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)