{"id":1131,"date":"2021-04-14T09:25:36","date_gmt":"2021-04-14T03:55:36","guid":{"rendered":"https:\/\/wordpress-505192-1602719.cloudwaysapps.com\/?p=1131"},"modified":"2021-04-14T09:47:57","modified_gmt":"2021-04-14T04:17:57","slug":"samacheer-kalvi-8th-maths-term-1-chapter-3-ex-3-4","status":"publish","type":"post","link":"https:\/\/samacheerkalviguru.com\/samacheer-kalvi-8th-maths-term-1-chapter-3-ex-3-4\/","title":{"rendered":"Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.4"},"content":{"rendered":"

Students can Download Maths Chapter 3 Algebra Ex 3.4 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions<\/a> Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.<\/p>\n

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.4<\/h2>\n

8th Standard Maths Exercise 3.4 Question 1.<\/strong>
\nFactorise the following by taking out the common factor
\n(i) 18xy – 12yz
\n(ii) 9x5<\/sup>y3<\/sup> + 6x3<\/sup>y2<\/sup> – 18x2<\/sup>y
\n(iii) x(b – 2c) + y(b – 2c)
\n(iv) (ax + ay) + (bx + by)
\n(v) 2x2<\/sup>(4x – 1) – 4x + 1
\n(vi) 3y(x – 2)2<\/sup> – 2(2 – x)
\n(vii) 6xy – 4y2<\/sup> + 12xy – 2yzx
\n(viii) a3<\/sup> – 3a2<\/sup> + a – 3
\n(ix) 3y3<\/sup> – 48y
\n(x) ab2<\/sup> – bc2<\/sup> – ab + c2<\/sup>
\nSolution:
\n(i) 18xy – 12yz = (2 \u00d7 3 \u00d7 3 \u00d7 y \u00d7 x) – (2 \u00d7 2 \u00d7 3 \u00d7 y \u00d7 z)
\nTaking out the common factors 2, 3, y, we get
\n= 2 \u00d7 3 \u00d7 y (3x – 2z) = 6y (3x- 2z)<\/p>\n

i) 9x5<\/sup>y3<\/sup> + 6x3<\/sup>y2<\/sup> – 18x2<\/sup>y = (3 \u00d7 3 \u00d7 x2<\/sup> \u00d7 x3<\/sup> \u00d7 y \u00d7 y2<\/sup>) + (2 \u00d7 3 \u00d7 x2<\/sup> \u00d7 x \u00d7 y \u00d7 y)
\nTaking out the common factors 3, x2<\/sup>, y, we get
\n= 3 \u00d7 x2<\/sup> \u00d7 y (3x3<\/sup> y2<\/sup> + 2xy – 6)
\n= 3x2<\/sup>y (3x3<\/sup> y2<\/sup> + 2xy – 6)<\/p>\n

(iii) x(b – 2c) + y(b – 2c)
\nTaking out the binomial factor (b – 2c) from each term, we have
\n= (b – 2c)(x + y)<\/p>\n

(iv) (ax + ay) + (bx + by)
\nTaking at \u2018a\u2019 from the first term and \u2018b\u2019 from the second term we have
\n(ax + ay) + (bx + by) = a (x + y) + b (x + y)
\nNow taking out the binomial factor (x + y) from each term
\n= (x + y)(a + b)<\/p>\n

(v) 2x2<\/sup>(4x – 1) – 4x + 1
\nTaking out -1 from last two terms
\n2x2<\/sup> (4x – 1) – 4x + 1 = 2x2<\/sup> (4x – 1) – 1 (4x- 1)
\nTaking out the binomial factor 4x – 1, we get
\n= (4x – 1)(2x2<\/sup> – 1)<\/p>\n

(vi) 3y(x – 2)2<\/sup> – 2(2 – x)
\n3y(x – 2)2<\/sup> – 2(2 – x) = 3y(x – 2)(x – 2)-2(-1) (x – 2) [\u2235 Taking out -1 from 2 – x]
\n= 3y (x – 2) (x – 2) + 2 (x – 2)
\nTaking out the binomial factor x – 2 from each term, we get
\n= (x – 2) [3y (x – 2) + 2]<\/p>\n

(vii) 6xy – 4y2<\/sup> + 12xy – 2yzx
\n= 6xy + 12xy – 4y2<\/sup> – 2yzx [\u2235 Addition is commutative]
\n= (6 \u00d7 x \u00d7 y) + (2 \u00d7 6 \u00d7 x \u00d7 y) + (-1) (2) (2) y + y) + ((-1) (2) (y) (z) (x))
\nTaking out 6 \u00d7 x \u00d7 y from first two terms and (-1) \u00d7 2 \u00d7 y from last two terms we get
\n= 6 \u00d7 x \u00d7 y(1 + 2) + (-1)(2)y[2y + zx]
\n= 6xy (3) -2y(2y + zx)
\n= (2 \u00d7 3 \u00d7 3 \u00d7 x \u00d7 y) – 2xy (2y + zx)
\nTaking out 2y from two terms
\n= 2y (9x – (2y + zx)) = 2y (9x – 2y – xz)<\/p>\n

(viii) a2<\/sup> – 3a2<\/sup> + a – 3 = a2<\/sup> (a – 3) + 1 (a – 3) [\u2235 Grouping the terms suitably]
\n= (a – 3) (a2<\/sup> + 1)<\/p>\n

(ix) 3y2<\/sup> – 48y = 3 \u00d7 y \u00d7 y2<\/sup> – 3 \u00d7 16 \u00d7 y
\nTaking out 3 \u00d7 y = 3y (y2<\/sup> – 16) = 3y (y2<\/sup> – 42<\/sup>)
\nComparing y2<\/sup> – 42<\/sup> with a2<\/sup> – b2<\/sup>
\na = y, b = 4
\na2<\/sup> – b2<\/sup> = (a + b) (a – b)
\ny2<\/sup> – 42<\/sup> = (y + 4) (y – 4)
\n\u2234 3y (y2<\/sup> – 16) = 3y (y + 4) (y – 4)<\/p>\n

(x) ab2<\/sup> – bc2<\/sup> – ab + c2<\/sup>
\nGrouping suitably
\nab2<\/sup> – bc2<\/sup> – ab + c2<\/sup> = b ((ab – c2<\/sup>) – 1(ab – c2<\/sup>)
\nTaking out the binomial factor ab – c2<\/sup> = (ab – c2<\/sup>) (b – 1)<\/p>\n

Samacheer Kalvi Guru 8th Maths Question 2.<\/strong>
\nFactorise the following expressions
\n(i) x2<\/sup> + 14x + 49
\n(ii) y2<\/sup> – 10y + 25
\n(iii) c2<\/sup> – 4c – 12
\n(iv) m2<\/sup> + m – 72
\n(v) 4x2<\/sup> – 8x + 3
\nSolution:
\nx2<\/sup> + 14x + 49 = x2<\/sup> + 14x + 72
\nComparing with a2<\/sup> + 2ab + b2<\/sup> = (a + b)2<\/sup> we have a = x and b = 7
\n\u21d2 x2<\/sup> + 2(x) (7) + 72<\/sup> = (x + 7)2<\/sup>
\n\u2234 x2<\/sup> + 14x + 49 = (x + 7)2<\/sup><\/p>\n

(ii) y2<\/sup> – 10y + 25 = y2<\/sup> – 10y + 52<\/sup>
\nComparing with a2<\/sup> – 2ab + b2<\/sup> = (a – b)2<\/sup> we get a = y ; b = 5
\n\u21d2 y2<\/sup> – 2(y) (5) + 52<\/sup> = (y – 5)2<\/sup>
\n\u2234 y2<\/sup> – 10y + 25 = (y – 5)2<\/sup><\/p>\n

(iii) c2<\/sup> – 4c – 12
\nThis is of the form ax2<\/sup> + bx + c
\nWhere a = 1,b = – 4 c = – 12, x = c
\nNow the product ac = 1 \u00d7 – 12 = – 12 and the sum b = -4
\n\"Samacheer<\/p>\n

(iv) m2<\/sup> + m – 72
\n\"Samacheer
\nThis is of the form ax2<\/sup> + bx + c
\nwhere a = 1, b = 1, c = -12
\nProduct a \u00d7 c = 1 \u00d7 -72 = -72
\nSum b = 1
\nThe middle term m can be written as 9m – 8m
\nm2<\/sup> + m – 72 = m2<\/sup> + 9m – 8m – 12
\n= m (m + 9) – 8 (m + 9)
\nTaking out (m + 9)
\n= (m + 9) (m – 8)
\n\u2234 m2<\/sup> + m – 72 = (m + 9) (m – 8)<\/p>\n

(v) 4x2<\/sup> – 8x + 3
\nThis is of the form ax2<\/sup> + bx + c with a = 4 b = -8 c = 3
\nProduct ac = 4 \u00d7 3 = 12
\nSum b = -8
\n\"Samacheer<\/p>\n

Samacheer Kalvi Guru Maths 8th Question 3.<\/strong>
\nFactorize the following expressions using a3<\/sup> + b3<\/sup> = (a + b)(a2<\/sup> – ab + b2<\/sup>) identity
\n(i) h3<\/sup> + k2<\/sup>
\n(ii) 2a3<\/sup> + 16
\n(iii) x3<\/sup>y3<\/sup> + 27
\n(iv) 64m3<\/sup> + n3<\/sup>
\n(v) r4<\/sup> + 27p3<\/sup>r
\nSolution:
\n(i) h3<\/sup> + k3<\/sup>
\nComparing h3<\/sup> + k3<\/sup> with a3<\/sup> + b3<\/sup> = (a + b)(a2<\/sup> – ab + b2<\/sup>) we have a = h, b = k
\n\u2234 h3<\/sup> + k3<\/sup> = (h + k)(h2<\/sup> – hk + k2<\/sup>)<\/p>\n

(ii) 2a2<\/sup> + 16
\n(2 \u00d7 a3<\/sup>) + (2 \u00d7 8) = 2(a3<\/sup> + 8) = 2 (a3<\/sup> + 23<\/sup>)
\n\u2234 2a3<\/sup> + 16 = 2 (a3<\/sup> + 23<\/sup>)
\nComparing with a3<\/sup> + b3<\/sup> we have a = a and b = 2
\na3<\/sup> + b3<\/sup> = (a + b)(a2<\/sup> – ab + b2<\/sup>)
\n2(a3<\/sup> + 23<\/sup>) = 2[(a + 2) (a2<\/sup> – (a) (2) + 22<\/sup>)] = 2[(a + 2) (a2<\/sup> – 2a + 4)]
\n2a3<\/sup> +16 = 2 (a + 2) (a2<\/sup> – 2a + 4)<\/p>\n

(iii) x3<\/sup> y3<\/sup> + 27 = (xy)3<\/sup> + 33<\/sup>
\nComparing with a3<\/sup> + b3<\/sup> we have a = xy ;b = 3
\na3<\/sup> + b3<\/sup> = (a + b) (a2<\/sup> – ab + b2<\/sup>)
\n(xy)3<\/sup> + 33<\/sup> = (xy + 3) ((xy)2<\/sup> – (xy) (3) + 32<\/sup>) = (xy + 3) (x2<\/sup>y2<\/sup> – 3xy + 9)
\n\u2234 x3<\/sup>y3<\/sup> + 27 = (xy + 3)(x2<\/sup>y2<\/sup> – 3xy + 9)<\/p>\n

(iv) 64m3<\/sup> + n3<\/sup> = (43<\/sup>m3<\/sup>) + n3<\/sup>
\n= (4m)3<\/sup> + n3<\/sup>
\nComparing this with a3<\/sup> + b3<\/sup> we have a = 4m; b = n
\na3<\/sup> + b3<\/sup> = (a + b) (a2<\/sup> – ab + b2<\/sup>)
\n(4m)3<\/sup> + n3<\/sup> = (4m + n) [(4m)2 – (4m) (n) + n2]
\n= (4m + n) [42<\/sup>m2<\/sup> – 4mn + n2<\/sup>]
\n= (4m + n) [ 16m2<\/sup> – 4mn + n2<\/sup>]
\n64m3<\/sup> + n3<\/sup> = (4m + n) (16m2<\/sup> – 4mn + n2<\/sup>)<\/p>\n

(v) r3<\/sup> + 27p3<\/sup>r = r (r3<\/sup> + 27p3<\/sup>) = r [r3<\/sup> + 33<\/sup>p3<\/sup>]
\nComparing r3<\/sup> + (3p)3<\/sup> with a3<\/sup> + b3<\/sup> we have a = r; b = 3p
\na3<\/sup> + b3<\/sup> = (a + b)(a2<\/sup> – ab + b2<\/sup>)
\nr[r3<\/sup> + (3p)3<\/sup>] = r[(r + 3p)(r2<\/sup> – r(3p) + (3p)2<\/sup>)]
\n= r[(r+ 3p) (r2<\/sup> – 3rp + 32<\/sup>p2<\/sup>]
\n= r(r + 3p) (r2<\/sup> – 3rp + 9p2<\/sup>)
\nr4<\/sup> + 27p3<\/sup>r = r(r + 3p) (r2<\/sup> – 3rp + 9p2<\/sup>)<\/p>\n

11th Maths Exercise 3.4 Samacheer Kalvi Question 4.<\/strong>
\nFactorize the following expressions using a3<\/sup> – b3<\/sup> = (a – b)(a2<\/sup> + ab + b2<\/sup>) identity
\n(i) y3<\/sup> – 27
\n(ii) 3b3<\/sup> – 192c3<\/sup>
\n(iii) -16y3<\/sup> + 2x3<\/sup>
\n(iv) x3<\/sup> y3<\/sup> – 73<\/sup>
\n(v) c3<\/sup> – 27b3<\/sup> a3<\/sup>
\nSolution:
\n(i) y3<\/sup> – 27 = y3<\/sup> – 33<\/sup>
\nComparing this with a3<\/sup> – b3<\/sup> , we have a = y and b = 3
\na3<\/sup> – b3<\/sup> = (a + b) (a2<\/sup> + ab + b2<\/sup> )
\ny3<\/sup> – 33<\/sup> = (y – 3)(y2<\/sup> + (y) (3) + 32<\/sup> ) = (y – 3) (y2<\/sup> + 3y + 9)
\ny3<\/sup> – 27 = (y – 3) (y2<\/sup> + 3y + 9)<\/p>\n

(ii) 3b3<\/sup> + 192c3<\/sup> = (3 \u00d7 b3<\/sup>) – (3 \u00d7 4 \u00d7 4 \u00d7 4 \u00d7 c3<\/sup>) = 3(b3<\/sup> – 43<\/sup> c3<\/sup>)
\n= 3(b3<\/sup> – (4c)3<\/sup> )
\nComparing b3<\/sup> – (4c)3<\/sup> with a3<\/sup> – b3<\/sup> we have a = b and b = 4c
\na3<\/sup> – b3<\/sup> = (a – b) (a2<\/sup> + ab + b2<\/sup>)
\n3(b3<\/sup> – (4c)3<\/sup>) = 3[(b – 4c)(b2<\/sup> + (b)(4c) + (4c)2<\/sup>)]
\n= 3[(b – 4c)(b2<\/sup> + 4bc + 42<\/sup> c2<\/sup>)]
\n3b3<\/sup> – 192c3<\/sup> = 3 [(b – 4c) (b2<\/sup> + 4bc + 16c2<\/sup>)]<\/p>\n

(iii) -16y3<\/sup> + 2x3<\/sup> = 2x3<\/sup> – 16y3<\/sup> [\u2235 Addition is commatative]
\n= 2(x3<\/sup> – 8y3<\/sup>) = 2(x3<\/sup> – 23y3<\/sup>)
\n= 2(x3<\/sup> – (2y)3<\/sup>)
\nComparing x3<\/sup> – (2y)3<\/sup> with a3<\/sup> – b3<\/sup> we have a = x and b = 2y
\na3<\/sup> – b3<\/sup> = (a – b)(a2<\/sup> + ab + b2<\/sup>)
\n2[x3<\/sup> – (2y)3<\/sup>] = 2[(x – 2y) (x2<\/sup> + (x) (2y) + (2y)2<\/sup>)]
\n= 2[(x – 2y) (x2<\/sup> + 2xy + 22<\/sup> y2<\/sup>)]
\n-16y3<\/sup> + 2x3<\/sup> = 2[(x – 2y)(x2<\/sup> + 2xy + 4y2<\/sup>)]<\/p>\n

(iv) x3<\/sup>y3<\/sup> – 73<\/sup> = (xy)3<\/sup> – 73<\/sup>
\nComparing with a3<\/sup> – b3<\/sup> we have a = xy and b = 7
\na3<\/sup> – b3<\/sup> = (a – b)(a2<\/sup> + ab + b2<\/sup>)
\n(xy)3<\/sup> – 73<\/sup> = (xy – 7) ((xy)2<\/sup> + (xy) (7) + 72<\/sup>)
\nx3<\/sup>y3<\/sup> – 73<\/sup> = (xy – 7) (x2<\/sup>y2<\/sup> + 7xy + 49)<\/p>\n

(v) c3<\/sup> – 27 b3<\/sup> a3<\/sup> = c3<\/sup> – 33<\/sup>b3<\/sup> a3<\/sup> = c3<\/sup> – (3ba)3<\/sup>
\nComparing this with a3<\/sup> – b3<\/sup> we have a = x and b = 3ba
\na3<\/sup> – b3<\/sup> = (a – b)(a2<\/sup> + ab + b2<\/sup>)
\n\u2234 c3<\/sup> – (3ba)3<\/sup> = (c – 3ba) (c2<\/sup> + (c) (3ba) + (3ba)2<\/sup>)
\n= (c – 3ba) (c2<\/sup> + 3bac + 32<\/sup> b2<\/sup>a2<\/sup>)
\nc3<\/sup> – 27b3<\/sup>a3<\/sup> = (c – 3ab) (c2<\/sup> + 3bac + 9a2<\/sup>b2<\/sup>)<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 3 Algebra Ex 3.4 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations. Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.4 8th Standard …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":"","footnotes":""},"categories":[4],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/1131"}],"collection":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/comments?post=1131"}],"version-history":[{"count":0,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/posts\/1131\/revisions"}],"wp:attachment":[{"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/media?parent=1131"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/categories?post=1131"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerkalviguru.com\/wp-json\/wp\/v2\/tags?post=1131"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}