Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.2

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) x4 + 3x3 – x – 3, x3 + x2 – 5x + 3
Answer:
p(x) = x4 + 3x3 – x – 3
g(x) = x3 + x2 – 5x + 3
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.2 1

3x2 + 6x – 9 = 3(x2 + 2x – 3)
Now dividing g(x) = x3 + x2 – 5x + 3
by the new remainder
(leaving the constant 3)
we get x2 + 2x – 3
G.C.F. = x2 + 2x – 3
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.2 2

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.2

(ii) x4 – 1, x3 – 11x2 + x – 11
p(x) = x4 – 1
g(x) = x3 – 11x2 + x – 11
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.2 3

120x2 + 120 = 120 (x2 + 1)
Now dividing g(x) = x3 – 11x2 + x – 11 by the new remainder (leaving the constant) we get x2 + 1
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.2 4
G.C.D. = x2 + 1

(iii) 3x4 + 6x3 – 12x2 – 24x, 4x4 + 14x3 + 8x2 – 8x
Answer:
p(x) = 3x4 + 6x3 – 12x2 – 24x
= 3x (x3 + 2x2 – 4x – 8)
g(x) = 4x4 + 14x3 + 8x2 – 8x
= 2x (2x3 + 7x2 + 4x – 4)
G.C.D. of 3x and 2x = x
Now g(x) is divide by p(x) we get
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.2 5

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.2

3x2 + 12x + 12 = 3 (x2 + 4x + 4)
Now dividing p(x) = x3 + 2x2 – 4x – 8
by the new remainder
(leaving the constant)
x2 + 4x + 4
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.2 6
G.C.D. = x(x2 + 4x + 4) [Note x is common for p(x) and g(x)]

(iv) 3x3 + 3x2 + 3x + 3, 6x3 + 12x2 + 6x+12
p(x) = 3x3 + 3x2 + 3x + 3
= 3(x3 + x2 + x + 1)
g(x) = 6x3 + 12x2 + 6x + 12
= 6(x3 + 2x2 + x + 2)
G.C.D. of 3 and 6 = 3
Now g(x) is divided by p(x)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.2 7
Now dividing p(x) by the remainder x2 + 1
we get x + 1
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.2 8
∴ G.C.D. = 3(x2 + 1) [3 is the G.C.D. of 3 and 6]

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.2

Question 2.
Find the LCM of the given polynomials
(i) 4x2y, 8x3y2
Answer:
4x2 y = 2 × 2 × x2 × y
8 x3 y2 = 2 × 2 × 2 × x3 × y2
L.C.M. = 23 × x3 × y2
= 8x3y2

Aliter: L.C.M of 4 and 8 = 8
L.C.M. of x2y and x3y2 = x3y2
∴ L.C.M. = 8x3y2

(ii) -9a3b2, 12a2b2c
Answer:
-9a3b2 = -(32 × a3 × b2)
12a2b2c = 22 × 3 × a2 × b2 × c
L.C.M. = -(22 × 32 × a3 × b2 × c)
= -36 a3b2c

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.2

(iii) 16m, -12m2n2, 8n2
Answer:
16m = 24 × m
-12 m2n2 = -(22 × 3 × m2 × n2)
8n2 = 23 × n2
L.C.M. = -(24 × 3 × m2 × n2)
= -48 m2n2

(iv) p2 – 3p + 2, p2 – 4
Answer:
P2 – 3p + 2 = p2 – 2p – p + 2
= p(p – 2) – 1 (p – 2)
= (p – 2) (p – 1)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.2 9
p2 – 4 = p2 – 22 (using a2 – b2 = (a + b) (a – b)]
= (p + 2) (p – 2)
L.C.M. = (p – 2) (p + 2) (p – 1)

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.2

(v) 2x2 – 5x – 3,4.x2 – 36
Answer:
2x2 – 5x – 3 = 2x2 – 6x + x – 3
= 2x (x – 3) + 1 (x – 3)
= (x – 3) (2x + 1)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.2 10
= 4x2 – 36 = 4 [x2 – 9]
= 4 [x2 – 32]
= 4(x + 3) (x – 3)
L.C.M. = 4(x – 3) (x + 3) (2x + 1)

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.2

(vi) (2x2 – 3xy)2,(4x – 6y)3,(8x3 – 27y3)
Answer:
(2x2 – 3xy)2 = x2 (2x – 3y)2
(4x – 6y)3 = 23 (2x – 3y)3
= 8 (2x – 3y)3
8x3 – 27y3 = (2x)3 – (3y)3
= (2x – 3y) [(2x)2 + 2x × 3y + (3y2)]
[using a3 – b3 = (a – b) (a2 + ab + b2)
(2x – 3y) (4x2 + 6xy + 9y2)
L.C.M. = 8x2 (2x – 3y)3 (4x2 + 6xy + 9y)2

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