# 10th Maths Chapter 3 Algebra Exercise 3.2 Guide

10th Standard Maths Chapter 3 Algebra Samacheer Kalvi Guide Exercise 3.2 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.

### 1. Find the GCD of the given polynomials

- (i) x4 + 3×3 – x – 3, x3 + x2 – 5x + 3
- (ii) x4 – 1, x3 – 11×2 + x – 11
- (iii) 3×4 + 6×3 – 12×4 – 24x, 4×4 + 14×3 + 8×2 – 8x
- (iv) 3×3 + 3×2 + 3x + 3, 6×3 + 12×2 + 6x + 12

**Solution:
**x4 + 3×3 – x – 3, x3 + x2 – 5x + 3

Let f(x) = x4 + 3×3 – 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 + 3×3 – x – 3), (x3 + x2 – 5x + 3) is (x2 + 2x – 3)

**(ii) x4 – 1, x3 – 11×2 + x – 11**

**(iii) 3×4 + 6×3 – 12×2 – 24x, 4×4 + 14×3 + 8×2 – 8x
**4×4 + 14×3 + 8×2 – 8x = 2 (2×4 + 7×3 + 4×2 -4x)

Let us divide

(2×4 + 7×3 + 4×2 + 4x) by x4 + 2×3 – 4×2 – 8x

(x3 + 4×3 + 4x) ≠ 0

Now let us divide

x4 + 2×3 – 4×2 – 8x by x3 + 4×2 + 4x

∴ x3 + 4×2 + 4x is the G.C.D of 3×4 + 6×3 -12×2 – 24x, 4×4 + 14×3 + 8×2 -8x

∴ Ans x (x2 + 4x + 4)

**(iv) f(x) = 3×3 + 3×2 + 3x + 3 = 3(x3 + x2 + x + 1)
**g(x) = 6×3 + 12×2 + 6x + 12

= 6(x3 + 2×2 + x + 2)

= 2 × 3 (x3 + 2×2 + 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) 2×2 – 5x – 3, 4×2 – 36
- (vi) (2×2 – 3xy)2, (4x – 6y)3, 8×3 – 27y3

**Solution:
**

**(i) 4x2y, 8x3y2**

4x2y =

__2__×

__2__x2y

8x3y2 =

__2__×

__2__×

__2__x3y2

L.C.M. = 2 × 2 × 2 x3y2

= 8×3 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) 2×2 – 5x – 3, 4×2 – 36
**2×2 – 5x – 3 =

__(x – 3)__(2x + 1)

4×2 – 36 = 4(x + 3)

__(x – 3)__

L.C.M. = 4(x + 3)(x – 3)(2x + 1)

(vi) (2×2 – 3xy)2 = (x(2x – 3y))2

(4x – 6y)3 = (2(2x – 3y))3

8×3 – 27y3= (2x)3 – (3y)3

= (2x – 3y) (4×2 + 6xy + 9y2)

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