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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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