# Factorization – EXERCISE 3.2.2 – Class 9

1. Factorize the following:

(i) 9x2 – 16y2

Solution:

= (3x)2 – (4y)2

Using a2 – b2 = (a + b) (a – b)

= (3x + 4y) (3x – 4y)

(ii) 5x2 – 7y2 [ Hint : ( 𝟓 )2 = 5 ]

Solution:

( 5 x)2 – ( 7 y)2

Using a2 – b2 = (a + b) (a – b)

= ( 5 x + 7 y) ( 5 x – 7 y)

(iii) 100 – 9x2

Solution:

= 102 – (3x)2

= (10 + 3x) (10 – 3x) [∵a2 – b2 = (a + b) (a – b)]

(iv) (x + 4y)2 – 4z2

Solution:

= (x + 4y)2 – (2z)2 [∵a2 – b2 = (a + b) (a – b)]

= (x + 4y + 2z) (x + 4y – 2z)

(v) x – 64xy4

Solution:

= x (1 – 64y4)

= x (1 + (8y2)2) [∵a2 – b2 = (a + b) (a – b)]

= x (1 + 8y2) (1 – 8y2)

(vi) a2 + 2ab + b2 – 9c2

Solution:

= (a + b)2 – (3c)2 [∵(a + b)2 = a2 + 2ab + b2 ;a2 – b2 = (a + b) (a – b)]

= (a + b + 3c) (a + b – 3c)

(vii) 4x2 – 9y2 – 2x – 3y

Solution:

= (2x)2 – (3y)2 – (2x + 3y)

= [(2x + 3y) (2x – 3y)] – (2x – 3y) [∵a2 – b2 = (a + b) (a – b)]

= (2x + 3y) (2x – 3y – 1)

(viii) a2 + b2 – a + b

Solution:

= a2 + b2 – (a – b) [∵a2 – b2 = (a + b) (a – b)]

= (a + b) (a – b) – (a – b)

= (a – b) (a + b – 1)

(ix) x4 – 625

Solution:

= (x2)2 – 252 [∵a2 – b2 = (a + b) (a – b)]

= (x + 25) (x2 – 25)

= (x2 + 25) (x2 – 52)

= (x2 + 25) (x + 5) (x – 5)

(x) 36 (5x + y)2 – 25 (4x – y)2

Solution:

= [6 (5x + y)]2 – [5 (4x – y)]2 [∵a2 – b2 = (a + b) (a – b)]

= (30x + 6y + 20x – 5y) (30x + 6y – 20x + 5y)

= (50x + y) (10x + 11y)

(xi) (a + 𝟏/𝐚 )2 – 4 (x – 𝟏/𝐱 )2

Solution:

= (a + 𝟏/𝐚 )2 – [2 (x – 𝟏/𝐱 )]2

= [a+ 𝟏/𝐚 + 2( x− 𝟏/𝐱)][ a+ 𝟏/𝐚 – 2(x− 𝟏/𝐱 )]

(xii) 12m2 – 75n2

Solution:

= ( √12 m2)2 – (√75 n2)2 [∵a2 – b2 = (a + b) (a – b)]

= (2 √3m2)2 – (5√3 n2)2 [ 12 =2√3 and 75 = 5√3

= (2√3m2 + 5√3 n2) (2√3 m2 – 5√3 n2)

= √3 (2m2 + 5n2) (2m – 5n) (2m + 5n)

1. Factorize by adding and subtracting appropriate quantity.

(i) x2 + 6x + 3

Solution:

= x2 + 2. x. 3 + 32 – 32 + 3

= (x + 3)2 – 9 + 3

= (x + 3)2 – 6 [∵a2 – b2 = (a + b) (a – b)]

= (x + 3)2 – 62

= (x + 3 + 6 ) (x + 3 – 6 )

(ii) x2 + 10x + 8

Solution:

= x2 + 2. x. 5 + 52 – 52 + 8

= (x + 5)2 – 25 + 8

= (x + 5)2 – 17 [∵a2 + 2ab + b2 = (a + b)2]

= (x + 5)2 – 172

= (x + 5 + 17 ) (x + 5 – 17 )

(iii) x2 + 10x + 20

Solution:

= x2 + 2. x. 5 + 52 – 52 + 20 [∵a2 + 2ab + b2 = (a + b)2]

= (x + 5)2 – 25 + 20

= (x + 5)2 – 5

= (x + 5)2 – 52 [∵a2 – b2 = (a + b) (a – b)]

= (x + 5 + 5 ) (x + 5 – 5 )

(iv) x2 + 2x – 1

Solution:

= x2 + 2. x. 1 + 12 – 12 + 1 [∵a2 + 2ab + b2 = (a + b)2]

= (x + 1)2 – 1 + 1

= (x + 1)2 – 2

= (x + 1)2 – 22 [∵a2 – b2 = (a + b) (a – b)]

= (x + 1 + 2 ) (x + 1 – 2 )

1. prove that (1 + 𝐱/𝐲 ) (1 – 𝐱/𝐲 ) (1 + 𝐱𝟐/𝐲𝟐 ) (1 + 𝐱𝟒/𝐲𝟒 ) + 𝐱𝟖/𝐲𝟖 = 1

Solution:

L.H.S = (1 + 𝐱/𝐲 ) (1 – 𝐱/𝐲 ) (1 + 𝐱𝟐/𝐲𝟐) (1 + 𝐱𝟒/𝐲𝟒 ) + 𝐱𝟖/𝐲𝟖

[∵a2 – b2 = (a + b) (a – b)]

= (1 – 𝐱𝟐/𝐲𝟐 ) (1 + 𝐱𝟐/𝐲𝟐 ) (1 + 𝐱𝟒/𝐲𝟒) + 𝐱𝟖/𝐲𝟖

= (1 – 𝐱𝟒/𝐲𝟒) (1 + 𝐱𝟒/𝐲𝟒 ) + 𝐱𝟖/𝐲𝟖

= (1)2 – (𝐱𝟒/𝐲𝟒)2 + 𝐱𝟖/𝐲𝟖

= 1 – 𝐱𝟖/𝐲𝟖 + 𝐱𝟖/𝐲𝟖

= 1 R.H.S

1. If x = 𝐚+𝐛/𝐚−𝐛 and y = 𝐚−𝐛/𝐚+𝐛 find x2 – y2.

Solution:

x2 – y2 = (x + y) (x – y)

= (a+b/a−b+a−b/a+b)(a+b/a−b a−b/a+b)

= [[(a+b)2+ (a−b)2 ]/(a2− b2)][[(a+b)2− (a−b)2]/(a2− b2)]

= {(a2+ 2ab + b2+ a2+ b2− 2ab)/[(a2− b2)2]} x (a2 + 2ab + b2 – a2 – b2 + 2ab)

= [(2a2+ 2b2)4ab]/[(a2− b2)2]

= [2(a2+ b2)4ab]/(a2− b2)2

= [8ab(a2+ b2)]/(a2− b2)2

1. If p = x – y, q = y – z and r = z – x, simplify r2 – p + 2pq – q2

Solution:

(z – x)2 – (x – y)2 + 2 (x – y) (y – z) – (y – z)2

= z2 – 2xy + x2 – (x2 – 2xy + y2) + 2 (xy – y2 – xz + yz) – (y2 + z2 – 2yz)

= z2 + x2 – 2xz – x2 + 2xy – y2 + 2xy – 2y2 – 2xz + 2yz – y2 – z2 + 2yz

= 4xy – 4xz + 4yz – 4y2

= 4[xy – xz + yz – y2]

= 4[x (y – z) – y (y – z)]

= 4 (x – y) (y – z)

= 4pq

1. The radius of a circle is 13cm in which a chord of 10cm is drawn. Find the distance of the chord from the centre of the circle.

[Hint: use Pythagoras rule and the fact that the perpendicular line drawn from the centre of the circle to any o the chords bisects the chord]

Solution: Let AC = 10cm is the chord of the circle with radius 13cm then using Pythagoras theory

OA2 = OB2 + AB2

132 = x2 + 52

x2 = 132 – 52

= (13 + 5) (13 – 5)

= 18 (8)

x2 = 144 (∵ the perpendicular line drawn from the centre of the circle to a chord bisects the chord)

x = 144

x = 12 cm

1. Is it possible to factorize x2 + 4x + 20? Give reason

Solution:

x2 + 4x + 20

= x2 + 2. x. 2 + 22 – 22 + 20

= x2 + 4x + 4 – 4 + 20

= (x + 4)2 + 16

= (x + 4)2 + 42

We cannot factorize their further since this is nit of the form a2 – b2

1. Rajiv says: