{"id":387,"date":"2023-10-17T11:45:45","date_gmt":"2023-10-17T06:15:45","guid":{"rendered":"https:\/\/tnboardsolutions.com\/?p=387"},"modified":"2023-11-10T14:40:19","modified_gmt":"2023-11-10T09:10:19","slug":"samacheer-kalvi-10th-maths-guide-chapter-3-ex-3-3","status":"publish","type":"post","link":"https:\/\/tnboardsolutions.com\/samacheer-kalvi-10th-maths-guide-chapter-3-ex-3-3\/","title":{"rendered":"Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.3"},"content":{"rendered":"

Students can download Maths Chapter 3 Algebra Ex 3.3 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide<\/a> 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.<\/p>\n

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.3<\/h2>\n

Question 1.
\nFind the LCM and GCD for the following and verify that f(x) \u00d7 g(x) = LCM \u00d7 GCD<\/p>\n

(i) 21x2<\/sup>y, 35xy2<\/sup>
\nAnswer:
\np(x) = 21 x2<\/sup>y = 3 \u00d7 7 \u00d7 x2<\/sup> \u00d7 y
\ng(x) = 35xy2<\/sup> = 5 \u00d7 7 \u00d7 x \u00d7 y2<\/sup>
\nG.C.D = 7 xy
\nL.C.M = 3 \u00d7 5 \u00d7 7 x2<\/sup> \u00d7 y2<\/sup>
\n= 105 x2<\/sup>y2<\/sup>
\nL.C.M \u00d7 G.C.D = 105x2<\/sup>y2<\/sup> \u00d7 7xy
\n= 735 x3<\/sup>y3<\/sup> ….(1)
\np(x) \u00d7 g(x) = 21x2<\/sup>y \u00d7 35xy2<\/sup>
\n= 735x3<\/sup>y3<\/sup> ….(2)
\nFrom (1) and (2) we get
\nL.C.M \u00d7 G.C.D. = p(x) \u00d7 g(x)<\/p>\n

\"Samacheer<\/p>\n

(ii) (x3<\/sup> – 1)(x + 1),(x3<\/sup> + 1)
\nAnswer:
\np(x) = (x3<\/sup> – 1) (x + 1) = (x – 1) (x2<\/sup> + x + 1) (x + 1)
\ng(x) = x3<\/sup> + 1 = (x + 1) (x2<\/sup> – x + 1)
\nG.C.D = (x + 1)
\nL.C.M = (x + 1) (x – 1) (x2<\/sup> + x + 1) (x2<\/sup> – x + 1)
\nL.C.M \u00d7 G.C.D = (x + 1) (x – 1)(x2<\/sup> + x + 1)(x2<\/sup> – x + 1)x(x + 1)
\n= (x + 1)2<\/sup> (x – 1) (x2<\/sup> + x + 1) (x2<\/sup> – x + 1) ……….(1)
\np(x) \u00d7 g(x) = (x – 1) (x2<\/sup> + x + 1) (x + 1) (x + 1) (x2<\/sup> – x + 1)
\n= (x + 1)2<\/sup> (x – 1) (x2<\/sup> + x + 1) (x2<\/sup> – x + 1) ……….(2)
\nFrom (1) and (2) we get
\nL.C.M \u00d7 G.C.D. = p(x) \u00d7 g(x)<\/p>\n

(iii) (x2<\/sup>y + xy2<\/sup>), (x2<\/sup> + xy)
\nAnswer:
\np(x) = x2<\/sup>y + xy2<\/sup> = xy(x + y)
\ng(x) = x2<\/sup> + xy = x(x + y)
\nG.C.D = x(x+y)
\nL.C.M = xy (x +y).
\nL.C.M \u00d7 G.C.D = xy(x + y) \u00d7 x(x + y)
\n= x2<\/sup>y(x + y)2<\/sup> …..(1)
\np(x) \u00d7 g(x) = xy(x + y) \u00d7 x(x + y)
\n= x2<\/sup>y(x + y)2<\/sup>
\nFrom (1) and (2) we get
\nL.C.M \u00d7 G.C.D. = p(x) \u00d7 g(x)<\/p>\n

\"Samacheer<\/p>\n

Question 2.
\nFind the LCM of each pair of the following polynomials
\n(i) a2<\/sup> + 4a – 12, a2<\/sup> – 5a + 6 whose GCD is a – 2
\nAnswer:
\np(x) = a2<\/sup> + 4a – 12
\n= a2<\/sup> + 6a – 2a – 12
\n= a (a + 6) – 2(a + 6)
\n= (a + 6) (a – 2)
\n\"Samacheer
\ng(x) = a2<\/sup> – 5a + 6
\n= a2<\/sup> – 3a – 2a + 6
\n\"Samacheer
\n= a(a – 3) – 2 (a – 3)
\n= (a – 3) (a – 2)
\n\"Samacheer<\/p>\n

(ii) x4<\/sup> – 27a3<\/sup>x, (x – 3a)2<\/sup> whose GCD is (x – 3a)
\nAnswer:
\np(x) = x4<\/sup> – 27a3<\/sup>x = x[x3<\/sup> – 27a3<\/sup>]
\n= x[x3<\/sup> – (3a)3<\/sup>]
\n= x(x – 3a) (x2<\/sup> + 3ax + 9a2<\/sup>)
\ng(x) = (x – 3a)2<\/sup>
\nG.C.D. = x – 3a
\n\"Samacheer
\nL.C.M. = x (x – 3a)2<\/sup> (x2<\/sup> + 3ax + 9a2<\/sup>)<\/p>\n

\"Samacheer<\/p>\n

Question 3.
\nFind the GCD of each pair of the following polynomials
\n(i) 12(x4<\/sup> – x3<\/sup>), 8(x4<\/sup> – 3x3<\/sup> + 2x2<\/sup>) whose LCM is 24x3<\/sup> (x – 1)(x – 2)
\nAnswer:
\np(x) = 12(x4<\/sup> – x3<\/sup>)
\n= 12x3<\/sup>(x- 1)
\ng(x) = 8(x4<\/sup> – 3x3<\/sup> + 2x2<\/sup>)
\n\"Samacheer
\n= 8x2<\/sup>(x2<\/sup> – 3x + 2)
\n= 8x2<\/sup>(x – 2)(x – 1)
\nL.C.M. = 24x3<\/sup> (x – 1) (x – 2)
\n\"Samacheer<\/p>\n

(ii) (x3<\/sup> + y3<\/sup>), (x4<\/sup> + x2<\/sup>y2<\/sup> + y4<\/sup>) whose LCM is (x3<\/sup> + y3<\/sup>) (x2<\/sup> + xy + y2<\/sup>)
\nAnswer:
\np(x) = x3<\/sup> + y3<\/sup>
\n= (x + y)(x2<\/sup> – xy + y2<\/sup>)
\ng(x) = x4<\/sup> + x2<\/sup>y2<\/sup> + y4<\/sup> = [x2<\/sup> + y2<\/sup>]2<\/sup> – (xy)2<\/sup>
\n= (x2<\/sup> + y2<\/sup> + xy) (x2<\/sup> + y2<\/sup> – xy)
\nL.C.M. = (x3<\/sup> + y3<\/sup>) (x2<\/sup> + xy + y2<\/sup>)
\n(x + y) (x2<\/sup> – xy + y2<\/sup>) (x2<\/sup> + xy + y2<\/sup>)
\n\"Samacheer
\nG.C.D. = x2<\/sup> – xy + y2<\/sup><\/p>\n

\"Samacheer<\/p>\n

Question 4.
\nGiven 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
\n\"Samacheer
\nAnswer:
\nL.C.M. = a3<\/sup> – 10a2<\/sup> + 11a + 70
\n= (a – 7) (a2<\/sup> – 3a – 10)
\n= (a – 7) (a – 5) (a + 2)
\n\"Samacheer
\nG.C.D. = (a – 7)
\np(x) = a2<\/sup> -12a + 35
\n= (a – 5)(a – 7)
\n\"Samacheer
\nq(x) = \\(\\frac{\\mathrm{LCM} \\times \\mathrm{GCD}}{p(x)}\\)
\n\"Samacheer<\/p>\n

\"Samacheer<\/p>\n

(ii) L.C.M (x2<\/sup> + y2<\/sup>)(x4<\/sup> + x2<\/sup>y2<\/sup> + y4<\/sup>)
\n(x2<\/sup> + y2<\/sup>)[(x2<\/sup> + y2<\/sup>)2<\/sup>-(xy)2<\/sup>]
\n(x2<\/sup> + y2<\/sup>) (x2<\/sup> + y2<\/sup> + xy) (x2<\/sup> + y2<\/sup> – xy)
\nG.C.D. = x2<\/sup> – y2<\/sup>
\n(x + y)(x – y)
\nq(x) = (x4<\/sup> – y4<\/sup>) (x2<\/sup> + y2<\/sup> – xy)
\n= [(x2<\/sup>)2<\/sup> – (y2<\/sup>)2<\/sup>](x2<\/sup> + y2<\/sup> – xy)
\n= (x2<\/sup> + y2<\/sup>) (x2<\/sup> – y2<\/sup>) (x2<\/sup> + y2<\/sup> – xy)
\n(x2<\/sup> + y2<\/sup>) (x + y) (x – y) (x2<\/sup> + y2<\/sup> – xy)
\nP(x) = x2<\/sup> + y2<\/sup> + xy
\n\"Samacheer<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can download Maths Chapter 3 Algebra Ex 3.3 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 3.3 Question …<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/posts\/387"}],"collection":[{"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/comments?post=387"}],"version-history":[{"count":1,"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/posts\/387\/revisions"}],"predecessor-version":[{"id":43479,"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/posts\/387\/revisions\/43479"}],"wp:attachment":[{"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/media?parent=387"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/categories?post=387"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tnboardsolutions.com\/wp-json\/wp\/v2\/tags?post=387"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}