# Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.9

## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.9

Question 1.
Find the GCD for the following:
(i) p5, p11, p9
(ii) 4x3, y3, z3
(iii) 9 a2 b2 c3, 15 a3 b2 c4
(iv) 64x8, 240x6
(v) ab2 c3, a2 b3 c, a3 bc2
(vi) 35x5 y3 z4, 49x2 yz3, 14xy2 z2
(vii) 25 ab3 c, 100 a2 bc, 125 ab
(viii) 3abc, 5xyz, 7pqr
Solution:
(i) p5, p11, p9
p5 = p5
p9 = p5 × p4
p11 = p5 × p6
G.C.D. of p5, p11, p9 = p5

(ii) 4x3, y3, z3
G.C.D. of 4x3 = 1 × 4x3
y3 = 1 × y3
z3 = 1 × z3
∴ G.C.D. = 1

(iii) 9 a2 b2 c3, 15 a3 b2 c4
9 a2 b2 c3 = 3 × 3a2 b2 c3
15 a3 b2 c4 = 3 × 5a3 b2 c4
G.C.D. = 9 a2 b2 c3, 15 a3 b2 c4 = 3 × a2 b2 c3 = 3 a2 b2 c3

(iv) 64x8, 240x6
64x8 = 2 × 2 × 2 × 2 × 2 × 2 × x6 × x8
240x6 = 2 × 2 × 2 × 2 × 3 × 5 x6
G.C.D. of 64x8, 240x6 = 2 × 2 × 2 × 2 x6 = 16x6

(v) ab2 c3, a2 b3 c, a3 bc2
ab2 c3 = a × b × b × c × c × c
a2 b3 c = a × a × b × b × b × c
a3 bc2 = a × a × a × b × c × c
G.C.D. of ab2 c3, a2 b3 c, a3 bc2 = a × b × c = abc

(vi) 35x5 y3 z4, 49x2 yz3, 14xy2 z2
35x5 y3 z4 = 5 × 7 x5 y3 z4
49x2 yz3 = 7 × 7 x2 y z3
14xy2 z2 = 2 × 7 x y2 z2
G.C.D. is = 7 x y z2

(vii) 25 ab3 c, 100 a2 bc, 125 ab
25 ab3 c = 5 × 5 ab3c
100 a2 bc = 2 × 5 × 2 × 5 a2 b c
125 ab = 5 × 5 × 5 ab
G.C.D is = 5 × 5 ab = 25ab

(viii) 3abc, 5xyz, 7pqr
3abc = 1 × 3 a bc
5xyz = 1 × 5 x y z
7pqr = 1 × 7 p q r
G.C.D is 1 Question 2.
Find the GCD of the following
(i) (2x + 5), (5x + 2)
(ii) am + 1, am + 2, am + 3
(iii) 2a2 + a, 4a2 – 1
(iv) 3a2, 5b3, 7c4
(v) x4 – 1, x2 – 1
(vi) a3 – 9ax2, (a – 3x)2
Solution:
(i) (2x + 5), (5x + 2)
(2x + 5) = 1 × (2x + 5)
(5x + 2) = 1 × (5x + 2)
∴ G.C.D = 1

(ii) am + 1, am + 2, am + 3
am + 1 = am × a1
am + 2 = am × a1 × a1
am + 3 = am × a1 × a1 × a1
G.C.G = am × a1 = am + 1

(iii) 2a2 + a, 4a2 – 1
2a2 + a = a (2a + 1)
4a2 – 1 = (2a)2 – 12 = (2a + 1) (2a -1)
G.C.D = (2a + 1)

(iv) 3a2, 5b3, 7c4
3a2 = 1 × 3a × a
5b3 = 1 × 5 b × b × b
7c4 = 1 × 7 × c × c × c × c
∴ G.C.D = 1

(v) x4 – 1, x2 – 1
x4 – 1 = (x2)2 – 12 = (x2 + 1) (x2 – 1)
= (x2 + 1) (x2 – 12) = (x2 + 1) (x + 1 ) (x – 1)
x2 – 1 = x2 – 12 = (x + 1) (x – 1)
G.C.D = (x + 1) (x – 1) = x2 – 1

(vi) a3 – 9ax2, (a – 3x)2
a3 – 9ax2 = a (a2 – (3x)2) = a (a + 3x) (a – 3x)
(a – 3x)2 = (a – 3x) (a – 3x)
G.C.D = (a – 3x)