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## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.5

**9th Class Math Exercise 3.5 Solution Question 1.**

Factorise the following expressions:

(i) 2a^{2} + 4a^{2}b + 8a^{2}c

(ii) ab – ac – mb + mc

Solution:

(i) 2a^{2} + 4a^{2}b + 8a^{2}c = 2a^{2} [ 1 + 2b + 4c]

(ii) ab – ac – mb + mc = a(b – c) – m (b – c) = (b – c) (a – m)

**9th Maths Exercise 3.5 Question 2.**

Factorise the following:

(i) x^{2} + 4x + 4

(ii) 3a^{2} – 24ab + 48b^{2}

(iii) x^{5} – 16x

(iv) \(m^{2}+\frac{1}{m^{2}}\) – 23

(v) 6 – 216x^{2}

(vi) \(a^{2}+\frac{1}{a^{2}}\) – 18

Solution:

(i) x^{2} + 4x + 4 = (x + 2) (x + 2) = (x + 2)^{2}

∵ (a + b)^{2} = a^{2} + 2ab + b^{2}

(ii) 3a^{2} – 24ab + 48b^{2} = 3[a^{2} – 8ab + 16 b^{2}]

= 3 [a – 4b]^{2} (∵ (a – b)^{2} = a^{2} – 2ab + b^{2})

(iii) x^{5} – 16x = x[x^{4} – 16] = x [(x^{2})^{2} – 4^{2}]

= x (x^{2} + 4) (x^{2} – 4)= x (x^{2} + 4) (x + 2) (x – 2)

**Class 9 Maths Exercise 3.5 Solutions Question 3.**

Factorise the following:

(i) 4x^{2} + 9y^{2} + 25z^{2} + 12xy + 30yz + 20xz

(ii) 25x^{2} + 4y^{2} + 9z^{2} – 20xy + 12yz – 30xz

Solution:

(i) 4x^{2} + 9y^{2} + 25z^{2} + 12 xy + 30 yz + 20 xz

= (2x)^{2} + (3y)^{2} + (5z)^{2} + 2 (2x) (3y) + 2 (3y) + (5z) + 2 × 3y × 5z

= (2x + 3y + 5z)^{2}

∵ (a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca

(ii) 25x^{2} + 4y^{2} + 9z^{2} – 20xy + 12 yz – 30xz

= (5x)^{2} + (-2y)^{2} + (-3z)^{2} + 2(5x) (-2y) + 2 (-2y) (-3z) + 2 (-3z) (5x)

= (5x – 2y – 3z)^{2}

**Exercise 3.5 Class 9 Question 4.**

Factorise the following

(i) 8x^{3} + 125y^{3}

(ii) 27x^{3} – 8y^{3}

(iii) a^{6} – 64

Solution:

(i) 8x^{3} + 125y^{3} (2x)^{3} + (5y)^{3}

∴ a^{3} – b^{3} = (a + b) (a^{2} – ab + b^{2})

= (2x + 5y) [(2x)^{2} – (2x)(5y) + (5y^{2}]

= (2x + 5y )^{2} (4x^{2} – 10xy + 25y^{2})

(ii) 27x^{3} – 8y^{3} = (3x)^{3} – (2y)^{2}

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

= (3x- 2y) (9x^{3} + 6xy + 4xy + 4y^{3})

(iii) a^{6} – 64 = (a^{2})^{3} – 4^{3} (a^{3} – b^{3} = (a – b) (a^{2} + ab + b^{2})

= (a^{2} – 4) (a^{4} + 4a^{2} + 4^{2})

= (a + 2) (a – 2) (a^{2} + 4 – 2a) (a^{2} – 4 + 2a)

**9th Maths Algebra Exercise 3.5 Question 5.**

Factorise the following:

(i) x^{3} + 8y^{3} + 6xy – 1

(ii) l^{3} – 8m^{3} – 27n^{3} – 18lmn

Solution:

(i) x^{3} + 8y^{3} + 6xy – 1 = x^{3} + (2y)^{3} + (-1)^{3} – 3 (x) (2y) (-1)

= (x + 2y – 1) (x^{2} + 4y^{2} + 1 – 2xy + 2y + x)

(ii) l^{3} – 8m^{3} – 27n^{3} – 18lmn = l^{3} + (-2m)^{3}+ (-3n)^{3} -3 (l) {-2m) (-3n)

= (l – 2m – 3n) (l^{2} + (-2m)^{2} + (-3n)^{3} – l × – 2m – (-2m × -3n) – (-3n × l))

= (l – 2m – 3n) (l^{2} + 4m^{2} + 9n^{2} + 2lm – 6mn + 3nl)