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• RD Sharma Class 9 Solutions

# Class 9 RD Sharma Solutions – Chapter 5 Factorisation of Algebraic Expressions- Exercise 5.1

### Question 1: Factorize x3 + x – 3x2 – 3

Solution:

x3 + x – 3x2 – 3

Here x is common factor in x3 + x and – 3 is common factor in – 3x2 – 3

x3 – 3x2 + x – 3

x2 (x – 3) + 1(x – 3)

Taking (x – 3) common

(x – 3) (x2 + 1)

Therefore, x3 + x – 3x2 – 3 = (x – 3) (x2 + 1)

### Question 2: Factorize a(a + b)3 – 3a2b(a + b)

Solution:

a(a + b)3 – 3a2b(a + b)

Taking (a + b) as common factor

= a(a + b) {(a + b)2 – 3ab}

= a(a + b) {a2 + b2 + 2ab – 3ab}

= a(a + b) (a2 + b2 – ab)

### Question 3: Factorize x(x3 – y3) + 3xy(x – y)

Solution:

x(x3 – y3) + 3xy(x – y)

= x(x – y) (x2 + xy + y2) + 3xy(x – y)

Taking x(x – y) as a common factor

= x(x – y) (x2 + xy + y2 + 3y)

= x(x – y) (x2 + xy + y2 + 3y)

Question 4: Factorize a2x2 + (ax2 + 1)x + a

Solution:

a2x2 + (ax2 + 1)x + a

= a2x2 + a + (ax2 + 1)x

= a(ax2 + 1) + x(ax2 + 1)

= (ax2 + 1) (a + x)

### Question 5: Factorize x2 + y – xy – x

Solution:

x2 + y – xy – x

= x2 – x – xy + y

= x(x – 1) – y(x – 1)

= (x – 1) (x – y)

Question 6: Factorize x3 – 2x2y + 3xy2 – 6y3

Solution:

x3 – 2x2y + 3xy2 – 6y3

= x2(x – 2y) + 3y2(x – 2y)

= (x – 2y) (x2 + 3y2)

### Question 7: Factorize 6ab – b2 + 12ac – 2bc

Solution:

6ab – b2 + 12ac – 2bc

= 6ab + 12ac – b2 – 2bc

Taking 6a common from first two terms and –b from last two terms

= 6a(b + 2c) – b(b + 2c)

Taking (b + 2c) common factor

### Question 8: Factorize (x2 + 1/x2) – 4(x + 1/x) + 6

Solution:

(x2 + 1/x2) – 4(x + 1/x) + 6

= x2 + 1/x2 – 4x – 4/x + 4 + 2

= x2 + 1/x2 + 4 + 2 – 4/x – 4x

= (x2) + (1/x)2+ (-2)2 + 2x(1/x) + 2(1/x)(-2) + 2(-2)x

As we know, x2 + y2 + z2 + 2xy + 2yz + 2zx = (x + y + z)2

So, we can write;

= (x + 1/x + (-2))2

or (x + 1/x – 2)2

Therefore, x2 + 1/x2) – 4(x + 1/x) + 6 = (x + 1/x – 2)2

### Question 9: Factorize x(x – 2) (x – 4) + 4x – 8

Solution:

x(x – 2) (x – 4) + 4x – 8

= x(x – 2) (x – 4) + 4(x – 2)

= (x – 2) [x(x – 4) + 4]

= (x – 2) (x2 – 4x + 4)

= (x – 2) [x2 – 2 (x)(2) + (2)2]

= (x – 2) (x – 2)2

= (x – 2)3

### Question 10: Factorize (x + 2) (x2 + 25) – 10x2 – 20x

Solution:

(x + 2) (x2 + 25) – 10x(x + 2)

Take (x + 2) as common factor;

= (x + 2)(x2 + 25 – 10x)

= (x + 2) (x2 – 10x + 25)

Expanding the middle term of (x2 – 10x + 25)

= (x + 2) (x2 – 5x – 5x + 25)

= (x + 2){x (x – 5) – 5 (x – 5)}

= (x + 2)(x – 5)(x – 5)

= (x + 2)(x – 5)2

Therefore, (x + 2) (x2 + 25) – 10x (x + 2) = (x + 2)(x – 5)2

### Question 11: Factorize 2a2 + 2√6 ab + 3b2

Solution:

2a2 + 2√6 ab + 3b2

Above expression can be written as (√2a)2 + 2 × √2a × √3b + (√3b)2

As we know, (p + q)2 = p2 + q2 + 2pq

Here p = √2a and q = √3b

= (√2a + √3b)2

Therefore, 2a2 + 2√6 ab + 3b2 = (√2a + √3b)2

### Question 12: Factorize (a – b + c)2 + (b – c + a)2 + 2(a – b + c) (b – c + a)

Solution:

(a – b + c)2 + (b – c + a)2 + 2(a – b + c) (b – c + a)

{Because p2 + q2 + 2pq = (p + q)2}

Here p = a – b + c and q = b – c + a

= [a – b + c + b – c + a]2

= (2a)2

= 4a2

### Question 13: Factorize a2 + b2 + 2(ab + bc + ca)

Solution:

a2 + b2 + 2ab + 2bc + 2ca

As we know, p2 + q2 + 2pq = (p + q)2

We get,

= (a + b)2 + 2bc + 2ca

= (a + b)2 + 2c(b + a)

Or (a + b)2 + 2c(a + b)

Take (a + b) as common factor;

= (a + b)(a + b + 2c)

Therefore, a2 + b2 + 2ab + 2bc + 2ca = (a + b)(a + b + 2c)

### Question 14: Factorize 4(x – y)2 – 12(x – y)(x + y) + 9(x + y)2

Solution:

Consider (x – y) = p, (x + y) = q

= 4p2 – 12pq + 9q2

Expanding the middle term, -12 = -6 -6 also 4 × 9 = -6 × -6

= 4p2 – 6pq – 6pq + 9q2

= 2p(2p – 3q) – 3q(2p – 3q)

= (2p – 3q) (2p – 3q)

= (2p – 3q)2

Substituting back p = x – y and q = x + y;

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

= (2x – 3x – 2y – 3y)2

= [-x – 5y]2

= [(-1)(x + 5y)]2

= (x + 5y)2

Therefore, 4(x – y)2 – 12(x – y)(x + y) + 9(x + y)2 = (x + 5y)2

### Question 15: Factorize a2 – b2 + 2bc – c2

Solution :

a2 – b2 + 2bc – c2

As we know, (a – b)2 = a2 + b2 – 2ab

= a2 – (b – c)2

Also, we know, a2 – b2 = (a + b)(a – b)

= (a + b – c)(a – (b – c))

= (a + b – c)(a – b + c)

Therefore, a2 – b2 + 2bc – c2 =(a + b – c)(a – b + c)

### Question 16: Factorize a2 + 2ab + b2 – c2

Solution:

a2 + 2ab + b2 – c2

= (a2 + 2ab + b2) – c2

= (a + b)2 – (c)2

We know, a2 – b2 = (a + b) (a – b)

= (a + b + c) (a + b – c)

Therefore, a2 + 2ab + b2 – c2 = (a + b + c) (a + b – c)

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