# 10th Maths Book Back Algebra Ex 3.2

## Samacheer Kalvi 10th Maths Book Back Solution:

Tamil Nadu 10th Maths Book Back Answers Unit 3 – Algebra Ex 3.2 are provided on this page. Samacheer Kalvi Maths Book Back Solutions/ Guide available for all Units. TN Samacheer Kalvi 10th Maths Book consists of 8 Units and each unit book back solutions given below topics wise with Questions and Answers. The complete Samacheer Kalvi Books Back Answers/Solutions are available on our site.

The Samacheer kalvi 10th Maths solutions are useful to enhance your skills. Candidates who prepared for the Competitive and board exams 10th Maths Book Back Answers in English and Tamil Medium. The 10th Maths Unit 3 Algebra consists of 19 units. Each Unit Book Back Answers provide topic-wise on this page. 10th Maths Book Back Answers are prepared according to the latest syllabus. The 10th Maths Book Back Algebra Ex 3.2 Answers in English.

### 10th Maths Book Back Answers/Solutions:

TN Samacheer Kalvi 10th Maths Unit 3 Chapter 2 Book Back Exercise has given below. The 10th Maths Book Back Solutions Guide is uploaded below,

### 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

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

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

(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

(x3 + 4x3 + 4x) ≠ 0
Now let us divide
x4 + 2x3 – 4x2 – 8x by x3 + 4x2 + 4x

∴ 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

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

(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)