{"id":3999,"date":"2021-02-25T17:08:33","date_gmt":"2021-02-25T11:38:33","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=3999"},"modified":"2021-09-15T11:40:40","modified_gmt":"2021-09-15T06:10:40","slug":"ap-board-9th-class-maths-solutions-chapter-2-ex-2-3","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-board-9th-class-maths-solutions-chapter-2-ex-2-3\/","title":{"rendered":"AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.3"},"content":{"rendered":"

AP State Syllabus\u00a0AP Board 9th Class Maths Solutions<\/a> Chapter 2 Polynomials and Factorisation Ex 2.3 Textbook Questions and Answers.<\/p>\n

AP State Syllabus 9th Class Maths Solutions 2nd Lesson Polynomials and Factorisation Exercise 2.3<\/h2>\n

Question 1.
\nFind the remainder when
\nx3<\/sup> + 3x2<\/sup> + 3x + 1 is divided by the following Linear polynomials i) x + 1 Each
\nSolution:
\nLet f(x) = x3<\/sup> + 3x2<\/sup> + 3x + 1
\nBy remainder theorem, the remainder is f (- 1)
\nf (- 1) = (- 1)3<\/sup> + 3(- 1)2<\/sup> + 3(- 1) + 1
\n= – 1 + 3 – 3 + 1 = 0<\/p>\n

\"AP<\/p>\n

ii) \\(x-\\frac{1}{2}\\)
\nf(x) = x3<\/sup> + 3x2<\/sup> + 3x + 1
\nBy remainder theorem, the remainder is \\(\\mathrm{f}\\left(\\frac{1}{2}\\right)\\)
\n\"AP<\/p>\n

iii) x
\nSolution:
\nf(x) = x3<\/sup> + 3x2<\/sup> + 3x + 1
\nThe remainder is f(0)
\n\u2234 f(0) = 03<\/sup> + 3(0)2<\/sup> + 3(0) + 1 = 1<\/p>\n

iv) x + \u03c0
\nSolution:
\nf(x) = x3<\/sup> + 3x2<\/sup> + 3x + 1
\nBy remainder theorem, the remainder is f(- \u03c0)
\nf(- \u03c0) = (- \u03c0)3<\/sup> + 3(-\u03c0)2<\/sup> + 3 (-\u03c0) + 1 .
\n= – \u03c03<\/sup> + 3\u03c02<\/sup> – 3\u03c0 + 1<\/p>\n

v) 5 + 2x
\nf(x) = x3<\/sup> + 3x2<\/sup> + 3x + 1
\nThe remainder is \\(\\mathrm{f}\\left(\\frac{-5}{2}\\right)\\)
\n\"AP<\/p>\n

\"AP<\/p>\n

Question 2.
\nFind the remainder when x3<\/sup> – px2<\/sup> + 6x – p is divided by x – p.
\nSolution:
\nLet f(x) = x3<\/sup> – px2<\/sup> + 6x – p
\n(x – a) = x – p)
\nBy Remainder theorem, the remainder is f(p)
\n\u2234 f(P) = P3<\/sup> – P(P)2<\/sup> + 6p – p
\n= p3<\/sup> – p3<\/sup> + 5p = 5p<\/p>\n

Question 3.
\nFind the remainder when 2x2<\/sup> – 3x + 5 is divided by 2x – 3. Does it exactly divide the polynomial ? State reason.
\nSolution:
\nLet f(x) = 2x2<\/sup> – 3x + 5 and
\nx – a = 2x – 3 = x – \\(\\frac{3}{2}\\)
\nBy Remainder theorem f(x) when divided by (x – \\(\\frac{3}{2}\\) ) leaves a remainder \\(\\mathrm{f}\\left(\\frac{3}{2}\\right)\\)
\n\"AP
\nAs the remainder is 5 we say that (2x – 3) is not a factor of f(x).<\/p>\n

Question 4.
\nFind the remainder when 9x3<\/sup> – 3x2<\/sup> + x – 5 is divided by x – \\(\\frac{2}{3}\\)
\nSolution:
\nLet f(x) = 9x3<\/sup> – 3x2<\/sup> + x – 5
\nx-a = x – \\(\\frac{2}{3}\\)
\nRemainder theorem the remainder is \\(\\mathrm{f}\\left(\\frac{2}{3}\\right)\\)
\n\"AP<\/p>\n

\"AP<\/p>\n

Question 5.
\nIf the polynomials 2x3<\/sup> + ax2<\/sup> + 3x – 5 and x3<\/sup> + x2<\/sup> – 4x + a leave the same remainder, when divided by x – 2, find the value of a.
\nSolution:
\nLet f(x) = 2x3<\/sup> + ax2<\/sup> + 3x – 5
\ng(x) = x3<\/sup> + x2<\/sup> – 4x + a
\nGiven that f(x) and g(x) divided x – 2
\ngive same remainder.
\ni e., f(2) = g(2)
\nBy Remainder theorem.
\nBut f(2) = 2(2)3<\/sup> + a(2)2<\/sup> + 3(2) – 5
\n= 2 x 8 + 4a + 6 – 5
\n= 17 +4a
\ng(2) = 23<\/sup> + 22<\/sup> – 4(2) + a .
\n= 8 + 4 – 8 + a = 4 + a
\ni.e., 4 + a = 17 + 4a
\n\u2234 a – 4a = 17 – 4
\n– 3a = 13
\na = -13\/3<\/p>\n

Question 6.
\nIf the polynomials x3<\/sup> + ax2 <\/sup>+ 5 and x3<\/sup> – 2x2<\/sup> + a are divided by (x + 2) leave the same remainder, find the value of a.
\nSolution:
\nLet f(x) = x3<\/sup> + ax2<\/sup> + 5
\ng(x) = x3<\/sup> – 2x2<\/sup> + a
\nGiven that when f(x) and g(x) divided by (x + 2) leaves the same remainder.
\ni.e.,f(-2) = g(-2)
\nBy Remainder theorem
\nf(- 2) = (- 2)3<\/sup> + a(- 2)2<\/sup> + 5
\n= -8 + 4a + 5 = 4a – 3
\ng(- 2) = (- 2)3<\/sup> – 2(- 2)2<\/sup> + a
\n= -8 – 8 + a = a – 16
\nBy problem,
\n4a – 3 = a – 16
\n4a – a = – 16 + 3
\n\u21d2 3a = – 13 \u21d2 a = -13\/3<\/p>\n

\"AP<\/p>\n

Question 7.
\nFind the remainder when f(x) = x4<\/sup> – 3x2<\/sup> + 4 is divided by g(x) = x – 2 and verify the result by actual division.
\nSolution:
\nGiven f(x) = x4<\/sup> – 3x2<\/sup> + 4
\ng(x) = x – 2
\nThe remainder when f(x) is divided by g(x) is f(2).
\nf(2) = 24<\/sup> – 3(2)2<\/sup> + 4
\n= 16 – 12 + 4
\n= 8
\nActual division
\n\"AP
\n\u2234 The remainder either by Remainder theorem or by actual division is the same.<\/p>\n

Question 8.
\nFind the remainder when p(x) = x3<\/sup> – 6x2<\/sup> + 14x – 3 is divided by g(x) = 1 – 2x and verify the result by long division method.
\nSolution:
\nGiven p(x) = x3<\/sup> – 6x2<\/sup> + 14x – 3
\ng(x) = 1 – 2x
\nBy Remainder theorem when p(x) is divided by g(x) is p(1\/2).
\n\"AP<\/p>\n

\"AP<\/p>\n

Question 9.
\nWhen a polynomial 2x3<\/sup> + 3x2<\/sup> + ax + b is divided by (x – 2) leaves remainder 2, and (x + 2) leaves remainder – 2. Find a and b.
\nSolution:Let f(x) = 2x3<\/sup> + 3x2<\/sup> + ax + b
\nThe remainder when f(x) is divided by (x – 2) is 2.
\ni.e., f(2) = 2
\n\u21d2 2(2)3<\/sup> + 3(2)2<\/sup>+ a(2) + b = 2
\n\u21d2 16 + 12 + 2a +b = 2
\n\u21d2 2a + b = – 26 …………………..(1)<\/p>\n

Also the remainder when f(x) is divided by (x + 2) is – 2.
\ni.e., f(- 2) = – 2
\n\u21d2 2(- 2)3<\/sup> + 3(- 2)2<\/sup> + a (- 2) + b = – 2
\n\u21d2 -16 + 12 – 2a + b = – 2
\n– 2a + b = 2 ………………..(2)
\nSolving (1) and (2),
\n\"AP
\nb = – 12
\nand 2a – 12 = – 26
\n2a = -26+ 12
\na = -14\/2 = -7,
\na = -7, b = – 12<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Syllabus\u00a0AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.3 Textbook Questions and Answers. AP State Syllabus 9th Class Maths Solutions 2nd Lesson Polynomials and Factorisation Exercise 2.3 Question 1. Find the remainder when x3 + 3×2 + 3x + 1 is divided by the following Linear polynomials i) x …<\/p>\n","protected":false},"author":4,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[5],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/3999"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=3999"}],"version-history":[{"count":0,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/3999\/revisions"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=3999"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=3999"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=3999"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}