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)

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2. 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 )

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3. prove that (1 + ^𝐱/_𝐲 ) (1 – ^𝐱/_𝐲 ) (1 + 𝐱^𝟐/𝐲^𝟐 ) (1 + 𝐱^𝟒/𝐲^𝟒 ) + 𝐱^𝟖/𝐲^𝟖 = 1

Solution:

L.H.S = (1 + ^𝐱/_𝐲 ) (1 – ^𝐱/_𝐲 ) (1 + 𝐱^𝟐/𝐲^𝟐) (1 + 𝐱^𝟒/𝐲^𝟒 ) + 𝐱^𝟖/𝐲^𝟖

[∵a² – b² = (a + b) (a – b)]

= (1 – 𝐱^𝟐/𝐲^𝟐 ) (1 + 𝐱^𝟐/𝐲^𝟐 ) (1 + 𝐱^𝟒/𝐲^𝟒) + 𝐱^𝟖/𝐲^𝟖

= (1 – 𝐱^𝟒/𝐲^𝟒) (1 + 𝐱^𝟒/𝐲^𝟒 ) + 𝐱^𝟖/𝐲^𝟖

= (1)² – (𝐱^𝟒/𝐲^𝟒)² + 𝐱^𝟖/𝐲^𝟖

= 1 – 𝐱^𝟖/𝐲^𝟖 + 𝐱^𝟖/𝐲^𝟖

= 1 R.H.S

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4. 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

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5. 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

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6. 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

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7. 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

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