Students can download Maths Chapter 3 Algebra Ex 3.2 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

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

Question 1.

Find the GCD of the given polynomials by Division Algorithm

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

Answer:

p(x) = x^{4} + 3x^{3} – x – 3

g(x) = x^{3} + x^{2} – 5x + 3

3x^{2} + 6x – 9 = 3(x^{2} + 2x – 3)

Now dividing g(x) = x^{3} + x^{2} – 5x + 3

by the new remainder

(leaving the constant 3)

we get x^{2} + 2x – 3

G.C.F. = x^{2} + 2x – 3

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

p(x) = x^{4} – 1

g(x) = x^{3} – 11x^{2} + x – 11

120x^{2} + 120 = 120 (x^{2} + 1)

Now dividing g(x) = x^{3} – 11x^{2} + x – 11 by the new remainder (leaving the constant) we get x^{2} + 1

G.C.D. = x^{2} + 1

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

Answer:

p(x) = 3x^{4} + 6x^{3} – 12x^{2} – 24x

= 3x (x^{3} + 2x^{2} – 4x – 8)

g(x) = 4x^{4} + 14x^{3} + 8x^{2} – 8x

= 2x (2x^{3} + 7x^{2} + 4x – 4)

G.C.D. of 3x and 2x = x

Now g(x) is divide by p(x) we get

3x^{2} + 12x + 12 = 3 (x^{2} + 4x + 4)

Now dividing p(x) = x^{3} + 2x^{2} – 4x – 8

by the new remainder

(leaving the constant)

x^{2} + 4x + 4

G.C.D. = x(x^{2} + 4x + 4) [Note x is common for p(x) and g(x)]

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

p(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)

G.C.D. of 3 and 6 = 3

Now g(x) is divided by p(x)

Now dividing p(x) by the remainder x^{2} + 1

we get x + 1

∴ G.C.D. = 3(x^{2} + 1) [3 is the G.C.D. of 3 and 6]

Question 2.

Find the LCM of the given polynomials

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

Answer:

4x^{2} y = 2 × 2 × x^{2} × y

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

L.C.M. = 2^{3} × x^{3} × y^{2}

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

Aliter: L.C.M of 4 and 8 = 8

L.C.M. of x^{2}y and x^{3}y^{2} = x^{3}y^{2}

∴ L.C.M. = 8x^{3}y^{2}

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

Answer:

-9a^{3}b^{2} = -(3^{2} × a^{3} × b^{2})

12a^{2}b^{2}c = 2^{2} × 3 × a^{2} × b^{2} × c

L.C.M. = -(2^{2} × 3^{2} × a^{3} × b^{2} × c)

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

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

Answer:

16m = 24 × m

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

8n^{2} = 2^{3} × n^{2}

L.C.M. = -(2^{4} × 3 × m^{2} × n^{2})

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

(iv) p^{2} – 3p + 2, p^{2} – 4

Answer:

P^{2} – 3p + 2 = p^{2} – 2p – p + 2

= p(p – 2) – 1 (p – 2)

= (p – 2) (p – 1)

p^{2} – 4 = p^{2} – 2^{2} (using a^{2} – b^{2} = (a + b) (a – b)]

= (p + 2) (p – 2)

L.C.M. = (p – 2) (p + 2) (p – 1)

(v) 2x^{2} – 5x – 3,4.x^{2} – 36

Answer:

2x^{2} – 5x – 3 = 2x^{2} – 6x + x – 3

= 2x (x – 3) + 1 (x – 3)

= (x – 3) (2x + 1)

= 4x^{2} – 36 = 4 [x^{2} – 9]

= 4 [x^{2} – 3^{2}]

= 4(x + 3) (x – 3)

L.C.M. = 4(x – 3) (x + 3) (2x + 1)

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

Answer:

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

(4x – 6y)^{3} = 2^{3} (2x – 3y)^{3}

= 8 (2x – 3y)^{3}

8x^{3} – 27y^{3} = (2x)^{3} – (3y)^{3}

= (2x – 3y) [(2x)^{2} + 2x × 3y + (3y^{2})]

[using a^{3} – b^{3} = (a – b) (a^{2} + ab + b^{2})

(2x – 3y) (4x^{2} + 6xy + 9y^{2})

L.C.M. = 8x^{2} (2x – 3y)^{3} (4x^{2} + 6xy + 9y)^{2}