 - 24
⇒ t² + 22t + 96
It can not be factorize 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. 96 x 1 = 96
• Factors of 96 = (1 x 96), (2 x 48), (3 x 32), (4 x 24), (6 x 16), (8 x 12)
• Thus two terms are 16 & 6 as 16 + 6 = 22
• For addition, as middle term has plus sign so both terms will have plus sign i.e (+16t & +6t).

⇒ t² + 16t + 6t + 96
⇒ t(t + 16) + 6(t + 16)
⇒ (t + 16)(t + 6)
[∵ t = x² + 7x]
⇒ (x² + 7x + 16)(x² + 7x + 6)
We can not factorize 1^st Bracket but 2^nd Bracket can be factorize by breaking middle term as:

ROUGH WORK for 2^nd Bracket:

• 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. 6 x 1 = 6
• Factors of 6 = (1 x 6), (2 x 3)
• Thus two terms are 6 & 1 as 6 + 1 = 7
• For addition, as middle term has plus sign so both terms will have plus sign i.e (+6x & +x).

⇒ (x² + 7x + 16){(x² + 6x + x + 6)}
⇒ (x² + 7x + 16){(x(x + 6) + 1(x + 6)}
⇒ (x² + 7x + 16)(x + 6)(x + 1) Ans

(iii) (x - 1)(x - 2)(x - 3)(x - 4) - 99
Solution:
⇒ (x - 1)(x - 2)(x - 3)(x - 4) - 99
Here 1 + 4 = 2 + 3 = 5
Therefore, by arranging the factors,we get:
⇒ (x - 1)(x - 4)(x - 2)(x - 3) - 99
⇒ {x (x - 4) - 1(x - 4)}{x(x - 3) - 2(x - 3)} - 99
⇒ {(x² - 4x - x + 4)}{(x² - 3x - 2x + 6)} - 99
⇒ (x² - 5x + 4)(x² - 5x + 6) - 99
Let x² - 5x = t
⇒ (t + 4)(t + 6) - 99
⇒ {t(t + 6) + 4(t + 6)} - 99
⇒ (t² + 6t + 4t + 24) - 99
⇒ t² + 10t + 24 - 99
⇒ t² + 10t - 75
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.75 x 1 = 75
• Factors of 75 = (1 x 75), (3 x 25), (5 x 15)
• Thus two terms are 15 & 5 as 15 - 5 = 10
• For difference, as middle term has plus sign so the greater number has plus sign (+15t) & smaller number has minus sign (-5t).

⇒ t² + 15t - 5t - 75
⇒ t(t + 15) - 5(t + 15)
⇒ (t + 15)(t - 5)
[∵ t = x² - 5x]
⇒ (x² - 5x + 15)(x² - 5x - 5) Ans

(iv) (x - 3)(x - 5)(x - 7)(x - 9) + 15
Solution:
⇒ (x - 3)(x - 5)(x - 7)(x - 9) + 15
Here 3 + 9 = 5 + 7 = 12
Therefore, by arranging the factors,we get:
⇒ (x - 3)(x - 9)(x - 5)(x - 7) + 15
⇒ {x (x - 9) - 3(x - 9)}{x(x - 7) - 5(x - 7)} + 15
⇒ {(x² - 9x - 3x + 27)}{(x² - 7x - 5x + 35)} + 15
⇒ (x² - 12x + 27)(x² - 7x + 35) + 15
Let x² - 12x = t
⇒ (t + 27)(t + 35) + 15
⇒ {t(t + 35) + 27(t + 35)} + 15
⇒ (t² + 35t + 27t + 945) + 15
⇒ t² + 62t + 945 + 15
⇒ t² + 62t + 960
It can be factorize 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. 960 x 1 = 960
• Factors of 960 = (1 x 960), (2 x 480), (3 x 320), (4 x 240), (5 x 192), (6 x 160), (8 x 120), (10 x 96), (12 x 80), (15 x 64), (16 X 60), (20 x 48), (24 x 40), (30 x 32)
• Thus two terms are 30 & 32 as 30 + 32 = 62
• For addition, as middle term has plus sign so both terms will have plus sign i.e (+30t & +32t).

⇒ t² + 32t + 30t + 960
⇒ t(t + 32) + 30(t + 32)
⇒ (t + 32)(t + 30)
[∵ t = x² - 12x]
⇒ (x² - 12x + 32)(x² - 12x + 30)
We can factorize 1^st Bracket by breaking middle term but 2^nd Bracket can not be factorize:

ROUGH WORK For 1^st Bracket:

• 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. 32 x 1 = 32
• Factors of 32 = (1 x 32), (2 x 16), (4 x 8)
• Thus two terms are 8 & 4 as 8 + 4 = 12
• For addition, as middle term has minus sign so both terms will have minus sign i.e (-8x & -4x).

⇒ (x² - 8x - 4x + 32)(x² - 12x + 30)
⇒ {(x(x - 8) - 4(x - 8)}(x² - 12x + 30)
⇒ (x - 8)(x - 4)(x² - 12x + 30) Ans

(v) (x - 1)(x - 2)(x - 3)(x - 4) - 224
Solution:
⇒ (x - 1)(x - 2)(x - 3)(x - 4) - 224
Here 1 + 4 = 2 + 3 = 5
Therefore, by arranging the factors,we get:
⇒ (x - 1)(x - 4)(x - 2)(x - 3) - 224
⇒ {x (x - 4) - 1(x - 4)}{x(x - 3) - 2(x - 3)} - 224
⇒ {(x² - 4x - x + 4)}{(x² - 3x - 2x + 6)} - 224
⇒ (x² - 5x + 4)(x² - 5x + 6) - 224
Let x² - 5x = t
⇒ (t + 4)(t + 6) - 224
⇒ {t(t + 6) + 4(t + 6)} - 224
⇒ (t² + 6t + 4t + 24) - 224
⇒ t² + 10t + 24 - 224
⇒ t² + 10t - 200
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. 200 x 1 = 200
• Factors of 200 = (1 x 200), (2 x 100), (4 x 50), (5 x 40), (8 x 25), (10 x 20)
• Thus two terms are 20 & 10 as 20 - 10 = 10
• For difference, as middle term has plus sign so the greater number has plus sign (+20t) & smaller number has minus sign (-10t).

⇒ t² + 20t - 10t - 200
⇒ t(t + 20) - 10(t + 20)
⇒ (t + 20)(t - 10)
[∵ t = x² - 5x]
⇒ (x² - 5x + 20)(x² - 5x - 10) Ans

(vi) (x - 2)(x - 3)(x - 4)(x - 5) - 255
Solution:
⇒ (x - 2)(x - 3)(x - 4)(x - 5) - 255
Here 2 + 5 = 3 + 4 = 7
Therefore, by arranging the factors,we get:
⇒ (x - 2)(x - 5)(x - 3)(x - 4) - 255
⇒ {x (x - 5) - 2(x - 5)}{x(x - 3) - 4(x - 3)} - 255
⇒ {(x² - 5x - 2x + 10)}{(x² - 3x - 4x + 12)} - 255
⇒ (x² - 7x + 10)(x² - 7x + 12) - 255
Let x² - 7x = t
⇒ (t + 10)(t + 12) - 255
⇒ {t(t + 12) + 10(t + 12)} - 255
⇒ (t² + 12t + 10t + 120) - 255
⇒ t² + 22t + 120 - 255
⇒ t² + 22t - 135
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. 135 x 1 = 135
• Factors of 135 = (1 x 135), (3 x 45), (5 x 27), (9 x 15)
• Thus two terms are 5 & 27 as 27 - 5 = 22
• For difference, as middle term has plus sign so the greater number has plus sign (+27t) & smaller number has minus sign (-5t).

⇒ t² + 27t - 5t - 135
⇒ t(t + 27) - 5(t + 27)
⇒ (t + 27)(t - 5)
[∵ t = x² - 7x]
⇒ (x² - 7x + 27)(x² - 7x - 5) Ans

2. Find the factors of:
(i) (x - 2)(x - 3)(x + 2)(x + 3) - 2x²
Solution:
⇒ (x - 2)(x - 3)(x + 2)(x + 3) - 2x²
By arranging the factors
⇒ (x - 2)(x + 2)(x - 3)(x + 3) - 2x²
[∵ (a + b)(a - b) = a² - b²]
⇒ {(x)² - (2)²}{(x)² - (3)²} - 2x²
⇒ (x² - 4)(x² - 9) - 2x²
⇒ {x²(x² - 9) - 4(x² - 9)} - 2x²
⇒ (x⁴ - 9x² - 4x² + 36) - 2x²
⇒ x⁴ - 13x² - 2x² + 36
⇒ x⁴ - 15x² + 36
It can be factorize 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. 36 x 1 = 36
• Factors of 36 = (1 x 36), (2 x 18), (3 x 12), (4 x 9), (6 x 6)
• Thus two terms are 12 & 3 as 12 + 3 = 15
• For addition, as middle term has minus sign so both terms will have minus sign i.e (- 12x² & -3x²).

⇒ x⁴ - 12x² - 3x² + 36
⇒ x²(x² - 12) - 3(x² - 12)
⇒ (x² - 12)(x² - 3)
It can be further factorize by difference of two squares
i.e. [∵ a² - b² = (a + b)(a - b)]
⇒ {(x)² - (12)²}{(x)² - (3)²}
⇒ (x - √12 )(x + √12 )((x - √3 )(x + √3 )
⇒ (x - √2 x 2 x 3 )(x + √2 x 2 x 3 )((x - √3 )(x + √3 )
⇒ (x - 2√3 )(x + 2√3 )((x - √3 )(x + √3 ) Ans


(ii) (x - 1)(x + 1)(x + 3)(x - 3) - 3x² - 23
Solution:
⇒ (x - 1)(x + 1)(x + 3)(x - 3) - 3x² - 23
[∵ (a + b)(a - b) = a² - b²]
⇒ {(x)² - (1)²}{(x)² - (3)²} - 3x² - 23
⇒ (x² - 1)(x² - 9) - 3x² - 23
⇒ {x²(x² - 9) - 1(x² - 9)} - 3x² - 23
⇒ (x⁴ - 9x² - x² + 9) - 3x² - 23
⇒ x⁴ - 10x² - 3x² + 9 - 23
⇒ x⁴ - 13x² - 14
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. 14 x 1 = 14
• Factors of 14 = (1 x 14), (2 x 7)
• Thus two terms are 14 & 1 as 14 - 1 = 13
• For difference, as middle term has minus sign so the greater number has minus sign (-14x²) & smaller number has plus sign (+x²).

⇒ x⁴ - 14x² + x² - 14
⇒ x²(x² - 14) + 1(x² - 14)
⇒ (x² - 14)(x² + 1)
1^st bracket can be factorize further by difference of two squares
i.e. [∵ a² - b² = (a + b)(a - b)]
⇒ {(x)² - (14)²}(x² + 1)
⇒ (x - √14 )(x + √14 )((x² + 1) Ans
(Note: In Book answer is incorrect because (x² + 1) can not be factorize further)

(iii) (x - 1)(x + 1)(x - 3)(x + 3) - 4x²
Solution:
⇒ (x - 1)(x + 1)(x - 3)(x + 3) + 4x²
[∵ (a + b)(a - b) = a² - b²]
⇒ {(x)² - (1)²}{(x)² - (3)²} + 4x²
⇒ (x² - 1)(x² - 9) - 4x²
⇒ {x²(x² - 9) - 1(x² - 9)} + 4x²
⇒ (x⁴ - 9x² - x² + 9) + 4x²
⇒ x⁴ - 10x² + 4x² + 9
⇒ x⁴ - 6x² + 9
It can be factorize by perfect square
i.e. [∵ a² - 2ab + b² = (a - b)²]
⇒ (x²)² - 2(x²)(3) + (3)²
⇒ (x² - 3)² Ans

(iv) (x - 2)(x + 2)(x - 4)(x + 4) - 14x²
Solution:
⇒ (x - 2)(x + 2)(x - 4)(x + 4) - 14x²
[∵ (a + b)(a - b) = a² - b²]
⇒ {(x)² - (2)²}{(x)² - (4)²} - 14x²
⇒ (x² - 4)(x² - 16) - 14x²
⇒ {x²(x² - 16) - 4(x² - 16)} - 14x²
⇒ (x⁴ - 16x² - 4x² + 64) - 14x²
⇒ x⁴ - 20x² - 14x² + 64
⇒ x⁴ - 34x² + 64
It can be factorize 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. 64 x 1 = 64
• Factors of 64 = (1 x 64), (2 x 32), (4 x 16), (8 x 8)
• Thus two terms are 32 & 2 as 32 + 2 = 34
• For addition, as middle term has minus sign so both terms will have minus sign i.e (- 32x² & -2x²).

⇒ x⁴ - 32x² - 2x² + 64
⇒ x²(x² - 32) - 2(x² - 32)
⇒ (x² - 32)(x² - 2)
It can be further factorize by difference of two squares
i.e. [∵ a² - b² = (a + b)(a - b)]
⇒ {(x)² - (32)²}{(x)² - (2)²}
⇒ (x - √32 )(x + √32 )((x - √2 )(x + √2 )
⇒ (x - √2 x 2 x 2 x2 x 2 )(x + √2 x 2 x 2 x 2 x 2 )((x - √2 )(x + √2 )
⇒ (x - 4√2 )(x + 4√2 )((x - √2 )(x + √2 ) Ans


(v) (x + 5)(x + 2)(x - 5)(x - 2) + 4x²
Solution:
⇒ (x + 5)(x + 2)(x - 5)(x - 2) + 4x²
By arranging the factors
⇒ (x + 2)(x - 2)(x + 5)(x - 5) + 4x²
[∵ (a + b)(a - b) = a² - b²]
⇒ {(x)² - (2)²}{(x)² - (5)²} + 4x²
⇒ (x² - 4)(x² - 25) + 4x²
⇒ {x²(x² - 25) - 4(x² - 25)} + 4x²
⇒ (x⁴ - 25x² - 4x² + 100) + 4x²
⇒ x⁴ - 29x² + 4x² + 100
⇒ x⁴ - 25x² + 100
It can be factorize 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. 100 x 1 = 100
• Factors of 100 = (1 x 100), (2 x 50), (4 x 25), (5 x 20), (10 x 10)
• Thus two terms are 20 & 5 as 20 + 5 = 25
• For addition, as middle term has minus sign so both terms will have minus sign i.e (-20x² & -5x²).

ROUGH WORK:

• Step 1: Find prime factors of 100 = 2 x 2 x 5 x 5
• Step 2: As the last term has plus sign so middle term will be obtained by adding two numbers from prime factors of 100.
  i.e (2 x 2 x 5) + 5 = 20 + 5 = 25
• Step 3: As mid term has minus sign so both values has minus sign as (-20x²) & (-5x²).

⇒ x⁴ - 20x² - 5x² + 100
⇒ x²(x² - 20) - 5(x² - 20)
⇒ (x² - 20)(x² - 5)
It can be further factorize by difference of two squares
i.e. [∵ a² - b² = (a + b)(a - b)]
⇒ {(x)² - (20)²}{(x)² - (5)²}
⇒ (x - √20 )(x + √20 )((x - √5 )(x + √5 )
⇒ (x - √2 x 2 x 5 )(x + √2 x 2 x 5 )((x - √5 )(x + √5 )
⇒ (x - 2√5 )(x + 2√5 )((x - √5 )(x + √5 ) Ans


(vi) (x² - x - 12)(x² - x - 12) - x²
Solution:
⇒ (x² - x - 12)(x² - x - 12) - x²
⇒ (x² - x - 12)² - x²
[∵ (a + b)(a - b) = a² - b²]
⇒ (x² - x - 12)² - (x)²
⇒ (x² - x - 12 - x)(x² - x - 12 + x)
⇒ (x² - 2x - 12)(x² - 12)
1^st Bracket can not be factorize further.
2^nd Bracket can be further factorize by difference of two squares
i.e. [∵ a² - b² = (a + b)(a - b)]
⇒ (x² - 2x - 12){(x)² - (12)²}
⇒ (x² - 2x - 12)(x - √12 )(x + √12 )
⇒ (x² - 2x - 12)(x - √2 x 2 x 3 )(x + √2 x 2 x 3 )
⇒ (x² - 2x - 12)(x - 2√3 )(x + 2√3 ) Ans