# FACTORIZATION EXERCISE 3.2.4 – Class 9

1. Resolve into factors

(i) x4 + y4 – 7x2 y2

Solution:

= x4 + y4 – 2x2 y2 – 2x2 y2 – 7x2 y2

= x4 + y4 + 2x2 y2 – 9x2 y2

= (x2 + y2)2 – (3xy)2

= (x2 + y2 + 3xy) (x2 + y2 – 3xy)

(ii) 4x4 + 25y4 + 10x2y2

Solution:

Take a = √2x ;b = √5y

a4 + b4 + a2 b2 = (a2 + b2 + ab) (a2 + b2 – ab)

4x4 + 25y4 + 10x2y2 = (√2 x)4 + (√5 y)4 + 2 x2 5 y2

= [( 2 x2) + ( 5 y2) + (√10 xy] [( 2 x2) + ( 5 y2) – (√10 xy]

= (2x2 +5y2 + 10 xy) (2x2 +5y2 – 10 xy)

(iii)9a4+100b4+30a2b2

Solution:

Take a = √3a ;b = √10b

a4 + b4 + a2 b2 = (a2 + b2 + ab) (a2 + b2 – ab)

9a4+100b4+30a2b2 = (√3a)4+(√10b)4+3a2.10b2

= (3a2+10b2+30a2b2) (3a2+10b2– 30a2b2)

(iv)81a4 +9a2b2+b4

Solution:

Take a = 3a ;b = b

a4 + b4 + a2 b2 = (a2 + b2 + ab) (a2 + b2 – ab)

81a4 +9a2b2+b4 = (3a)4+(3a)2(b)2+b4

= (9a2+b2+3ab)(9a2+b2-3ab)

(v) x4 – 6x2y2 + y4

= x4 + y4 + 2x2y2 – 2x2y2 – 6x2y2

= (x2 + y2)2 – ( 8 xy)2

= (x2 + y2 + 8 xy) (x2 + y2 – 8 xy)

1. vi) m4 + n4 – 18m2n2

= m4 + n4 + 2m2n2 – 2m2n2 – 18m2n2

= (m2 + n2)2 – 20 m2n2

= (m2 + n2 + √20 mn) (m2 + n2 – √20 mn)

(vii) 4m4 + 9n4 – 24m2n2

= 4m4 + 9n4 + 12 m2n2 – 12 m2n2 – 24m2n2

= (2m2 + 3n2)2 – 36mn

= (2m2 + 3n2 + 6mn) (2m2 + 3n2 – 6mn)

(viii) 9x4 + 4y4 + 11x2y2

= 9x4 + 4y4 + 12x2y2 – 12x2y2 + 11x2y2

= (3x2)2 + (2y2)2 + 2 (3x)2 (2y)2 – x2y2

= (3x2 + 2y2)2 – x2y2

= (3x2 + 2y2 + xy) (3x2 + 2y2 – xy)

1. Find the factors of the following

(i) x4 + 9x2+81

Solution:

= x4+9x2+81+9x2-9x2

= (x2)2+18x2+81-9a2

=(x2+9)2-(3a)2

=( x2+9 +3a)( x2+9 – 3a)

(ii) a4 + 4a2 + 16

Solution:

= a4 + 4a2 + 16 + 4a2 – 4a2

= (a2)2 + 8a2 + (4)2 – 4a2

= (a2 + 4)2 – (2a)2

= (a2 + 4 – 2a) (a2 + 4 + 2a)

1. Factorise the following

(i) 64a4 + 1

Solution:

Adding and subtracting 16a2 we get

64a4 + 1 + 16a2 – 16a2

= (8a2)2 + 1 + 16a2 – 16a2

= (8a2 + 1)2 – (4a)2

= (8a2 + 1 + 4a) (8a2 + 1 – 4a)

(ii) 3x4 + 12y4

Solution:

3(x4 + 4y4)

3[(x2)2 + (2y2)2]

By adding and subtracting 2ab i. e.

2 x x2 x 2y = 4x2y2

= 3 [(x2)2 + 2(y2)2 + 4x2y2 – 4x2y2]

= 3 [(x2 + 2y2)2 – (2xy)2]

= 3 [(x2 + 2y2 + 2xy) (x2 + 2y2 – 2xy)]

(iii) 4x4 + 81y4

Solution:

(2x2)2 + (9y2)2

By adding and subtracting 2ab i. e.

2 x x2 x 9y2 = 36x2y2

= (2x2)2 + (9y2)2 + 36x2y2 – 36x2y2

= (2x2 + 9y2)2 – (6xy)2

= (2x2 + 9y2 + 6xy) (2x2 + 9y2 – 6xy)

(iv) a8 – 16b8

Solution:

(a4)2 – (4b4)2

Using a2 – b2 = (a + b) (a – b)

= (a4 + 4b4) (a4 – 4b4)

= (a4 + 4b4) [(a2)2 – (2b2)2]

= (a4 + 4b4) (a2 + 2b2) (a2 – 2b2)