Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.5

You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

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

Question 1.
Factorise the following expressions:
(i) 2a2 + 4a2b + 8a2c
(ii) ab – ac – mb + mc
Solution:
(i) 2a2 + 4a2b + 8a2c = 2a2 [ 1 + 2b + 4c]
(ii) ab – ac – mb + mc = a(b – c) – m (b – c) = (b – c) (a – m)

Question 2.
Factorise the following:
(i) x2 + 4x + 4
(ii) 3a2 – 24ab + 48b2
(iii) x5 – 16x
(iv) \(m^{2}+\frac{1}{m^{2}}\) – 23
(v) 6 – 216x2
(vi) \(a^{2}+\frac{1}{a^{2}}\) – 18
Solution:
(i) x2 + 4x + 4 = (x + 2) (x + 2) = (x + 2)2
∵ (a + b)2 = a2 + 2ab + b2
(ii) 3a2 – 24ab + 48b2 = 3[a2 – 8ab + 16 b2]
= 3 [a – 4b]2 (∵ (a – b)2 = a2 – 2ab + b2)
(iii) x5 – 16x = x[x4 – 16] = x [(x2)2 – 42]
= x (x2 + 4) (x2 – 4)= x (x2 + 4) (x + 2) (x – 2)
Samacheer Kalvi 9th Maths Chapter 3 Algebra Ex 3.5 1

Question 3.
Factorise the following:
(i) 4x2 + 9y2 + 25z2 + 12xy + 30yz + 20xz
(ii) 25x2 + 4y2 + 9z2 – 20xy + 12yz – 30xz
Solution:
(i) 4x2 + 9y2 + 25z2 + 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 = a2 + b2 + c2 + 2ab + 2bc + 2ca

(ii) 25x2 + 4y2 + 9z2 – 20xy + 12 yz – 30xz
= (5x)2 + (-2y)2 + (-3z)2 + 2(5x) (-2y) + 2 (-2y) (-3z) + 2 (-3z) (5x)
= (5x – 2y – 3z)2

Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.5

Question 4.
Factorise the following
(i) 8x3 + 125y3
(ii) 27x3 – 8y3
(iii) a6 – 64
Solution:
(i) 8x3 + 125y3 (2x)3 + (5y)3
∴ a3 – b3 = (a + b) (a2 – ab + b2)
= (2x + 5y) [(2x)2 – (2x)(5y) + (5y2]
= (2x + 5y )2 (4x2 – 10xy + 25y2)

(ii) 27x3 – 8y3 = (3x)3 – (2y)2
= (3x – 2y ) ((3x)2 + 3x × 2y + (2y)3)
= (3x- 2y) (9x3 + 6xy + 4xy + 4y3)

(iii) a6 – 64 = (a2)3 – 43 (a3 – b3 = (a – b) (a2 + ab + b2)
= (a2 – 4) (a4 + 4a2 + 42)
= (a + 2) (a – 2) (a2 + 4 – 2a) (a2 – 4 + 2a)

Question 5.
Factorise the following:
(i) x3 + 8y3 + 6xy – 1
(ii) l3 – 8m3 – 27n3 – 18lmn
Solution:
(i) x3 + 8y3 + 6xy – 1 = x3 + (2y)3 + (-1)3 – 3 (x) (2y) (-1)
= (x + 2y – 1) (x2 + 4y2 + 1 – 2xy + 2y + x)

(ii) l3 – 8m3 – 27n3 – 18lmn = l3 + (-2m)3+ (-3n)3 -3 (l) {-2m) (-3n)
= (l – 2m – 3n) (l2 + (-2m)2 + (-3n)3 – l × – 2m – (-2m × -3n) – (-3n × l))
= (l – 2m – 3n) (l2 + 4m2 + 9n2 + 2lm – 6mn + 3nl)

Leave a Comment

Your email address will not be published. Required fields are marked *