This document contains solutions to examples finding the greatest common divisor (GCD) of various algebraic expressions. It provides step-by-step workings for expressions involving variables, exponents, and coefficients. Examples include finding the GCD of polynomials, expressions with common factors, and expressions where the GCD is 1. The document demonstrates multiple methods for determining the GCD algebraically and factorizing expressions.

1. PEDAGOGY OF
MATHEMATICS – PART II
BY
Dr. I. UMA MAHESWARI
Principal
Peniel Rural College of Education,Vemparali,
Dindigul District
iuma_maheswari@yahoo.co.in

2. STD IX
CHAPTER 3 – ALGEBRA
Ex – 3.9

9. (i) P5, P11, P3
Solution:
p5 = p5
p11 = p11
P9 = P9
G.C.D. is p5 (Highest common power
is 5)
(ii) 4x3, y3, z3
Solution:
4x3 = 2 × 2 × x3
y3 = y3
z3 = z3
G.C.D. of 4x3, y3 and z3 =
1

10. (iii) 9a²b²c3, 15a3b2c4
Solution:
9a²b²c3 = 3 × 3 × a² × b² × c3
15a3b²c3 = 3 × 5 × a3 × b2 × c4
G.C.D = 3 × a2 × b2 × c3
= 3a2b2c3
(iv) 64x8, 240x6
Solution:
64x8 = 2 × 2 × 2 × 2 × 2 × 2 ×
x8
240x6 = 24 × 3 × 5 × x6
G.C.D = 24 × x6
= 16x6

11. (v) ab²c3, a²b3c,
a3ac²
Solution:
ab²c3 = a × b² × c3
a²b3c = a² × b3 × c
a3bc² = a3 × b × c²
G.C.D. = abc
(vi) 35x5y3z4, 49x2yz3,
14xy2z2
Solution:
35x5y3z4 = 5 × 7 × x5 × y3 ×
z4
49x²yz3 = 7 × 7 × x2 × z3
14xy²z² = 2 × 7 × x × y² × z²
G.C.D. = 7 × x × y × z²
= 7xyz²

12. (vii) 25ab3c, 100a²bc, 125 ab
Solution:
25ab3c = 5 × 5 × a × b3 × c
100a²be = 2 × 2 × 5 × 5 × a² × b
× c
125ab = 5 × 5 × 5 × a × b
G.C.D. = 5 × 5 × a × b
= 25ab
(viii) 3abc, 5xyz,
7pqr
Solution:
3abc = 3 × a × b × c
5xyz = 5 × x × y × z
7pqr = 7 × p × q × r
G.C.D. = 1

14. Solution:
(i) (2x + 5) = 2x + 5
5x + 2 = 5x + 2
G.C.D. = 1
(ii) am+1 = am × a1
am+2 = am × a2
am+3 = am × a3
G.C.D.= am × a
= am+1
(iii) 2a² + a = a(2a + 1)
4a² – 1 = (2a)2 – 1
(Using a² – b² = (a + b)(a – b)
= (2a + 1)(2a – 1)
G.C.D. = 2a + 1
(iv) 3a² = 3 × a²
5b3 = 5 × b3
7c4 = 7 × c4
G.C.D. = 1

15. (v) x4 – 1 = (x²)² – 1
= (x² + 1 ) (x² – 1)
= (x² + 1 ) (x + 1 ) (x – 1 )
x² – 1 = (x + 1 ) (x – 1 )
G.C.D. = (x + 1 ) (x – 1 )
(vi) a3 – 9ax2 = a(a2 –
9x2)
= a[a2 – (3x)2]
= a(a + 3x)(a – 3x)
(a – 3x)2 = (a – 3x)2
G.C.D. = a – 3x