{"id":5357,"date":"2021-03-05T10:48:25","date_gmt":"2021-03-05T05:18:25","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=5357"},"modified":"2021-09-15T11:39:29","modified_gmt":"2021-09-15T06:09:29","slug":"ap-ssc-10th-class-maths-solutions-chapter-3-optional-exercise","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-ssc-10th-class-maths-solutions-chapter-3-optional-exercise\/","title":{"rendered":"AP SSC 10th Class Maths Solutions Chapter 3 Polynomials Optional Exercise"},"content":{"rendered":"

AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions<\/a> Chapter 3 Polynomials Optional Exercise Textbook Questions and Answers.<\/p>\n

AP State Syllabus SSC 10th Class Maths Solutions 3rd Lesson Polynomials Optional Exercise<\/h2>\n

10th Class Maths 3rd Lesson Polynomials Optional Exercise Textbook Questions and Answers<\/h3>\n

Question 1.
\nVerify that the numbers given along-side the cubic polynomials below are their zeroes. Also verify the relation-ship between the zeroes and the coefficients in each case:
\ni) 2x3<\/sup> + x2<\/sup> – 5x + 2; (\\(\\frac{1}{2}\\), 1,-2)
\nii) x3<\/sup> + 4x2<\/sup> + 5x – 2 ; (1, 1, 1)
\nAnswer:
\ni) Given polynomial 2x3<\/sup> + x2<\/sup> – 5x + 2
\nComparing the given polynomial with ax3<\/sup> + bx2<\/sup> + cx + d,
\nwe get a = 2, b = 1, c = – 5 and d = 2
\n\"AP
\nP(1) = 2(1)3<\/sup> + (1)2<\/sup> – 5(1) – 2
\n= 2 + 1 – 5 + 2 = 0
\np(-2) = 2(-2)3<\/sup> + (-2)2<\/sup> – 5(-2) + 2
\n= 2(-8) + 4 + 10 + 2
\n= – 16 + 16 = 0
\n\u2234 \\(\\frac{1}{2}\\), 1 and – 2 are the zeroes of 2x3<\/sup> + x2<\/sup> – 5x + 2
\nSo, \u03b1 = \\(\\frac{1}{2}\\), \u03b2 = 1 and \u03b3 = – 2 Therefore,
\n\"AP<\/p>\n

\"AP<\/p>\n

ii) Given polynomial x3<\/sup> + 4x2<\/sup> + 5x – 2
\nComparing the given polynomial with ax3<\/sup> + bx2<\/sup> + cx + d,
\nwe get a = 1, b = 4, c = 5 and d = – 2.
\nGiven zeroes are (1, 1, 1)
\np(1) = (1)3<\/sup> + 4(1)2<\/sup> + 5(1) – 2
\n= 1 + 4 + 5 – 2
\n= 10 – 2 = 8
\n\u2234 (1, 1, 1) are not zeroes of the given polynomial p(x).<\/p>\n

Question 2.
\nFind a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, -7, -14 respectively.
\nAnswer:
\nLet the cubic polynomial be
\nax3<\/sup> + bx2<\/sup> + cx + d, and its zeroes be \u03b1, \u03b2 and \u03b3.
\nThen,
\n\u03b1 + \u03b2 + \u03b3 = 2 = \\(\\frac{-(-2)}{1}\\) = \\(\\frac{-b}{a}\\)
\n\u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1 = -7 = \\(\\frac{-7}{1}\\) = \\(\\frac{c}{a}\\)
\n\u03b1\u03b2\u03b3 = – 14 = \\(\\frac{-14}{1}\\) = \\(\\frac{-d}{a}\\)
\na = 1, then b = -2, c = -7 and d = 14.
\nSo, one cubic polynomial which satisfies the given conditions will be x3<\/sup> – 2x2<\/sup> – 7x + 14.<\/p>\n

Question 3.
\nIf the zeroes of the polynomial x3<\/sup> – 3x2<\/sup> + x + 1 are a – b, a, a + b, find a and b.
\nAnswer:
\nGiven polynomial x3<\/sup> – 3x2<\/sup> + x + 1
\nSince, (a – b), a, (a + b) are the zeroes of the polynomial x3<\/sup> – 3x2<\/sup> + x + 1.
\nTherefore, sum of the zeroes
\n= (a – b) + a + (a + b) = \\(\\frac{-(-3)}{1}\\) = 3
\nSo, 3a = 3 \u21d2 a = 1
\n\u2234 Sum of the products of its zeroes taken two at a time.
\n= a(a – b) + a(a + b) + (a + b) (a – b) = \\(\\frac{1}{1}\\) = 1
\n\u21d2 a2<\/sup> – ab + a2<\/sup> + ab + a2<\/sup> – b2<\/sup> = 1
\n\u21d2 3a2<\/sup> – b2<\/sup> = 1
\nSo, 3(1)2<\/sup> – b2<\/sup> = 1 \u21d2 3 – b2<\/sup> = 1
\n\u21d2 b2<\/sup> = 2
\n\u21d2 b = \u221a2 = \u00b1 \u221a2
\nHere, a = 1 and b = \u00b1 \u221a2<\/p>\n

\"AP<\/p>\n

Question 4.
\nIf two zeroes of the polynomial x4<\/sup> – 6x3<\/sup> – 26x2<\/sup> + 138x – 35 are 2 \u00b1 \u221a3 , find other zeroes.
\nAnswer:
\nLet the other, two zeroes are \u03b1, \u03b2.
\nThen the sum of zeroes of given polynomial = 2 + \u221a3 + 2 – \u221a3 + \u03b1 + \u03b2
\n= \\(\\frac{-b}{a}\\) = \\(\\frac{-(-6)}{1}\\) = 6
\n4 + \u03b1 + \u03b2 = 6
\n\u21d2 \u03b1 + \u03b2 = 2 ….. (1)
\nNow product of zeroes is
\n(2 + \u221a3 ) (2 – \u221a3) (\u03b1) (\u03b2)= \\(\\frac{e}{a}\\) = \\(\\frac{-35}{1}\\)
\n(4 – 3) (\u03b1\u03b2) = – 35
\n\u21d2 \u03b1\u03b2 = – 35 …… (2)
\nNow (\u03b1 – \u03b2)2<\/sup> = (\u03b1 + \u03b2)2<\/sup> – 4\u03b1\u03b2
\n= (2)2<\/sup> – 4(-35) = 4 + 140 = 144
\n\u21d2 (\u03b1 – \u03b2) = \u00b1 12 ….. (3)
\nNow solving (1) & (3) we get
\n\"AP
\n\u21d2 \u03b1 = 7; then \u03b1 + \u03b2 = 7 + \u03b2 = 2
\n\u21d2 \u03b2 = -5
\nThe remaining zeroes are \u03b1, \u03b2
\n= 7, -5
\nSo total zeroes of given polynomial are 2 + \u221a3, 2 – \u221a3, 7, – 5.<\/p>\n

\"AP<\/p>\n

Question 5.
\nIf the polynomial x4<\/sup> – 6x3<\/sup> – 16x2<\/sup> + 25x + 10 is divided by another polynomial x2<\/sup> – 2x + k, the remainder comes out to be x + a, find k and a.
\nAnswer:
\nGiven polynomial x4<\/sup> – 6x3<\/sup> – 16x2<\/sup> + 25x + 10 and another polynomial is x2<\/sup> – 2x + k.
\nRemainder is x + a
\nLet us divide
\nx4<\/sup> – 6x3<\/sup> – 16x2<\/sup> + 25x + 10 by x2<\/sup> – 2x + k
\n\"AP
\n\u2234 Remainder
\n= x(4k + 25 – 2k – 48) + 10 + k(k + 24)
\n= x(2k – 23) + (k2<\/sup> + 24k + 10)
\nGiven remainder is x + a
\non comparing the coefficients of x and constant terms on both sides
\n2k – 23 = 1 ……. (1)
\n2k = 1 + 23 = 24
\n\u21d2 k = \\(\\frac{24}{2}\\) = 12
\nk2<\/sup> + 24k + 10 = a …….. (2)
\nSubstitute ‘k’ value in equation (2)
\n(12)2<\/sup> + 24(12) + 10 = a
\n144 + 288 + 10 = a
\n\u21d2 a = 442
\n\u2234 Required k = 12 and ‘a’ = 442<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions Chapter 3 Polynomials Optional Exercise Textbook Questions and Answers. AP State Syllabus SSC 10th Class Maths Solutions 3rd Lesson Polynomials Optional Exercise 10th Class Maths 3rd Lesson Polynomials Optional Exercise Textbook Questions and Answers Question 1. Verify that the numbers given along-side the cubic polynomials …<\/p>\n","protected":false},"author":5,"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:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/5357"}],"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\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=5357"}],"version-history":[{"count":0,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/5357\/revisions"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=5357"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=5357"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=5357"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}