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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)
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
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)