Unit 4: Factorization - Solved Exercise 4.4 - Mathematics For Class IX (Science Group)

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Unit 4: Factorization
Solved Exercise 4.4

1. Factorize the following:
(i) b³ + 3b²c + 3bc² + c³
Solution:
⇒ b³ + 3b²c + 3bc² + c³
⇒ (b)³ + 3(b)²(c) + 3(b)(c)² + (c)³
By using formula:
[∵ (a³ + 3a²b + 3ab² + b³) = (a + b)³]
⇒ (b + c)³ Ans.

(ii) 8x³ + 12x²y + 6xy² + y³
Solution:
⇒ 8x³ + 12x²y + 6xy² + y³

 (Rough Work: 8 = 2 x 2 x 2 = 2³)

⇒ 2³x³ + 12x²y + 6xy² + c³
⇒ (2x)³ + 3(2x)²)(y) + 3(2x)(y)² + (y)³
By using formula:
[∵ (a³ + 3a²b + 3ab² + b³) = (a + b)³]
⇒ (2x + y)³ Ans.

(iv) 8x³ + 36x² + 54x + 27
Solution:
⇒ 8x³ + 36x² + 54x + 27

 (Rough Work: 8 = 2 x 2 x 2 = 2³ and  27 = 3 x 3 x 3 = 3³)

⇒ 2³x³ + 36x² + 54x + 3³
⇒ (2x)³ + 3(2x)²)(3) + 3(2x)(3)² + (3)³
By using formula:
[∵ (a³ + 3a²b + 3ab² + b³) = (a + b)³]
⇒ (2x + 3)³ Ans.

2. Find The Factors Of:
(i) d³ - 6d²c + 12dc² - 8c³
Solution:
⇒ d³ - 6d²c + 12dc² - 8c³

 (Rough Work: 8 = 2 x 2 x 2 = 2³)

⇒ (d)³ + 3(d)²(-2c) + 3(d)(-2c)² + (-2c)³
⇒ (d)³ - 3(d)²(2c) + 3(d)(2c)² - (2c)³
By using formula:
[∵ (a³ - 3a²b + 3ab² - b³) = (a - b)³]
⇒ (d - 2c)³ Ans.

(vi) 125z³ - 75z²y² + 15zy⁴ - y⁶
Solution:
⇒ 125z³ - 75z²y² + 15zy⁴ - y⁶

 Rough Work:
125 = 5 x 5 x 5 = 5³

⇒ 5³z³ - 75z²y² + 15zy⁴ + (-y²)³
⇒ (5z)³ + 3(5z)²)(-y²) + 3(5z)(-y²)² + (-y²)³
By using formula:
[∵ (a³ - 3a²b + 3ab² - b³) = (a - b)³]
⇒ (5z - y²)³ Ans.

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:

• Step 1: Find prime factors of 84 = 2 x 2 x 3 x 7
• Step 2: As the last term has minus sign so middle term will obtained by subtracting two numbers from prime factors of 84.
  i.e (7 x 3) - (2 x 2) = 21 - 4 = 17
• Step 3: As mid term has minus sign so greater value has minus sign (-21t) & smaller value 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 For 1^st Bracket:

• Step 1:Find prime factors of 21 = 3 x 7
• Step 2: 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 = 4
• Step 3:As mid term has minus sign so greater value has minus sign (-7x) & smaller value 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:

• Step 1: Find prime factors of 21 = 3 x 7
• Step 2: As the last term has plus sign so middle term will obtained by sum of two numbers from prime factors of 21.
  i.e 7 + 3 = 10
• Step 3: As mid term has plus sign so both values have plus sign (+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:

• Step 1: Find prime factors of 30 = 2 x 3 x 5
• Step 2: As the last term has minus sign so middle term will obtained by subtracting two numbers from prime factors of 30.
  i.e (5 x 2) - 3 = 10 - 3 = 7
• Step 3: As mid term has plus sign so greater value has plus sign (+10t) & smaller value has plus 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 For 2^nd Bracket:

• Step 1:Find prime factors of 3 = 3 x 1
• Step 2: As the last term has minus sign so middle term will obtained by subtracting two numbers from prime factors of 3.
  i.e 3 - 1 = 2
• Step 3:As mid term has minus sign so greater value has minus sign (-3x) & smaller value 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:

• Step 1: Find prime factors of 12 = 2 x 2 x 3
• Step 2: As the last term has minus sign so middle term will obtained by subtracting of two numbers from prime factors of 12.
  i.e (3 x 2) - 2 = 6 - 2 = 4
• Step 3: As mid term has plus sign so greater value has plus sign (+6t) & smaller value has plus 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:

• Step 1: Find prime factors of 96 = 2 x 2 x 2 x 2 x 2 x 3
• Step 2: As the last term has minus sign so middle term will obtained by subtracting of two numbers from prime factors of 96.
  i.e (2 x 2 x 2 x 2) - (2 x 3) = 16 - 6 = 10
• Step 3: As mid term has plus sign so greater value has plus sign (+16t) & smaller value has plus 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:

• Step 1: Find prime factors of 24 = 2 x 2 x 2 x 3
• Step 2: As the last term has minus sign so middle term will obtained by subtracting two numbers from prime factors of 24.
  i.e (3 x 2 x 2) - 2 = 12 - 2 = 10
• Step 3: As mid term has plus sign so greater value has plus sign (+12t) & smaller value has plus 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:

• Step 1: Find prime factors of 96 = 2 x 2 x 2 x 2 x 2 x 3
• Step 2: As the last term has plus sign so middle term will obtained by adding two numbers from prime factors of 96.
  i.e (2 x 2 x 2 x 2) + (2 x 3) = 16 + 6 = 22
• Step 3: As mid term has plus sign so both values 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:

• Step 1:Find prime factors of 6 = 6 x 1
• Step 2: As the last term has plus sign so middle term will obtained by adding two numbers from prime factors of 6.
  i.e 6 + 1 = 7
• Step 3:As mid term has plus sign so both values 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:

• Step 1: Find prime factors of 75 = 3 x 5 x 5
• Step 2: As the last term has minus sign so middle term will obtained by subtracting two numbers from prime factors of 24.
  i.e (3 x 5) - 5 = 15 - 5 = 10
• Step 3: As mid term has plus sign so greater value has plus sign (+15t) & smaller value has plus 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:

• Step 1: Find prime factors of 960 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 5
• Step 2: As the last term has plus sign so middle term will obtained by adding two numbers from prime factors of 960.
  i.e (2 x 2 x 2 x 2 x 2) + (2 x 3 x 5) = 32 + 30 = 62
• Step 3: As mid term has plus sign so both values have plus sign i.e. (+32t) & (+30t).

⇒ 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:

• Step 1:Find prime factors of 32 = 2 x 2 x 2 x 2 x 2
• Step 2: As the last term has plus sign so middle term will obtained by adding two numbers from prime factors of 32.
  i.e (2 x 2 x 2) + (2 x 2) = 8 + 4 = 12
• Step 3:As mid term has minus sign so both values 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:

• Step 1: Find prime factors of 200 = 2 x 2 x 2 x 5 x 5
• Step 2: 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 2 x 5) - (2 x 5) = 20 - 10 = 10
• Step 3: As mid term has plus sign so greater value has plus sign (+20t) & smaller value has plus 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:

• Step 1: Find prime factors of 135 = 3 x 3 x 3 x 5
• Step 2: As the last term has minus sign so middle term will obtained by subtracting two numbers from prime factors of 24.
  i.e (3 x 3 x 3) - 5 = 27 - 5 = 22
• Step 3: As mid term has plus sign so greater value has plus sign (+27t) & smaller value has plus 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:

• Step 1: Find prime factors of 36 = 2 x 2 x 3 x 3
• Step 2: As the last term has plus sign so middle term will obtained by adding two numbers from prime factors of 35.
  i.e (2 x 2 x 3) + 3 = 12 + 3 = 15
• Step 3: As mid term has minus sign so both values has minus sign as (-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:

• Step 1: Find prime factors of 14 = 1 x 2 x 7
• Step 2: As the last term has minus sign so middle term will obtained by subtracting two numbers from prime factors of 14.
  i.e (7 x 2) - 1 = 14 - 1 = 13
• Step 3: As mid term has minus sign so greater value has minus sign (-14x²) & smaller value 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:

• Step 1: Find prime factors of 64 = 2 x 2 x 2 x 2 x 2 x 2
• Step 2: As the last term has plus sign so middle term will obtained by adding two numbers from prime factors of 64.
  i.e (2 x 2 x 2 x 2 x 2) + 2 = 32 + 2 = 34
• Step 3: As mid term has minus sign so both values has minus sign as (-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:

• 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

Unit 4: Factorization - Explanation & Examples Of Exercise 4.3, 4.4 & 4.5 - Mathematics For Class IX (Science Group)

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Unit 4: Factorization
Example Of Exercise 4.3

Example Of Exercise 4.4

Example Of Exercise 4.5

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

Go To Index
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

Unit 4: Factorization - Explanation & Examples Of Exercise 4.2 - Mathematics For Class IX (Science Group)

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Unit 4: Factorization
Explanation Of Exercise 4.2

FACTORIZATION EXPRESSIONS:
Those algebraic expressions which are neither perfect square nor in the form of the difference of two squares. Such expressions are factorize as:

Type 1: a⁴ + a²b² + b⁴ OR a⁴ + 4b⁴
Such expression is factorize by Adding Or Subtracting Middle Term
Middle term is added or subtracted to make an expression a perfect square.
Examples:

Type 2: x² + px + q
Factorize The Expression by Breaking Middle Term

Type 3: ax² + bx + c, a ≠ 0
To factorize The expression ax² + bx + c, a ≠ 0, the following steps are needed:

1. Find the product ac, where a is coefficient of x² and c is constant.
2. Find the two numbers x₁ and x₂ such that x₁ + x₁ = b and x₁.x₂ = ac.

Unit 4: Factorization - Mathematics For Class IX (Science Group) - Solved Exercise 4.1

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Unit 4: Factorization
Solved Exercise 4.1

1. Factorize the following:
(i) 4x + 16y + 4z
Solution:
⇒ 4x + 16y + 4z [taking 4 as a common from the expressions]
⇒ 4(x + 4y + z) Ans.

ii) x² + 3x²y + 4x²y²z
Solution:
⇒ x² + 3x²y + 4x²y²z [taking x² as a common from the expressions]
⇒ x²(1 + 3y + 4y²z) Ans.

iii) 3pqr + 6qpt + 3pqs
Solution:
⇒ 3pqr + 6qpt + 3pqs [taking 3pq as a common from the expressions]
⇒ 3pq(r + 2t + s) Ans.

iv) 9qr(s² + t²) + 18q²r²(s² + t²)
Solution:
⇒ 9qr(s² + t²) + 18q²r²(s² + t²) [taking 9qr(s² + t²) as a common from the expressions]
⇒ 9qr(s² + t²)(1 + 2qr Ans).

vi) a(x - y) - a²b(x - y) + a²b²(x - y)
Solution:
⇒ a(x - y) - a²b(x - y) + a²b²(x - y) [taking a(x - y) as a common from the expressions]
⇒ a(x - y)(1 - ab + ab²) Ans

2. Factorize the following:
i) 7x + xz + 7z + z²
Solution:
⇒ 7x + xz + 7z + z² [Taking x in the first pair & z in the second pair as a common from the expressions]
⇒ x(7 + z) + z(7 + z)
⇒ (x + z)(7 + z) are required factors Ans.

ii) 9a²b + 18ab² - 6ac - 12bc
Solution:
⇒ 9a²b + 18ab² - 6ac - 12bc [Taking 9ab in the first pair & -6c in the second pair as a common from the expressions]
⇒ 9ab(a + 2b) - 6c(a + 2b)
⇒ (9ab - 6c)(a + 2b) are required factors Ans.

OR

⇒ 9a²b + 18ab² - 6ac - 12bc [Taking 3 as a common from the expressions]
⇒ 3(3a²b + 6ab² - 2ac - 4bc) [Now taking 3ab in the first pair & -2c in the second pair as a common from the expressions]
⇒ 3{3ab(a + 2b) - 2c(a + 2b)}
⇒ 3(3ab - 2c)(a + 2b)
⇒ (9ab - 6c)(a + 2b) are required factors Ans.

iii) 6t - 12p + 4tq - 8pq
Solution:
⇒ 6t - 12p + 4tq - 8pq [Taking 6 in the first pair & +4q in the second pair as a common from the expressions]
⇒ 6(t - 2p) + 4q(t - 2q)
⇒ (6 + 4q)(t - 2p) are required factors Ans.

(iv) r² + 9rs - 7rs - 63s²
Solution:
⇒ r² + 9rs - 7rs - 63s² [Taking r in the first pair & -7s in the second pair as a common from the expressions]
⇒ r(r + 9s) - 7s(r + 9s)
⇒ (r - 7s)(r + 9s) are required factors Ans.

3. Find the factors of:
i) 4a² + 12ab + 9b²
Solution:
By using formula: [∵ a² + 2ab + b² = (a + b)²] a perfect square
⇒ (2a)² + 2(2a)(3b) + (3b)²
⇒ (2a + 3b)² Ans

ii) 36x⁴ + 12x² + 1
Solution:
By using formula: [∵ a² + 2ab + b² = (a + b)²] a perfect square
⇒ (6x²)² + 2(6x²)(1) + (1)²
⇒ (6x² + 1)² Ans

iv) 81y² + 144yz + 64z²
Solution:
By using formula: [∵ a² + 2ab + b² = (a + b)²] a perfect square
⇒ (9y)² + 2(9y)(8z) + (8z)²
⇒ (9y + 8z)² Ans

v) 625 + 50a²b + a⁴b²
Solution:
By using formula: [∵ a² + 2ab + b² = (a + b)²] a perfect square
⇒ (25)² + 2(25)(a²b) + (a²b)²
⇒ (25 + a²b)² Ans

vi) a² + 0.4a + 0.04
Solution:
By using formula: [∵ a² + 2ab + b² = (a + b)²] a perfect square
⇒ (a)² + 2(a)(0.2) + (0.2)²
⇒ (a + 0.2)² Ans

4. Factorize:
i) b⁴ - 4b²c² + 4c⁴
Solution:
By using formula: [∵ a² - 2ab + b² = (a - b)²] a perfect square
⇒ (b²)² - 2(b²)(2c²) + (2c²)²
⇒ (b² - 2c²)² Ans

iii) 2a³b³ - 16a²b⁴ + 32ab⁵
Solution:
⇒ 2a³b³ - 16a²b⁴ + 32ab⁵ [By taking 2ab³ as a common from the expression]
⇒ 2ab³(a² - 8ab + 16b²)
By using formula: [∵ a² - 2ab + b² = (a - b)²] a perfect square
⇒ 2ab³{(a)² - 2(a)(4b) + (4b)²)}
⇒ 2ab³(a - 4b)² Ans

iv) 9(p+q)² - 6(p+q)r² + r⁴
Solution:
By using formula: [∵ a² - 2ab + b² = (a - b)²] a perfect square
⇒ {3(p+q)}² - 2{3(p+q)}(r²) + (r²)²
⇒ {3(p+q) - r²}² Ans

v) x²y² - 0.1xy + 0.0025
Solution:
By using formula: [∵ a² - 2ab + b² = (a - b)²] a perfect square
⇒ (xy)² - 2(xy)(0.05) + (0.05)²
⇒ (xy - 0.05)² Ans

vi) (a - b)² + 18(a - b) + 81
Solution:
By using formula: [∵ a² + 2ab + b² = (a + b)²] a perfect square
⇒ (a - b)² + 2(a - b)(9) + (9)²
⇒ {(a - b) - 9}² Ans

5. Factorize:
i) 4a² - 9b²
Solution:
By using formula: [∵ a² - b² = (a - b)(a + b)] difference of two squares
⇒ (2a)² - (3b)²
⇒ (2a - 3b)(2a + 3b) Ans

ii) 16x² - 25y²
Solution:
By using formula: [∵ a² - b² = (a - b)(a + b)] difference of two squares
⇒ (4x)² - (5y)²
⇒ (4x - 5y)(4x + 5y) Ans

iii) 100x²z² - y⁴
Solution:
By using formula: [∵ a² - b² = (a - b)(a + b)] difference of two squares
⇒ (10xz)² - (y²)²
⇒ (10xz - y²)(10xz + y²) Ans

6. Find the factors of:
i) (2x + z)² - (2x - z)²
Solution:
By using formula: [∵ a² - b² = (a - b)(a + b)] difference of two squares
⇒ (2x + z)² - (2x - z)²
⇒ {(2x + z) - (2x - z)} {(2x + z) + (2x - z)} Ans

ii) (4a - 9b)² - (2a + 5b)²
Solution:
By using formula: [∵ a² - b² = (a - b)(a + b)] difference of two squares
⇒ (4a - 9b)² - (2a + 5b)²
⇒ {(4a - 9b) - (2a + 5b)} {(4a - 9b) + (2a + 5b)} Ans

iii) 169x⁴ - (3t + 4)²
Solution:
By using formula: [∵ a² - b² = (a - b)(a + b)] difference of two squares
⇒ (13x²)² - (3t + 4)²
⇒ {13x² - (3t + 4)} {13x² + (3t + 4)]}Ans

iv) (9x² - 4y²)² - (4x² - y²)²
Solution:
By using formula: [∵ a² - b² = (a - b)(a + b)] difference of two squares
⇒ (9x² - 4y²)² - (4x² - y²)²
⇒ {(3x)² - (2y)²)}² - {(2x)² - (y²)}²
⇒ {(3x - 2y)(3x + 2y)}² - {(2x - y)(2x + y)}²
[again by using formula: a² - b² = (a - b)(a + b)]
⇒ {(3x - 2y)(3x + 2y) - (2x - y)(2x + y)} {(3x - 2y)(3x + 2y) + (2x - y)(2x + y)} Ans

OR

Solution:
By using formula: [∵ a² - b² = (a - b)(a + b)] difference of two squares
⇒ (9x² - 4y²)² - (4x² - y²)²
⇒ {(9x² - 4y²) - (4x² - y²)} {(9x² - 4y²) + (4x² - y²)}
[again by using formula: a² - b² = (a - b)(a + b)]
⇒ {(3x)² - (2y)²)} - {(2x)² - (y)²)} {(3x)² - (2y)²)} + {(2x)² - (y)²)}
⇒ {(3x - 2y)(3x + 2y) - (2x - y)(2x + y)} {(3x - 2y)(3x + 2y) + (2x - y)(2x + y)} Ans

7. Find the factors of:
i) (x² + 2xy + y²) - 9z⁴
Solution:
By using formula: [∵ a² + 2ab + b² = (a + b)²] a perfect square
⇒ (x)² + 2(x)(y) + (y)²) - (3z²)²
⇒  (x + y)² - (3z²)²
[again by using formula: a² - b² = (a - b)(a + b)] Difference of two squares
⇒ {(x + y) - 3z²} {(x + y) + 3z²}
⇒ (x + y - 3z²)(x + y + 3z²) Ans

ii) (4a² + 8ab² + 4b⁴) - 9c²
Solution:
By using formula: [∵ a² + 2ab + b² = (a + b)²] a perfect square
⇒ {(2a)² + 2(2a)(2b²) + (2b²)²)} - (3c)²
⇒ (2a + 2b²)² - (3c)²
[Again by using formula: a² - b² = (a - b)(a + b)] Difference of two squares
⇒ {(2a + 2b²) - 3c} {(2a + 2b²) + 3c}
⇒ (2a + 2b² - 3c)(2a + 2b² + 3c) Ans

iii) 16d⁴ - (c⁴ - 2c²d + d²)
Solution:
By using formula: [∵ a² - 2ab + b² = (a - b)²] a perfect square
⇒ (4d²)² - {(c²)² - 2(c²)(d) + (d)²)}
⇒ ∴ (4d²)² - (c² - d)²
[Again by using formula: a² - b² = (a - b)(a + b)] Difference of two squares
⇒ {4d² - (c² - d)} {4d² + (c² - d)
⇒ (4d² - c² + d)(4d² + c² - d) Ans

iv) 4(x² + 2xy² + y⁴) - 9y⁶
Solution:
By using formula: [∵ a² + 2ab + b² = (a + b)²] a perfect square
⇒ 4{(x)² + 2(x)(y²) + (y²)²} - (3y³)²
⇒ ∴ 4{(x + y²)²} - (3y³)²
[Again by using a formula: a² - b² = (a - b)(a + b)] Difference of two squares
⇒ {2(x + y²)}² - (3y³)²
⇒ {2(x + y²) + 3y³} {2(x + y²) - 3y³}Ans

v) x² - y² - 4x - 2y + 3
Solution:
[we rearrange the terms]
⇒ (x² - 4x + 3) - y² - 2y [Taking -1 as common from second expression]
⇒ (x² - 4x + 3) - (y² + 2y)
[Add 1 in both the terms expression]
⇒ (x² - 4x + 3 + 1) - (y² + 2y +1)
⇒ (x² - 4x + 4) - (y² + 2y +1)
By using formula: [∵ a² 土 2ab + b² = (a + b)²] a perfect square
⇒ {(x)² - 2(x)(2) + (2)²} - {(y)² + 2(y)(1) +(1)²}
⇒ (x - 2)² - (y + 1)²
[Again by using formula: a² - b² = (a - b)(a + b)] Difference of two squares
⇒ {(x - 2) + (y + 1)} {(x - 2) - (y + 1)} Ans

vi) 4x² - y² - 2y - 1
Solution:
⇒ 4x² - (y² + 2y + 1) [Taking -1 as common from second expression]
By using formula: [∵ a² + 2ab + b² = (a + b)²] a perfect square
⇒ 4x² - {(y)² + 2(y)(1) + (1)²}
⇒ 4x² - (y + 1)²
[Again by using formula: a² - b² = (a - b)(a + b)] Difference of two squares
⇒ (2x)² - (y + 1)²
⇒ {(2x - (y + 1)} {(2x + (y + 1)}
⇒ (2x - y - 1)(2x + y + 1) Ans

8. Factorize:
i) (√ ab)²z² - (√ c)²
Solution:
By using formula: [∵ a² - b² = (a - b)(a + b)] difference of two squares
⇒ (√ abz)² - (√ c)²
⇒ (√ abz - √ c)(√ abz + √ c) Ans

i) (√ 4x)² - (√ 9y)²
Solution:
By using formula: [∵ a² - b² = (a - b)(a + b)] difference of two squares
⇒ (√ 4x)² - (√ 9y)²
⇒ (√ 4x - √ 9y)(√ 4x + √ 9y) Ans