Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.2

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Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.2

Question 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

Question 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)