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

**Question 1: Factorize x**^{3} + x – 3x^{2} – 3

^{3}+ x – 3x

^{2}– 3

**Solution:**

x

^{3}+ x – 3x^{2}– 3Here x is common factor in x

^{3}+ x and – 3 is common factor in – 3x^{2}– 3x

^{3}– 3x^{2}+ x – 3x

^{2}(x – 3) + 1(x – 3)Taking (x – 3) common

(x – 3) (x

^{2}+ 1)Therefore, x

^{3}+ x – 3x^{2}– 3 = (x – 3) (x^{2}+ 1)

**Question 2: Factorize a(a + b)**^{3} – 3a^{2}b(a + b)

^{3}– 3a

^{2}b(a + b)

**Solution:**

a(a + b)

^{3}– 3a^{2}b(a + b)Taking (a + b) as common factor

= a(a + b) {(a + b)

^{2}– 3ab}= a(a + b) {a

^{2}+ b^{2 }+ 2ab – 3ab}= a(a + b) (a

^{2}+ b^{2}– ab)

**Question 3: Factorize x(x**^{3} – y^{3}**) + 3xy(x – y)**

^{3}– y

**Solution:**

x(x

^{3}– y^{3}) + 3xy(x – y)= x(x – y) (x

^{2}+ xy + y^{2}) + 3xy(x – y)Taking x(x – y) as a common factor

= x(x – y) (x

^{2}+ xy + y^{2}+ 3y)= x(x – y) (x

^{2}+ xy + y^{2}+ 3y)

**Question 4: Factorize a ^{2}x^{2} + (ax^{2} + 1)x + a**

**Solution:**

a

^{2}x^{2}+ (ax^{2}+ 1)x + a= a

^{2}x^{2}+ a + (ax^{2}+ 1)x= a(ax

^{2}+ 1) + x(ax^{2}+ 1)= (ax

^{2}+ 1) (a + x)

**Question 5: Factorize x**^{2}** + y – xy – x**

**Solution:**

x

^{2}+ y – xy – x= x

^{2}– x – xy + y= x(x – 1) – y(x – 1)

= (x – 1) (x – y)

**Question 6: Factorize x ^{3} – 2x2y + 3xy**

^{2}

**– 6y**

^{3}**Solution:**

x

^{3}– 2x2y + 3xy^{2}– 6y^{3}= x

^{2}(x – 2y) + 3y^{2}(x – 2y)= (x – 2y) (x

^{2}+ 3y^{2})

**Question 7: Factorize 6ab – b**^{2}** + 12ac – 2bc**

**Solution:**

6ab – b

^{2}+ 12ac – 2bc= 6ab + 12ac – b

^{2}– 2bcTaking 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 (x**^{2}** + 1/x**^{2}**) – 4(x + 1/x) + 6**

**Solution:**

(x

^{2}+ 1/x^{2}) – 4(x + 1/x) + 6= x

^{2}+ 1/x^{2 }– 4x – 4/x + 4 + 2= x

^{2}+ 1/x^{2}+ 4 + 2 – 4/x – 4x= (x

^{2}) + (1/x)^{2}+ (-2)^{2}+ 2x(1/x) + 2(1/x)(-2) + 2(-2)xAs we know, x

^{2}+ y^{2}+ z^{2}+ 2xy + 2yz + 2zx = (x + y + z)^{2}So, we can write;

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

^{2}or (x + 1/x – 2)

^{2}Therefore, x

^{2}+ 1/x^{2}) – 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) (x

^{2}– 4x + 4)= (x – 2) [x

^{2}– 2 (x)(2) + (2)^{2}]= (x – 2) (x – 2)

^{2}= (x – 2)

^{3}

**Question 10: Factorize (x + 2) (x**^{2}** + 25) – 10x**^{2}** – 20x**

**Solution:**

(x + 2) (x

^{2}+ 25) – 10x(x + 2)Take (x + 2) as common factor;

= (x + 2)(x

^{2}+ 25 – 10x)= (x + 2) (x

^{2}– 10x + 25)Expanding the middle term of (x

^{2}– 10x + 25)= (x + 2) (x

^{2}– 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) (x

^{2}+ 25) – 10x (x + 2) = (x + 2)(x – 5)^{2}

**Question 11: Factorize 2a**^{2}** + 2√6 ab + 3b**^{2}

**Solution:**

2a

^{2}+ 2√6 ab + 3b^{2}Above expression can be written as (√2a)

^{2}+ 2 × √2a × √3b + (√3b)^{2}As we know, (p + q)

^{2}= p^{2}+ q^{2}+ 2pqHere p = √2a and q = √3b

= (√2a + √3b)

^{2}Therefore, 2a

^{2}+ 2√6 ab + 3b^{2}= (√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 p

^{2}+ q^{2}+ 2pq = (p + q)^{2}}Here p = a – b + c and q = b – c + a

= [a – b + c + b – c + a]

^{2}= (2a)

^{2}= 4a

^{2}

**Question 13: Factorize a**^{2}** + b**^{2}** + 2(ab + bc + ca)**

**Solution:**

a

^{2}+ b^{2}+ 2ab + 2bc + 2caAs we know, p

^{2 }+ q^{2}+ 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, a

^{2}+ b^{2}+ 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

= 4p

^{2}– 12pq + 9q^{2}Expanding the middle term, -12 = -6 -6 also 4 × 9 = -6 × -6

= 4p

^{2}– 6pq – 6pq + 9q^{2}= 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 a**^{2}** – b**^{2}** + 2bc – c**^{2}

**Solution :**

a

^{2}– b^{2}+ 2bc – c^{2}As we know, (a – b)

^{2}= a^{2}+ b^{2}– 2ab= a

^{2}– (b – c)^{2}Also, we know, a

^{2}– b^{2}= (a + b)(a – b)= (a + b – c)(a – (b – c))

= (a + b – c)(a – b + c)

Therefore, a

^{2}– b^{2}+ 2bc – c^{2}=(a + b – c)(a – b + c)

**Question 16: Factorize a**^{2 }**+ 2ab + b**^{2}** – c**^{2}

**Solution:**

a

^{2}+ 2ab + b^{2}– c^{2}= (a

^{2}+ 2ab + b^{2}) – c^{2}= (a + b)

^{2}– (c)^{2}We know, a

^{2}– b^{2}= (a + b) (a – b)= (a + b + c) (a + b – c)

Therefore, a

^{2}+ 2ab + b^{2}– c^{2}= (a + b + c) (a + b – c)