You are currently viewing 10th Maths Guide Chapter 3 Algebra Exercise 3.3

10th Maths Guide Chapter 3 Algebra Exercise 3.3

10th Maths Guide Chapter 3 Algebra Exercise 3.3

10th Standard Maths Chapter 3 Algebra Samacheer Kalvi Guide Exercise 3.3 Book Back Answers Solutions. TN 10th SSLC Samacheer Kalvi Guide. 10th All Subject Guide – Click Here. Class 1 to 12 All Subject Book Back Answers – Click Here

10th Maths Chapter 3 Algebra Guide

1. Find the LCM and GCD for the following and verify that f(x) × g(x) = LCM × GCD.

  • (i) 21x2y, 35xy2
  • (ii) (x3 – 1)(x + 1), x3 + 1
  • (ii) (x3 – 1) (x + 1), (x3 – 1)
  • (iii) (x2y + xy2), (x2 + xy)

Solution:
(i) f(x) = 21x2y = 3 × 7x2y
g(x) = 35xy2 = 7 × 5xy2
G.C.D. = 7xy
L.C.M. = 7 × 3 × 5 × x2y2 = 105×2 × y2
L.C.M × G.C.D = f(x) × g(x)
105x2y2 × 7xy = 21x2y × 35xy2
735x3y3 = 735x3y3
Hence verified.

(ii) (x3 – 1)(x + 1) = (x – 1)(x2 + x + 1)(x + 1)
x3 + 1 = (x + 1) (x2 – x + 1)
G.C.D = (x+ 1)
L.C.M = (x – 1)(x + 1)(x2 + x + 1)(x2 – x + 1)
∴ L.C.M. × G.C.D = f(x) × g(x)
(x – 1)(x + 1)(x2 + x + 1) (x2 – x + 1) = (x – 1)
(x2 + x + 1) × (x + 1) (x2 – x + 1)
(x3 – 1)(x + 1)(x3 + 1) = (x3 – 1)(x + 1)(x3 + 1)
∴ Hence verified.

(iii) f(x) = x2y + xy2 = xy(x + y)
g(x) = x2 + xy = x(x + y)
L.C.M. = x y (x + y)
G.C.D. = x (x + y)
To verify:
L.C.M. × G.C.D. = xy(x + y) × (x + y)
= x2y (x + y)2 ……….. (1)
f(x) × g (x) = (x2y + xy2)(x2 + xy)
= x2y (x + y)2 …………… (2)
∴ L.C.M. × G.C.D = f(x) × g{x).
Hence verified.

 

 

2.Find the LCM of each pair of the following polynomials

  • (i) a2 + 4a – 12, a2 – 5a + 6 whose GCD is a – 2
  • (ii) x4 – 27a3x, (x – 3a)2 whose GCD is (x – 3a)

Solution:
(i) f(x) = a2 + 4a – 12 = (a + 6)(a – 2)

 

10th maths unit - 3 book back answer

(ii) f(x) = x4 – 27a3x = x(x3 – (3a)3)
g(x) = (x – 3a)2
G.C.D = (x – 3a)
L.C.M. × G.C.D = f(x) × g(x)
L. C.M = x(x3−(3a)3)×(x−3a)2(x−3a)
L.C.M = x(x3 – (3a)3) . (x – 3a)
= x(x – 3a)2 (x2 + 3ax + 9a2)

 

 

3. Find the GCD of each pair of the following polynomials

  • (i) 12(x4 – x3), 8(x4 – 3×3 + 2×2) whose LCM is 24×3 (x – 1)(x – 2)
  • (ii) (x3 + y3), (x4 + x2y2 + y4) whose LCM is (x3 + y3) (x2 + xy + y2)

Solution:
(i) f(x)= 12(x4 – x3)
g(x) = 8(x4 – 3×3 + 2×2)

 

L.C.M = 24×3 (x – 1)(x – 2)

10th maths unit - 3 book back answer

(ii) (x3 + y3), (x4 + x2y2 + y4)
L.C.M. = (x3 + y3)(x2 + xy + y2)
10th maths unit - 3 book back answer

4.Given the LCM and GCD of the two polynomials p(x) and q(x) find the unknown polynomial in the following table
10th maths unit - 3 book back answer

Solution:

10th maths unit - 3 book back answer

10th maths unit - 3 book back answer

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