{"id":4847,"date":"2021-03-02T14:48:47","date_gmt":"2021-03-02T09:18:47","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=4847"},"modified":"2021-09-15T11:40:35","modified_gmt":"2021-09-15T06:10:35","slug":"ap-ssc-10th-class-maths-solutions-chapter-3-ex-3-2","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-ssc-10th-class-maths-solutions-chapter-3-ex-3-2\/","title":{"rendered":"AP SSC 10th Class Maths Solutions Chapter 3 Polynomials Ex 3.2"},"content":{"rendered":"

AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions<\/a> Chapter 3 Polynomials Ex 3.2 Textbook Questions and Answers.<\/p>\n

AP State Syllabus SSC 10th Class Maths Solutions 3rd Lesson Polynomials Exercise 3.2<\/h2>\n

10th Class Maths 3rd Lesson Polynomials Ex 3.2 Textbook Questions and Answers<\/h3>\n

Question 1.
\nThe graphs of y = p(x) are given in the figure below, for some polynomials p(x). In each case, find the number of zeroes of p(x).
\n\"AP
\nAnswer:
\ni) There are no zeroes as the graph does not intersect the X – axis.
\nii) The number of zeroes is one as the graph intersects the X – axis at one point only.
\niii) The number of zeroes is three as the graph intersects the X – axis at three points.
\niv) The number of zeroes is two as the graph intersects the X – axis at two points.
\nv) The number of zeroes is four as the graph intersects the X – axis at four points.
\nvi) The number of zeroes is three as the graph intersects the X – axis at three points.<\/p>\n

\"AP<\/p>\n

Question 2.
\nFind the zeroes of the given polynomials,
\n(i) p(x) = 3x
\n(ii) p(x) = x2<\/sup> + 5x + 6
\n(iii) p(x) = (x + 2) (x + 3)
\n(iv) p(x) = x4<\/sup> – 16
\nAnswer:
\ni) Given p(x) = 3x
\nLet p(x) = 0
\nSo, 3x = 0
\nx = \\(\\frac{0}{3}\\) = 0,
\nZeroes of p(x) = 3x is zero.
\n\u2234 No. of zeroes is one.<\/p>\n

ii) Given p(x) = x2<\/sup> + 5x + 6 is a quadratic polynomial.
\nIt has atmost two zeroes.
\nTo find zeroes, let p(x) = 0
\n\u21d2 x2<\/sup> + 5x + 6 = 0
\n\u21d2 x2<\/sup> + 3x + 2x + 6 = 0
\n\u21d2 x(x + 3) + 2 (x + 3) = 0
\n\u21d2 (x + 3) (x + 2) = 0
\n\u21d2 x + 3 = 0 or x + 2 = 0
\n\u21d2 x = -3 or x = -2
\nTherefore the zeroes of the polynomial are -3 and -2.<\/p>\n

iii) Given p(x) = (x + 2) (x + 3)
\nIt is a quadratic polynomial.
\nIt has atmost two zeroes.
\nLet p(x) = 0
\n\u21d2 (x + 2) (x + 3) = 0
\n\u21d2 (x + 2) = 0 or (x + 3) = 0
\n\u21d2 x = -2 or x = -3
\nTherefore the zeroes of the polynomial are -2 and – 3.<\/p>\n

iv) Given p(x) = x4<\/sup> – 16 is a biquadratic polynomial. It has atmost two zeroes.
\nLet p(x) = 0
\n\u21d2 x4<\/sup> – 16 = 0
\n\u21d2 (x2<\/sup>)2<\/sup> – 42<\/sup> = 0
\n\u21d2 (x2<\/sup> – 4) (x2<\/sup> + 4) = 0
\n\u21d2 (x + 2) (x – 2) (x2<\/sup> + 4) = 0
\n\u21d2 (x + 2) = 0 or (x – 2) = 0 or (x2<\/sup> + 4) = 0
\n\u21d2 x = -2 (or) x = 2 (or) x2<\/sup> = -4
\nTherefore the zeroes of the polynomial are 2, – 2, we do not consider \u221a-4 since it is not real.<\/p>\n

\"AP<\/p>\n

Question 3.
\nDraw the graphs of the given polynomial and find the zeroes. Justify the answers,
\ni) p(x) = x2<\/sup> – x – 12
\nii) p(x) = x2<\/sup> – 6x + 9
\niii) p(x) = x2<\/sup> – 4x + 5
\niv) p(x) = x2<\/sup> + 3x – 4
\nv) p(x) = x2<\/sup> – 1
\nAnswer:
\ni) Given polynomial p(x) = x2<\/sup> – x – 12.
\nList of values of p(x):
\n\"AP
\nNow, let\u2019s locate the points listed above on a graph paper and draw the graph.
\n\"AP
\nResult: We observe that the graph cuts the X – axis at (-3, 0) and (4, 0).
\nSo, the zeroes of the polynomial are -3 and 4.
\nJustification:
\nGiven p(x) = x2<\/sup> – x – 12 = 0
\n\u21d2 x2<\/sup> – 4x + 3x – 12 = 0
\n\u21d2 x(x – 4) + 3(x – 4) = 0
\n\u21d2 (x – 4) (x + 3) = 0
\n\u21d2 x – 4 = 0 and x + 3 = 0
\nx = 4 and x = – 3<\/p>\n

ii) Given polynomial p(x) = x2<\/sup> – 6x + 9
\nList of values of p(x):
\n\"AP
\nNow, let\u2019s locate the points listed above on a graph paper and draw the graph.
\n\"AP
\nResult: We observe that the graph cuts the X – axis at (3, 0).
\nSo, the zeroes of the given polynomial are same i.e., 3.
\nJustification:
\nGiven p(x) = x2<\/sup> – 6x + 9
\n\u21d2 x2<\/sup> – 3x – 3x + 9 = 0
\n\u21d2 x(x – 3) – 3(x – 3) = 0
\n\u21d2 (x – 3) (x – 3) = 0
\n\u21d2 x – 3 = 0 and x – 3 = 0
\nx = 3 and x = 3<\/p>\n

\"AP<\/p>\n

iii) Given polynomial p(x) = x2<\/sup> – 4x + 5
\nList of values of p(x):
\n\"AP
\nNow, let\u2019s locate the points listed above on a graph paper and draw the graph.
\n\"AP
\nResult: We observe that the graph does not cut the X – axis at any point.
\nSo, the quadratic polynomial p(x) has no zeroes.
\nJustification: For the given p(x) = x2<\/sup> – 4x + 5 not possible to split in factors.<\/p>\n

iv) Given polynomial p(x) = x2<\/sup> + 3x – 4.
\nList of values of p(x):
\n\"AP
\nNow, let\u2019s locate the points listed above on a graph paper and draw the graph.
\n\"AP
\nResult: We observe that the graph cuts the X – axis at (-4, 0) and (1, 0).
\nSo, the zeroes of the polynomial are -4 and 1.
\nJustification:
\nGiven p(x) = x2<\/sup> + 3x – 4 = 0
\n\u21d2 x2<\/sup> + 4x – x – 4 = 0
\n\u21d2 x(x + 4)- 1(x + 4) = 0
\n\u21d2 (x + 4) (x – 1) = 0
\n\u21d2 x + 4 = 0 and x – 1 = 0
\nx = – 4 and x = 1<\/p>\n

\"AP<\/p>\n

v) Given polynomial p(x) = x2<\/sup> – 1
\nList of values of p(x):
\n\"AP
\nNow, let\u2019s locate the points listed above on a graph paper and draw the graph.
\n\"AP
\nResult: We observe that the graph cuts the X – axis at (-1, 0) and (1,0).
\nSo, the zeroes of the polynomial are – 1 and 1.
\nJustification:
\nGiven p(x) = x2<\/sup> – 1 = 0
\n\u21d2 p(x) = (x + 1) (x – 1) = 0 [\u2235 a2<\/sup> – b2<\/sup> = (a + b) (a – b)]
\n\u21d2 x + 1 = 0 and x – 1 = 0
\nx = -1 and x = 1<\/p>\n

Question 4.
\nWhy are \\(\\frac{1}{4}\\) and -1 zeroes of the polynomial p(x) = 4x2<\/sup> + 3x – 1 ?
\nAnswer:
\nGiven polynomial p(x) = 4x2<\/sup> + 3x – 1
\nGiven zeroes are \\(\\frac{1}{4}\\) and -1
\nLet x = \\(\\frac{1}{4}\\)
\n\"AP
\nLet x = -1
\n\u21d2 p(-1) = 4(-1)2<\/sup> + 3(-1)-1 = 4 – 3 – 1 = 4 – 4 = 0
\n\u2234 P(\\(\\frac{1}{4}\\)) = 0 and p(-1) = 0
\nSo these values are zeroes of the polynomial p(x).<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions Chapter 3 Polynomials Ex 3.2 Textbook Questions and Answers. AP State Syllabus SSC 10th Class Maths Solutions 3rd Lesson Polynomials Exercise 3.2 10th Class Maths 3rd Lesson Polynomials Ex 3.2 Textbook Questions and Answers Question 1. The graphs of y = p(x) are given in …<\/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\/4847"}],"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=4847"}],"version-history":[{"count":0,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/4847\/revisions"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=4847"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=4847"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=4847"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}