{"id":854,"date":"2023-11-07T04:17:45","date_gmt":"2023-11-06T22:47:45","guid":{"rendered":"https:\/\/wordpress-505192-1602719.cloudwaysapps.com\/?p=854"},"modified":"2023-11-08T17:09:30","modified_gmt":"2023-11-08T11:39:30","slug":"samacheer-kalvi-9th-maths-chapter-3-ex-3-8","status":"publish","type":"post","link":"https:\/\/samacheerguru.com\/samacheer-kalvi-9th-maths-chapter-3-ex-3-8\/","title":{"rendered":"Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.8"},"content":{"rendered":"

You can Download Samacheer Kalvi 9th Maths Book Solutions<\/a>\u00a0Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.<\/p>\n

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.8<\/h2>\n

Question 1.
\nFactorise each of the following polynomials using synthetic division:
\n(i) x3<\/sup> – 3x2<\/sup> – 10x + 24
\n(ii) 2x3<\/sup> – 3x2<\/sup> – 3x + 2
\n(iii) -7x + 3 + 4x3<\/sup>
\n(iv) x3<\/sup> + x2<\/sup> – 14x – 24
\n(v) x3<\/sup> – 7x + 6
\n(vi) x3<\/sup> – 10x2<\/sup> – x + 10
\nSolution:
\n(i) x3<\/sup> – 3x2<\/sup> – 10x + 24
\nLet p(x) = x3<\/sup> – 3x2<\/sup> – 10x + 24
\nSum of all the co-efficients = 1 – 3 – 10 + 24 = 25 – 13 = 12 \u2260 0
\nHence (x – 1) is not a factor.
\nSum of co-efficient of even powers with constant = -3 + 24 = 21
\nSum of co-efficients of odd powers = 1 – 10 = – 9
\n21 \u2260 -9
\nHence (x + 1) is not a factor.
\np (2) = 23<\/sup> – 3 (22<\/sup>) – 10 \u00d7 2 + 24 = 8 – 12 – 20 + 24
\n= 32 – 32 = 0 \u2234 (x – 2) is a factor.
\nNow we use synthetic division to find other factor
\n\"Samacheer
\nThqs (x – 2) (x + 3) (x – 4) are the factors.
\n\u2234 x3<\/sup> – 3x2<\/sup> – 10x + 24 = (x – 2) (x + 3) (x – 4)<\/p>\n

(ii) 2x2<\/sup> – 3x2<\/sup> – 3x + 2
\nLet p (x) = 2x3<\/sup> – 3x2<\/sup> – 3x + 2
\nSum of all the co-efficients are
\n2 – 3 – 3 + 2 = 4 – 6 = -2 \u2260 0
\n\u2234 (x – 1) is not a factor
\nSum of co-efficients of even powers of x with constant = -3 + 2 = – 1
\nSum of co-efficients of odd powers of x = 2- 3= -1
\n(-1) = (-1)
\n\u2234 (x + 1) is a factor
\nLet us find the other factors using synthetic division
\n\"Samacheer
\nQuotient is 2x2<\/sup> – 5x + 2 = 2x – 4x – x + 2 = 2x (x – 2) – 1 (x – 2)
\n= (x – 2) (2x – 1)
\n\u2234 2x3<\/sup> – 3x2<\/sup> – 3x + 2 = (x + 1) (x – 2) (2x – 1)<\/p>\n

(iii) -7x + 3 + 4x3<\/sup>
\nLet p(x) = 4x3<\/sup> + 0x2<\/sup> – 7x + 3
\nSum of the co-efficients are = 4 + 0 – 7 + 3
\n= 7 – 7 = 0
\n\u2234 (x- 1) is a factor
\nSum of co-efficients of even powers of x with constant = 0 + 3 = 3
\nSum of co-efficients of odd powers of x with constant = 4 – 7 = -3
\n-3 \u2260 -3
\n\u2234 (x + 1) is not a factor
\nUsing synthetic division, let us find the other factors.
\n\"Samacheer
\n\"Samacheer
\nQuotient is 4x2<\/sup> + 4x – 3
\n= 4x2<\/sup> + 6x – 2x – 3
\n= 2x (2x + 3) – 1 (2x + 3)
\n= (2x + 3) (2x – 1)
\n\u2234 The factors are (x – 1), (2x + 3) and (2x – 1)
\n\u2234 -7x + 3 + 4x3<\/sup> = (x + 1) (2x + 3) (2x – 1)<\/p>\n

\"Samacheer<\/p>\n

(iv) x3<\/sup> + x2<\/sup> – 14x – 24
\nLet p (x) = x3<\/sup> + x2<\/sup> – 14x – 24
\nSum of the co-efficients are = 1 + 1 – 14 – 24 = -36 \u2260 0
\n\u2234 (x – 1) is not a factor
\nSum of co-efficients of even powers of x with constant = 1 – 24 = -23
\nSum of co-efficients of odd powers of x = 1 – 14 = -3
\n-23 \u2260 -13
\n\u2234 (x + 1) is also not a factor
\np(2) = 23<\/sup> + 22<\/sup> – 14 (2) – 24 = 8 + 4 – 28 – 24
\n= 12 – 52 \u2260 0, (x – 2) is a not a factor
\np (-2) = (-2)3<\/sup> + (-2)2<\/sup> – 14 (-2) – 24
\n= -8 + 4 + 28 – 24 = 32 – 32 = 0
\n\u2234 (x + 2) is a factor
\nTo find the other factors let us use synthetic division.
\n\"Samacheer
\n\u2234 The factors are (x + 2), (x + 3), (x + 4)
\n\u2234 x3<\/sup> + x2<\/sup> – 14x – 24 = (x + 2) (x + 3) (x – 4)<\/p>\n

(v) x3<\/sup> – 7x + 6
\nLet p (x) = x3<\/sup> + 0x2<\/sup> – 7x + 6
\nSum of the co-efficients are = 1 + 0 – 7 + 6 = 7 – 7 = 0
\n\u2234(x- 1) is a factor
\nSum of co-efficients of even powers of x with constant = 0 + 6 = 6
\nSum of coefficient of odd powers of x = 1 – 7 = -7
\n6 \u2260 -7
\n\u2234 (x + 1) is not a factor
\nTo find the other factors, let us use synthetic division.
\n\"Samacheer
\n\u2234 The factors are (x – 1), (x – 2), (x + 3)
\n\u2234 x3<\/sup> + 0x2<\/sup> – 7x + 6 = (x – 1) (x – 2) (x + 3)<\/p>\n

(vi) x3<\/sup> – 10x2<\/sup> – x + 10
\nLet p (x) = x3<\/sup> – 10x2<\/sup> – x + 10
\nSum of the co-efficients = 1 – 0 – 1 + 10
\n= 11 – 11 = 0
\n\u2234 (x – 1) is a factor
\nSum of co-efficients of even powers of x with constant = -10 + 10 = 0
\nSum of co-efficients of odd powers of = 1 – 1 = 0
\n\u2234(x + 1) is a factor
\nSynthetic division
\n\"Samacheer
\n\u2234 x3<\/sup> + 10x2<\/sup> – x + 10 = (x – 1) (x + 1) (x – 10)<\/p>\n","protected":false},"excerpt":{"rendered":"

You can Download Samacheer Kalvi 9th Maths Book Solutions\u00a0Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations. Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.8 Question 1. Factorise each of the following polynomials using synthetic division: (i) x3 – 3×2 – 10x […]<\/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":[3],"tags":[],"blocksy_meta":"","jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/posts\/854"}],"collection":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/comments?post=854"}],"version-history":[{"count":1,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/posts\/854\/revisions"}],"predecessor-version":[{"id":55004,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/posts\/854\/revisions\/55004"}],"wp:attachment":[{"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/media?parent=854"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/categories?post=854"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/samacheerguru.com\/wp-json\/wp\/v2\/tags?post=854"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}