02 Apr 2022

Samacheer kalvi 10th Maths – Algebra Ex 3.2

10th Maths Book Back Question and Answers – Chapter 3 Exercise 3.2:

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 3.2 Solutions are available below. Check the complete Samacheer Kalvi 10th Maths – Algebra Ex 3.2 Book Back Answers below:

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Samacheer Kalvi 10th Maths Book Back Answers – Ex 3.2 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 3.2 Algebra

1. Find the GCD of the given polynomials
(i) x4 + 3x3 – x – 3, x3 + x2 – 5x + 3
(ii) x4 – 1, x3 – 11x2 + x – 11
(iii) 3x4 + 6x3 – 12x4 – 24x, 4x4 + 14x3 + 8x2 – 8x
(iv) 3x3 + 3x2 + 3x + 3, 6x3 + 12x2 + 6x + 12
Solution:
x4 + 3x3 – x – 3, x3 + x2 – 5x + 3
Let f(x) = x4 + 3x3 – x – 3
g(x) = x3 + x2 – 5x + 3
10th maths unit - 3 book back answer

Note that 3 is not a divisor of g(x). Now dividing g(x) = x3 + x2 – 5x + 3 by the new remainder x2 + 2x – 3 (leaving the constant factor 3) we get
10th maths unit - 3 book back answer

Here we get zero remainder
G.C.D of (x4 + 3x3 – x – 3), (x3 + x2 – 5x + 3) is (x2 + 2x – 3)

(ii) x4 – 1, x3 – 11x2 + x – 11

10th Maths unit - 3 book back

(iii) 3x4 + 6x3 – 12x2 – 24x, 4x4 + 14x3 + 8x2 – 8x
4x4 + 14x3 + 8x2 – 8x = 2 (2x4 + 7x3 + 4x2 -4x)
Let us divide
(2x4 + 7x3 + 4x2 + 4x) by x4 + 2x3 – 4x2 – 8x

10th maths unit - 3 book back answer

(x3 + 4x3 + 4x) ≠ 0
Now let us divide
x4 + 2x3 – 4x2 – 8x by x3 + 4x2 + 4x
10th maths unit - 3 maths book back answer

∴ x3 + 4x2 + 4x is the G.C.D of 3x4 + 6x3 -12x2 – 24x, 4x4 + 14x3 + 8x2 -8x
∴ Ans x (x2 + 4x + 4)

(iv) f(x) = 3x3 + 3x2 + 3x + 3 = 3(x3 + x2 + x + 1)
g(x) = 6x3 + 12x2 + 6x + 12
= 6(x3 + 2x2 + x + 2)
= 2 × 3 (x3 + 2x2 + x + 2)
f(x) ⇒ x3 + x2 + x + 1
10th maths unit - 3 book back answer

2. Find the LCM of the given expressions,
(i) 4x2y, 8x3y2
(ii) -9a3b2, 12a2b2c
(iii) 16m, -12m2n2, 8n2
(iv) p2 – 3p + 2, p2 – 4
(v) 2x2 – 5x – 3, 4x2 – 36
(vi) (2x2 – 3xy)2, (4x – 6y)3, 8x3 – 27y3
Solution:
(i) 4x2y, 8x3y2
4x2y = 2 × 2 x2y
8x3y2 = 2 × 2 × 2 x3y2
L.C.M. = 2 × 2 × 2 x3y2
= 8x3 y2

(ii) -9a3b2 = -3 × 3 a3b2
12a2b2c = 2 × 3 × 2a2b2c
L.C.M. = -3 × 3 × 2 × 2 a3b2c
= -36a3b2c

(iii) 16m, -12m2n2, 8n2
16 m = 2 × 2 × 2 × 2 × m
-12m2n2 = -2 × 2 × 3 × m2n2
8n2 = 2 × 2 × 2 × n2
L.C.M.= -2 × 2 × 2 × 2 × 3 m2n2
= -48 m2n2

10th maths unit - 3 book back answer

(v) 2x2 – 5x – 3, 4x2 – 36
2x2 – 5x – 3 = (x – 3)(2x + 1)
4x2 – 36 = 4(x + 3)(x – 3)
L.C.M. = 4(x + 3)(x – 3)(2x + 1)

(vi) (2x2 – 3xy)2 = (x(2x – 3y))2
(4x – 6y)3 = (2(2x – 3y))3
8x3 – 27y3= (2x)3 – (3y)3
= (2x – 3y) (4x2 + 6xy + 9y2)
L.C.M. = 23 × x2 (2x – 3y)3 (4x2 + 6xy + 9y2)

Other Important Links for 10th Maths Book Back Answers Solutions:

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