Chapter 3, Algebra, Ex 3.3,

10 Dec 2020 2:11 pm

Tamilnadu, Samacheer, Kalvi, 10th, sslc, Maths, Solutions, Chapter 3, Algebra, Ex 3.3,

(i) 21x2y, 35xy2
Answer:
p(x) = 21 x2y = 3 × 7 × x2 × y
g(x) = 35xy2 = 5 × 7 × x × y2
G.C.D = 7 xy
L.C.M = 3 × 5 × 7 x2 × y2
= 105 x2y2
L.C.M × G.C.D = 105x2y2 × 7xy
= 735 x3y3 ….(1)
p(x) × g(x) = 21x2y × 35xy2
= 735x3y3 ….(2)
From (1) and (2) we get
L.C.M × G.C.D. = p(x) × g(x)

(ii) (x3 – 1)(x + 1),(x3 + 1)
Answer:
p(x) = (x3 – 1) (x + 1) = (x – 1) (x2 + x + 1) (x + 1)
g(x) = x3 + 1 = (x + 1) (x2 – x + 1)
G.C.D = (x + 1)
L.C.M = (x + 1) (x – 1) (x2 + x + 1) (x2 – x + 1)
L.C.M × G.C.D = (x + 1) (x – 1)(x2 + x + 1)(x2 – x + 1)x(x + 1)
= (x + 1)2 (x – 1) (x2 + x + 1) (x2 – x + 1) ……….(1)
p(x) × g(x) = (x – 1) (x2 + x + 1) (x + 1) (x + 1) (x2 – x + 1)
= (x + 1)2 (x – 1) (x2 + x + 1) (x2 – x + 1) ……….(2)
From (1) and (2) we get
L.C.M × G.C.D. = p(x) × g(x)

(iii) (x2y + xy2), (x2 + xy)
Answer:
p(x) = x2y + xy2 = xy(x + y)
g(x) = x2 + xy = x(x + y)
G.C.D = x(x+y)
L.C.M = xy (x +y).
L.C.M × G.C.D = xy(x + y) × x(x + y)
= x2y(x + y)2 …..(1)
p(x) × g(x) = xy(x + y) × x(x + y)
= x2y(x + y)2
From (1) and (2) we get
L.C.M × G.C.D. = p(x) × g(x)

Question 2.
Find the LCM of each pair of the following polynomials
(i) a2 + 4a – 12, a2 – 5a + 6 whose GCD is a – 2
Answer:
p(x) = a2 + 4a – 12
= a2 + 6a – 2a – 12
= a (a + 6) – 2(a + 6)
= (a + 6) (a – 2)

g(x) = a2 – 5a + 6
= a2 – 3a – 2a + 6
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.3 2
= a(a – 3) – 2 (a – 3)
= (a – 3) (a – 2)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.3 3

(ii) x4 – 27a3x, (x – 3a)2 whose GCD is (x – 3a)
Answer:
p(x) = x4 – 27a3x = x[x3 – 27a3]
= x[x3 – (3a)3]
= x(x – 3a) (x2 + 3ax + 9a2)
g(x) = (x – 3a)2
G.C.D. = x – 3a

L.C.M. = x (x – 3a)2 (x2 + 3ax + 9a2)


Question 3.
Find the GCD of each pair of the following polynomials
(i) 12(x4 – x3), 8(x4 – 3x3 + 2x2) whose LCM is 24x3 (x – 1)(x – 2)
Answer:
p(x) = 12(x4 – x3)
= 12x3(x- 1)
g(x) = 8(x4 – 3x3 + 2x2)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.3 5
= 8x2(x2 – 3x + 2)
= 8x2(x – 2)(x – 1)
L.C.M. = 24x3 (x – 1) (x – 2)

(ii) (x3 + y3), (x4 + x2y2 + y4) whose LCM is (x3 + y3) (x2 + xy + y2)
Answer:
p(x) = x3 + y3
= (x + y)(x2 – xy + y2)
g(x) = x4 + x2y2 + y4 = [x2 + y2]2 – (xy)2
= (x2 + y2 + xy) (x2 + y2 – xy)
L.C.M. = (x3 + y3) (x2 + xy + y2)
(x + y) (x2 – xy + y2) (x2 + xy + y2)

G.C.D. = x2 – xy + y2


Question 4.
Given the L.C.M and G.C.D of the two polynomials p(x) and q(x) find the unknown polynomial in the following table


Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.3 8
Answer:
L.C.M. = a3 – 10a2 + 11a + 70
= (a – 7) (a2 – 3a – 10)
= (a – 7) (a – 5) (a + 2)


Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.3 9

G.C.D. = (a – 7)
p(x) = a2 -12a + 35
= (a – 5)(a – 7)


Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.3 10
q(x) = LCM×GCDp(x)
Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.3 11

(ii) L.C.M (x2 + y2)(x4 + x2y2 + y4)
(x2 + y2)[(x2 + y2)2-(xy)2]
(x2 + y2) (x2 + y2 + xy) (x2 + y2 – xy)
G.C.D. = x2 – y2
(x + y)(x – y)
q(x) = (x4 – y4) (x2 + y2 – xy)
= [(x2)2 – (y2)2](x2 + y2 – xy)
= (x2 + y2) (x2 – y2) (x2 + y2 – xy)
(x2 + y2) (x + y) (x – y) (x2 + y2 – xy)
P(x) = x2 + y2 + xy


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