FACTORIZATION EXERCISE 3.2.5 – Class 9

  1. Factorise:

(i) 8y3 – 1

Solution:

= (2y)3 – 13 [∵ a3 – b3 = (a – b) (a2 + ab + b2)]

= (2y – 1) [(2y)2 + 2y .1 + 12]

= (2y – 1) (4y2 + 2y + 1)

 

(ii) 27x3 – 8

Solution:

= (3x)3 – 23 [∵ a3 – b3 = (a – b) (a2 + ab + b2)]

= (3x – 2) [(3x)2 + 3x .2 + 22]

= (3x – 2) (9x2 + 6x + 4)

 

(iii) x3 + 8y3

Solution:

= x3 + (2y)3 [∵ a3 + b3 = (a + b) (a2 – ab + b2)]

= (x + 2y) [x2 – x.2y + (2y)2]

= (x + 2y) (x2 – 2xy + 4y)2

 

(iv) 1 – x3

Solution:

= 13 – x3 [∵a3 – b3 = (a – b) (a2 + ab + b2)]

= (1 – x) (12 + 1.x + x2)

= (1 – x) (1 + x + x2)

 

(v) a3 b3 + c3

Solution:

= (ab)3 + c3 [∵ a3 + b3 = (a + b) (a2 – ab + b2)]

= (ab + c) [(ab)2 – ab .c + c2]

= (ab + c) (a2b2 – abc + c2)

 

(vi) a3b – 𝐛/𝟔𝟒

Solution:

= b (a31/64)

= b [a3 – (1/4 )3]

= b (a – 1/4) [a2 + a. 1/4 + ( 1/4 )2]

= b (a – 14) (a2 + a/4 + 1/16 )

 

(vii) 𝐚3/𝟖 + 1

Solution:

= ( a/2 )3 + 13 [∵ a3 + b3 = (a + b) (a2 – ab + b2)]

= ( a/2 + 1) [( a/2 )2a/2 .1 + 12]

= ( a/2 + 1) [a2/4a/3+ 1]

 

(viii) 3a6 𝐛𝟔/𝟗

Solution:

= 3(a6− b6/27)

= 3[(a2)3− (b2/3)3]

= 3[(a2− b2/3) (a2)3− a2.b2/3 (b2/3)2

= 3(a2− b2/3)( a4+a2 . b2/3+ b4/3 )

 

(ix) 2a3 + 𝟏/𝟒

Solution:

= 2 (a3 + 𝟏/𝟒 )

= 2 (a3 + 𝟏/2 )3

= 2 (a + 𝟏/2 ) [ a2 – a. 𝟏/2 + (𝟏/𝟒 )2]

= 2 (a + 𝟏/2 ) (a2a/2 + 𝟏/𝟒 )

 

(x) x3 – 512

Solution:

= x3 – 83 [∵ a3 – b3 = (a – b) (a2 + ab + b2)]

= (x – 8) (x2 + 8x + 64)

 

(xi) 32x3 – 500

= 4 (8x3 – 125)

= 4 [(2x)3 – 53]

= 4 (2x – 5) [(2x)2 + 2x .5 – 52]

= 4 (2x – 5) (4x2 + 10 x + 25)

 

(xii) x7 + xy6

= x (x6 + y6)

= x [(x2)3 + (y2)3]

= x [(x2 + y2) { (x2)2 – x2 y2 + (y2)2}]

= x (x2 + y2) (x4 – x2 y2 + y2)

 

(xiii) 2a4 – 128a

= 2a (a3 – 64)

= 2a (a3 – 43)

= 2a (a – 4) (a2 + 4a + 42)

= 2a (a – 4) (a2 + 4a + 16)


  1. Factorise

(i) (1 – a)3 + (3a)3

Solution:

Using x3 + y3 = (x + y) (x2 – xy + y2)

(1 – a)3 + (3a)3

= (1 – a + 3a) [(1 – a)2 – (1 – a) 3a + (3a)2]

= (1 + 2a) [(1 – a)2 – 3a + 3a2 + 9a2]

= (1 + 2a) (1 + a2 – 2a – 3a + 12a2)

= (1 + 2a) (13a2 – 5a + 1)

 

(ii) 8x3 – 27y3

Solution:

= (2x)3 – ( 3y)3

= ( 2x – 3y) [(2x)2 + 2x .3y + (3y)2]

= (2x – 3y) (4x2 + 6xy + 9y2)

 

(iii) z4 x3 + 8y3 z4

Solution:

= z4 (x3 + 8y3)

= z4 [x3 + (2y)3]

= z4 (x + 2y) [x2 – x.2y + (2y)2]

= z4 (x + 2y) [x2 – 2xy + 4y2]

 

(iv) 3(x + y)3 + 𝟏/9 (xy)3

Solution:

=  (x + y)3 + 1/27 (xy)3

= (x + y)3 + ( xy/3 )3

= (x + y + xy/3 ) (x + y)2 – [ (x + y) ( xy/3 )+ ( xy/3)2]

= (x + y + xy/3 ) (x2 + y+ 2xy – (x + y)( xy/3) + x2y2/9 )

 

(v) x6 + y6

Solution:

= (x2)3 – (y2)3

= (x2 – y2) [(x2)3 + x2 y2– (y2)3]

= (x2 – y2) (x4 + x2 y2 + y4)

= (x + y) (x – y) (x2 + x y + y2) (x2 – xy + y2)

 

(vi) a3 – 2√𝟐 b3

Solution:

= a3 – (√2 b)3

= (a –√2b) [a2 + a.√2 b + √2 b2]

= (a –√𝟐 b) (a2 + √𝟐 ab + √𝟐b2)


  1. Factorize the following

(i) x6 – 26x3 – 27

Solution:

put x3 = a

a3 – 26a – 27

= a3 – 27a + a – 27

= a ( a – 27) + 1 (a – 27)

= (a – 27) ( a + 1)

= (x3 – 27) (x3 + 1)

= (x – 3) (x2 + 3x + 9) (x – 1) (x2 – x + 1)

 

(ii) z6 – 63z3 – 64

Solution:

put z3 = a

a3 – 63a – 64

= a3 – 64a + a – 64

= a (a – 64) + 1 (a – 64)

= (a – 64) (a + 1)

= (z3 – 64) (z3 + 1)

= (z – 4) (z2 + 4z + 16) (z + 1) (z2 – z + 1)

 

(iii) a3 – b3 – a + b

Solution:

= (a – b) (a2 + ab + b2) – (a – b)

= (a – b) [ a2 + ab + b2 – 1]

 

(iv) x6 + 7x3 – 8

Solution:

put x3 = a

a3 + 7a – 8

= a3 + 8a – a – 8

= a (a + 8) – 1 (a + 8)

= (a – 1) (a + 8)

= (x3 – 1) (x3 + 8)

= (x – 1) (x2 + x – 1) (x + 2) (x2 – 2x + 4)

 

(v) a3 𝟏/𝐚𝟑 – 2a + 𝟐𝐚

Solution:

= (a – 𝟏/𝐚) (a2 + a. 𝟏/𝐚+ 𝟏/𝐚2 ) – 2 (a – 𝟏/𝐚 )

= (a – 𝟏/𝐚 ) [a2 + 1 + 𝟏/𝐚2 – 2]

= (a – 𝟏/𝐚 ) (a2 + 𝟏/𝐚𝟐 – 1 )


 

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