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

**10th Maths Exercise 3.2 Samacheer Kalvi Question 1.**

Find the GCD of the given polynomials

(i) x^{4} + 3x^{3} – x – 3, x^{3} + x^{2} – 5x + 3

(ii) x^{4} – 1, x^{3} – 11x^{2} + x – 11

(iii) 3x^{4} + 6x^{3} – 12x^{4} – 24x, 4x^{4} + 14x^{3} + 8x^{2} – 8x

(iv) 3x^{3} + 3x^{2} + 3x + 3, 6x^{3} + 12x^{2} + 6x + 12

Solution:

x^{4} + 3x^{3} – x – 3, x^{3} + x^{2} – 5x + 3

Let f(x) = x^{4} + 3x^{3} – x – 3

g(x) = x^{3} + x^{2} – 5x + 3

Note that 3 is not a divisor of g(x). Now dividing g(x) = x^{3} + x^{2} – 5x + 3 by the new remainder x^{2} + 2x – 3 (leaving the constant factor 3) we get

Here we get zero remainder

G.C.D of (x^{4} + 3x^{3} – x – 3), (x^{3} + x^{2} – 5x + 3) is (x^{2} + 2x – 3)

(ii) x^{4} – 1, x^{3} – 11x^{2} + x – 11

(iii) 3x^{4} + 6x^{3} – 12x^{2} – 24x, 4x^{4} + 14x^{3} + 8x^{2} – 8x

4x^{4} + 14x^{3} + 8x^{2} – 8x = 2 (2x^{4} + 7x^{3} + 4x^{2} -4x)

Let us divide

(2x^{4} + 7x^{3} + 4x^{2} + 4x) by x^{4} + 2x^{3} – 4x^{2} – 8x

(x^{3} + 4x^{3} + 4x) ≠ 0

Now let us divide

x^{4} + 2x^{3} – 4x^{2} – 8x by x^{3} + 4x^{2} + 4x

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

∴ Ans x (x^{2} + 4x + 4)

(iv) f(x) = 3x^{3} + 3x^{2} + 3x + 3 = 3(x^{3} + x^{2} + x + 1)

g(x) = 6x^{3} + 12x^{2} + 6x + 12

= 6(x^{3} + 2x^{2} + x + 2)

= 2 × 3 (x^{3} + 2x^{2} + x + 2)

f(x) ⇒ x^{3} + x^{2} + x + 1

The quadratic equation is an equation where the highest exponent of the variable is square. The standard form of quadratic equations is ax² + bx + c = 0.

**Ex 3.2 Class 10 Samacheer Question 2.**

Find the LCM of the given expressions,

(i) 4x^{2}y, 8x^{3}y^{2}

(ii) -9a^{3}b^{2}, 12a^{2}b^{2}c

(iii) 16m, -12m^{2}n^{2}, 8n^{2}

(iv) p^{2} – 3p + 2, p^{2} – 4

(v) 2x^{2} – 5x – 3, 4x^{2} – 36

(vi) (2x^{2} – 3xy)^{2}, (4x – 6y)^{3}, 8x^{3} – 27y^{3}

Solution:

(i) 4x^{2}y, 8x^{3}y^{2}

4x^{2}y = 2 × 2 x^{2}y

8x^{3}y^{2} = 2 × 2 × 2 x^{3}y^{2}

L.C.M. = 2 × 2 × 2 x^{3}y^{2}

= 8x^{3} y^{2}

(ii) -9a^{3}b^{2} = -3 × 3 a^{3}b^{2}

12a^{2}b^{2}c = 2 × 3 × 2a^{2}b^{2}c

L.C.M. = -3 × 3 × 2 × 2 a^{3}b^{2}c

= -36a^{3}b^{2}c

(iii) 16m, -12m^{2}n^{2}, 8n^{2}

16 m = 2 × 2 × 2 × 2 × m

-12m^{2}n^{2} = -2 × 2 × 3 × m^{2}n^{2}

8n^{2} = 2 × 2 × 2 × n^{2}

L.C.M.= -2 × 2 × 2 × 2 × 3 m^{2}n^{2}

= -48 m^{2}n^{2}

(v) 2x^{2} – 5x – 3, 4x^{2} – 36

2x^{2} – 5x – 3 = (x – 3)(2x + 1)

4x^{2} – 36 = 4(x + 3)(x – 3)

L.C.M. = 4(x + 3)(x – 3)(2x + 1)

(vi) (2x^{2} – 3xy)^{2} = (x(2x – 3y))^{2}

(4x – 6y)^{3} = (2(2x – 3y))^{3}

8x^{3} – 27y^{3}= (2x)^{3} – (3y)^{3}

= (2x – 3y) (4x^{2} + 6xy + 9y^{2})

L.C.M. = 2^{3} × x^{2} (2x – 3y)^{3} (4x^{2} + 6xy + 9y^{2})