Students can download Maths Chapter 3 Algebra Ex 3.9 Questions and Answers, Notes, Samacheer Kalvi 9th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

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

Question 1.

Find the GCD for the following:

(i) P^{5}, P^{11}, P^{3}

Solution:

p^{5} = p^{5}

p^{11} = p^{11}

P^{9} = P^{9}

G.C.D. is p^{5} (Highest common power is 5)

(ii) 4x^{3}, y^{3}, z^{3}

Solution:

4x^{3} = 2 × 2 × x^{3}

y^{3} = y^{3}

z^{3} = z^{3}

G.C.D. of 4x^{3}, y^{3} and z^{3} = 1

(iii) 9a²b²c^{3}, 15a^{3}b^{2}c^{4}

Solution:

9a²b²c^{3} = 3 × 3 × a² × b² × c^{3}

15a^{3}b²c^{3} = 3 × 5 × a^{3} × b^{2} × c^{4}

G.C.D = 3 × a^{2} × b^{2} × c^{3}

= 3a^{2}b^{2}c^{3}

(iv) 64x^{8}, 240x^{6}

Solution:

64x^{8} = 2 × 2 × 2 × 2 × 2 × 2 × x^{8}

= 2^{6} × x^{8}

240x^{6} = 2^{4} × 3 × 5 × x^{6}

G.C.D = 2^{4} × x^{6}

= 16x^{6}

(v) ab²c^{3}, a²b^{3}c, a^{3}ac²

Solution:

ab²c^{3} = a × b² × c^{3}

a²b^{3}c = a² × b^{3} × c

a^{3}bc² = a^{3} × b × c²

G.C.D. = abc

(vi) 35x^{5}y^{3}z^{4}, 49x^{2}yz^{3}, 14xy^{2}z^{2}

Solution:

35x^{5}y^{3}z^{4} = 5 × 7 × x^{5} × y^{3} × z^{4}

49x²yz^{3} = 7 × 7 × x^{2} × z^{3}

14xy²z² = 2 × 7 × x × y² × z²

G.C.D. = 7 × x × y × z²

= 7xyz²

(vii) 25ab^{3}c, 100a²bc, 125 ab

Solution:

25ab^{3}c = 5 × 5 × a × b^{3} × 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

Question 2.

Find the GCD for the following:

(i) (2x + 5), (5x + 2)

(ii) a^{m+1}, a^{m+2}, a^{m+3}

(iii) 2a² + a, 4a² – 1

(iv) 3a², 5b^{3}, 7c^{4}

(v) x^{4} – 1, x² – 1

(vi) a^{3} – 9ax², (a – 3x)²

Solution:

(i) (2x + 5) = 2x + 5

5x + 2 = 5x + 2

G.C.D. = 1

(ii) a^{m+1} = a^{m} × a^{1}

a^{m+2} = a^{m} × a^{2}

a^{m+3} = a^{m} × a^{3}

G.C.D.= a^{m} × a

= a^{m+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

5b^{3} = 5 × b^{3}

7c^{4} = 7 × c^{4}

G.C.D. = 1

(v) x^{4} – 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) a^{3} – 9ax^{2} = a(a^{2} – 9x^{2})

= a[a^{2} – (3x)^{2}]

= a(a + 3x)(a – 3x)

(a – 3x)^{2} = (a – 3x)^{2}

G.C.D. = a – 3x