02 Apr 2022

Samacheer kalvi 10th Maths – Algebra Ex 3.3

10th Maths Book Back Question and Answers – Chapter 3 Exercise 3.3:

Samacheer Kalvi 10th Standard Maths Book Back Questions with Answers PDF uploaded and the same given below. Class-tenth candidates and those preparing for TNPSC exams can check the Maths Book Back Answers PDF below. Samacheer Kalvi Class 10th Std Maths Book Back Answers Chapter 3 Exercise 3.3 Solutions are available below. Check the complete Samacheer Kalvi 10th Maths – Algebra Ex 3.3 Book Back Answers below:

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Samacheer Kalvi 10th Maths Book Back Answers – Ex 3.3 Algebra

Samacheer Kalvi 10th Maths Book Subject One Mark, Two Mark, Five Mark Guide questions and answers are below. Check Maths Book Back Questions with Answers. Take the printout and use it for exam purposes.

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Chapter 3

Exercise 3.3 Algebra

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 = 105x2 × 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 – 3x3 + 2x2) whose LCM is 24x3 (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 – 3x3 + 2x2)

L.C.M = 24x3 (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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