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Showing posts with label Maths For Class IX (SC). Show all posts
Showing posts with label Maths For Class IX (SC). Show all posts

Saturday 5 October 2024

Unit 4: Factorization - Solved Exercise 4.6 - Mathematics For Class IX (Science Group)

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Unit 4: Factorization
Solved Exercise 4.6

1. Find the remainder byusing the remainder theorem when:
(i) x3 - 6x2 + 11x - 8 is divided by (x - 1)
Solution:
By the given condition:
⇒ x - 1 = 0
∴ x = 1
Here
⇒ P(x) = x3 - 6x2 + 11x - 8
By substituting value of x, we get:
⇒ P(1) = 13 - 6(12) + 11(1) - 8
⇒ P(1) = 1 - 6 + 11 - 8
⇒ P(1) = 1 + 11 - 6 - 8
⇒ P(1) = 12 - 14
⇒ P(1) = -2
Thus the Remainder is -2 Answer

(ii) x3 + 6x2 + 11x + 8 is divided by (x + 1)
Solution:
By the given condition:
⇒ x + 1 = 0
∴ x = -1
Here
⇒ P(x) = x3 + 6x2 + 11x + 8
By substituting value of x, we get:
⇒ P(-1) = (-1)3 + 6(-12) + 11(-1) + 8
⇒ P(-1) = - 1 + 6 - 11 + 8
⇒ P(-1) = - 1 - 11 + 6 + 8
⇒ P(-1) = - 12 + 14
⇒ P(-1) = 2
Thus the Remainder is 2 Answer

(iii) x3 - x2 - 26 + 40 is divided by (x - 2)
Solution:
By the given condition:
⇒ x - 2 = 0
∴ x = 2
Here
⇒ P(x) = x3 - x2 - 26 + 40
By substituting value of x, we get:
⇒ P(2) = (2)3 - (2)2 - 26 + 40
⇒ P(2) = 8 - 4 + 14
⇒ P(2) = 4 + 14
⇒ P(2) = 18
Thus the Remainder is 18 Answer

(iv) x3 - 3x2 + 4x - 14 is divided by (x + 2)
Solution:
By the given condition:
⇒ x + 2 = 0
∴ x = -2
Here
⇒ P(x) = x3 - 3x2 + 4x - 14
By substituting value of x, we get:
⇒ P(-2) = (-2)3 - 3(-2)2 + 4(-2) - 14
⇒ P(-2) = - 8 - 12 - 8 - 14
⇒ P(-2) = -42
Thus the Remainder is -42 Answer

(v) (2y + 1)3 + 6(3 + 4y) - 9 is divided by (2y + 1)

(vi) 4y3 - 4y + 3 is divided by (2y - 1)

(vii) (2y + 1)3 - 6(3 - 4y) - 10 is divided by (2y - 1)

(viii) x4 + x2y2 + y4 is divided by (x - y)
Solution:
By the given condition:
⇒ x - y = 0
∴ x = y
Here
⇒ P(x) = x4 + x2y2 + y4
By substituting value of x, we get:
⇒ P(y) = (y)4 + (y)2y2 + y4
⇒ P(y) = (y)4 + y2.y2 + y4
⇒ P(y) = y4 + y4 + y4
⇒ P(y) = 3y4
Thus the Remainder is 3y4 Answer

2. Find the value of m, if p(y) = my3 + 4y2 + 3y - 4 and q(y) = y3 - 4y + m leaves the same remainder when divided by (y - 3)
Solution:
By the given condition:
⇒ y - 3 = 0
∴ y = 3
Here
⇒ p(y) = my3 + 4y2 + 3y - 4
& q(y) = y3 - 4y + m
By substituting value of y, we get:
For p(y)
⇒ p(y) = m(3)3 + 4(3)2 + 3(3) - 4
⇒ P(y) = m(27) + 4(9) + 9 - 4
⇒ P(y) = 27m + 36 + 5
⇒ P(y) = 27m + 41 ..... (i)

For q(y)
⇒ q(y) = (3)3 - 4(3) + m
⇒ q(y) = 27 - 12 + m
⇒ q(y) = 15 + m ..... (ii)

Given that both reminders are same, so by equating equation (i) & (ii), we have:
⇒ 27m + 41 = 15 + m
(by subtracted m and 41 on both side)
⇒ 27m + 41 - m - 41 = 15 + m - m - 41
⇒ 27m - m + 41 - 41 = 15 - 41 + m - m
⇒ 26m = - 26
⇒ m = - 26 / 26
⇒ m = -1
Answer: Thus the value of m is -1

3. If the polynomial 4x3 - 7x2 + 6x - 3k is exactly divisible by (y + 2), find the value of k
Solution:
By the given condition:
⇒ x + 2 = 0
∴ x = -2
Here
⇒ p(x) = 4x3 - 7x2 + 6x - 3k
By substituting value of x, we get:
⇒ p(-2) = 4(-2)3 - 7(-2)2 + 6(-2) - 3k
⇒ p(-2) = 4(-8) - 7(4) - 12 - 3k
⇒ p(-2) = - 32 - 28 - 12 - 3k
⇒ p(-2) = - 72 - 3k ..... (i)
For exactly division
⇒ p(-2) = 0 .....(ii)
By equating equation (i) & (ii), we get:
⇒ - 72 - 3k = 0
⇒ - 72 + 72 - 3k = 0 + 72
⇒ - 72 + 72 - 3k = 0 + 72
⇒ -3k = 72 (Divide both side by -3)
⇒ -3k / -3 = 72 / -3
⇒ k = -24
Answer: Thus the value of k is -24

4. Find the value of r, if (y + 2) is a factor of the polynomial 3y2 - 4ry - 4r2.
Solution:
By the given condition:
⇒ y + 2 = 0
∴ x = -2
Here
⇒ p(y) = 3y2 - 4ry - 4r2
& p(-2) = 0
Then 0 = 3y2 - 4ry - 4r2
By substituting value of y, we get:
⇒ 0 = 3(-2)2 - 4r(-2) - 4r2
⇒ 0 = 3(4) + 8r - 4r2
⇒ 0 = 12 + 8r - 4r2
⇒ 0 = -4(-3 - 2r + r2)

(Divide both side by -4)
⇒ 0 / -4 = -4(-3 - 2r + r2) / -4
OR  r2 - 2r - 3 = 0 (by arranging in descending order)
(It can be factorize by breaking method)

 Rough Work:
 As last term has minus sign so middle term is obtained by subtracting two numbers which are factors of last term i.e. 3
 Factor of 3 = 1, 3
 ∴  3 - 1 = 2
 As middle term has minus sign so greater number contain minus sign
⇒ r2 - 3r + r - 3 = 0
⇒ r(r - 3) + 1(r - 3) = 0
⇒ (r - 3)(r + 1) = 0
Let
 ⇒ r - 3 = 0
 ⇒ r = 0 + 3
 ⇒ r = 3
 ⇒ r + 1 = 0
 ⇒ r = 0 - 1
 ⇒ r = -1

Answer: Thus the value of r is 3 and -1.


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Saturday 28 September 2024

Unit 4: Factorization - Solved Exercise 4.5 - Mathematics For Class IX (Science Group)

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Unit 4: Factorization
Solved Exercise 4.5

1. Factorize the following:
(i) x3 + 8y3
Solution:
⇒ x3 + 8y3
⇒ (x)3 + (2y)3
By Using Formula:
[a3 + b3 = (a + b)(a2 - ab + b2) ]
⇒ (x + 2y)[(x)2 - (x)(2y) + (2y)2]
⇒ (x + 2y)(x2 - 2xy + 4y2)
Therefore,
⇒ x3 + 8y3 = (x + 2y)(x2 - 2xy + 4y2) Ans.

(ii) a11 + a2b9
Solution:
⇒ a11 + a2b9
By Taking a2 as a common from the expression,
⇒ a2(a9 + b9)
⇒ a2{(a3)3 + (b3)3)
By Using Formula:
[a3 + b3 = (a + b)(a2 - ab + b2) ]
⇒ a2{(a3 + b3)[(a3)2 - (a3)(b3) + (b3)2]}
⇒ a2(a3 + b3)(a6 - a3b3 + b6)
⇒ a2{(a)3 + (b)3)}(a6 - a3b3 + b6)
Again By Using Formula:
[ a3 + b3 = (a + b)(a2 - ab + b2) ]
⇒ a2{(a + b)[(a)2 - (a)(b) + (b)2]}(a6 - a3b3 + b6)]
⇒ a2{(a + b)(a2 - ab + b2)(a6 - a3b3 + b6)
⇒ a2(a + b)(a2 - ab + b2)(a6 - a3b3 + b6)
Therefore,
⇒ a11 + a2b9 = a2(a + b)(a2 - ab + b2)(a6 - a3b3 + b6) Ans.

(iii) a6 + 1
Solution:
⇒ a6 + 1
⇒ {(a2)3 + (1)3}
By Using Formula:
[a3 + b3 = (a + b)(a2 - ab + b2) ]
⇒ (a2 +1)[(a2)2 - (a2)(1) + (1)2]
⇒ (a2 + 1)(a4 - a2 + 1
Therefore,
⇒ a6 + 1 = (a2 + 1)(a4 - a2 + 1 Ans

(iv) a3b3 + 512
Solution:
⇒ (ab)3 + (8)3
By Using Formula:
[a3 + b3 = (a + b)(a2 - ab + b2)]
⇒ (ab +8)[(ab)2 - (ab)(8) + (8)2]
⇒ (ab + 8)(a2b2 - 8ab + 64
Therefore,
⇒ a3b3 + 512 = (ab + 8)(a2b2 - 8ab + 64) Ans

(v) a3b3 + 27b6
Solution:
⇒ a3b3 + 27b6
By Taking b3 as a common from the expression,
⇒ b3(a3 + 27b3)
⇒ b3{(a)3 + (3b)3}
By Using Formula:
[a3 + b3 = (a + b)(a2 - ab + b2)]
⇒ b3{(a +3b)[(a)2 - (a)(3b) + (3b)2]}
⇒ b3(a + 3b)(a2 - 3ab + 9b2)
Therefore,
⇒ a3b3 + 27b6 = b3(a + 3b)(a2 - 3ab + 9b2) Ans


(vii) x9 + x3y6z9
Solution:
⇒ x9 + x3y6z9
By Taking x3 as a common from the expression,
⇒ x3(x6 + y6z9)
⇒ x3{(x2)3 + (y2z3)3}
By Using Formula:
[a3 + b3 = (a + b)(a2 - ab + b2)]
⇒ x3(x2 + y2z3)[(x2)2 - (x2)(y2z3) + (y2z3)2]
⇒ x3(x2 + y2z3)(x4 - x2y2z3 + y4z6)
Therefore,
⇒ x9 + x3y6z9 = x3(x2 + y2z3)(x4 - x2y2z3 + y4z6) Ans.


2. Find the factors of:
(i) x3 - 8y3
Solution:
⇒ x3 - 8y3
⇒ (x)3 - (2y)3
By Using Formula:
[a3 - b3 = (a - b)(a2 + ab + b2)]
⇒ (x - 2y)[(x)2 + (x)(2y) + (2y)2]
⇒ (x - 2y)(x2 + 2xy + 4y2)
Therefore,
⇒ x3 - 8y3 = (x - 2y)(x2 + 2xy + 4y2) Ans.

(ii) x9 - 8y9
Solution:
⇒ x9 - 8y9
⇒ (x3)3 - (2y3)3
By Using Formula:
[a3 - b3 = (a - b)(a2 + ab + b2)]
⇒ (x3 - 2y3)[(x3)2 + (x3)(2y3) + (2y3)2]
⇒ (x3 - 2y3)(x6 + 2x3y3 + 4y6)
Therefore,
⇒ x9 - 8y3 = (x3 - 2y3)(x6 + 2x3y3 + 4y6) Ans.


(iv) a6 - b6
Solution:
By Using Formula of difference of two squares
[a2 - b2 = (a - b)(a + b)]
⇒ (a3)2 - (b3)2
⇒ (a3 - b3)(a3 + b3)
By Using Formula:
[a3 - b3 = (a - b)(a2 + ab + b2)] & [a3 + b3 = (a + b)(a2 - ab + b2)]
⇒ {(a)3) - (b)3}{(a)3 + (b)3}
⇒ {(a - b)[(a)2 + (a)(b) + (b)2)]}{(a + b)[(a)2 - (a)(b) + (b)2)]}
⇒ (a - b)(a2 + ab + b2)(a + b)(a2 - ab + b2)
Therefore,
⇒ a6 - b6 = (a - b)(a2 + ab + b2)(a + b)(a2 - ab + b2) Ans.


(vi) a12 - b12
Solution:
By Using Formula of difference of two squares
[a2 - b2 = (a - b)(a + b)]
⇒ (a6)2 - (b6)2
⇒ (a6 - b6)(a6 + b6)
Again byy Using Formula of difference of two squares
[a2 - b2 = (a - b)(a + b)]
⇒ {(a3)2 - (b3)2}(a6 + b6)
⇒ (a3 - b3)(a3 + b3)(a6 + b6)
By Using Formula:
[a3 - b3 = (a - b)(a2 + ab + b2)] & [a3 + b3 = (a + b)(a2 - ab + b2)]
⇒ {(a)3) - (b)3}{(a)3 + (b)3}{(a2)3 + (b2)3}
⇒ {(a - b)[(a)2 + (a)(b) + (b)2)]}{(a + b)[(a)2 - (a)(b) + (b)2)]}{(a2 + b2)[(a2)2 - (a2)(b2) + (b2)2)]}
⇒ (a - b)(a2 + ab + b2)(a + b)(a2 - ab + b2)(a2 + b2)(a4 - a2b2 + b4)
Therefore,
⇒ a12 - b12 = (a - b)(a + b)(a2 + ab + b2)(a2 - ab + b2) (a2 + b2)(a4 - a2b2 + b4) Ans.
OR

(vi) a12 - b12
Solution:
⇒ a12 - b12
By Using Formula:
[a3 - b3 = (a - b)(a2 + ab + b2)]
⇒ (a4)3 - (b4)3
⇒ (a4 - b4){(a4)2 + (a4)(b4) + (b4)2}
⇒ (a4 - b4)(a8 + a4b4 + b8)
In Text Book this is answer but it can factorize more
By Using Formula of difference of two squares for first bracket
[a2 - b2 = (a - b)(a + b)] &
Add and subtract a4b4 in last bracket we get:
⇒ {(a2)2 - (b2)2}{(a8 + a4b4 + b8) - a4b4 + a4b4}
⇒ {(a2 - b2)(a2 + b2)}{(a8 + 2a4b4 + b8) - a4b4}
Again by Using Formula of difference of two squares for first bracket
[a2 - b2 = (a - b)(a + b)] &
And perfect square formula for last bracket
[a2 + 2ab + b2) = (a + b)2]
⇒ {(a)2 - (b)2)}(a2 + b2){(a4)2 + 2(a4)(b4) + (b4)2 - a4b4}
⇒ (a - b)(a + b)(a2 + b2){(a4 + b4)2 - (a2b2)2}
Again by Using Formula of difference of two squares for last bracket
[a2 - b2 = (a - b)(a + b)]
⇒ (a - b)(a + b)(a2 + b2)(a4 + b4 - a2b2)(a4 + b4 + a2b2)
⇒ (a - b)(a + b)(a2 + b2)(a4 - a2b2 + b4)(a4 + a2b2 + b4)
Add and subtract a2b2 in last bracket we get:
⇒ (a - b)(a + b)(a2 + b2)(a4 - a2b2 + b4){(a4 + a2b2 + b4) - a2b2 + a2b2}
⇒ (a - b)(a + b)(a2 + b2)(a4 - a2b2 + b4){(a4 + 2a2b2 + b4) - a2b2}
Again by Using perfect square formula for last bracket
[a2 + 2ab + b2) = (a + b)2]
⇒ (a - b)(a + b)(a2 + b2)(a4 - a2b2 + b4){[(a2)2 + 2(a2)(b2) + (b2)2] - (ab)2}
⇒ (a - b)(a + b)(a2 + b2)(a4 - a2b2 + b4){(a2 + b2)2 - (ab)2}
Again by Using Formula of difference of two squares for last bracket
[a2 - b2 = (a - b)(a + b)]
⇒ (a - b)(a + b)(a2 + b2)(a4 - a2b2 + b4)(a2 + b2 - ab)(a2 + b2 + ab)
⇒ (a - b)(a + b)(a2 + b2)(a4 - a2b2 + b4)(a2 - ab + b2)(a2 + ab + b2)
Therefore,
⇒ a12 - b12 = (a - b)(a + b)(a2 + b2)(a2 + ab + b2)(a2 - ab + b2) (a4 + a2b2 + b4) Ans.




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Wednesday 18 September 2024

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.




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Tuesday 3 September 2024

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:
* 1st Bracket: By breaking middle term &
* 2nd 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 1st Bracket but 2nd Bracket can be factorize by breaking middle term as:

ROUGH WORK WORK For 2nd 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 1st Bracket but 2nd Bracket can be factorize by breaking middle term as:

ROUGH WORK for 2nd 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 1st Bracket by breaking middle term but 2nd Bracket can not be factorize:

ROUGH WORK For 1st 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)
1st 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)
1st Bracket can not be factorize further.
2nd 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




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Sunday 1 September 2024

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

Go To Index
Unit 4: Factorization
Example Of Exercise 4.3



Example Of Exercise 4.4


Example Of Exercise 4.5

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Monday 19 August 2024

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:
  • 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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