{"id":3984,"date":"2021-02-25T15:42:00","date_gmt":"2021-02-25T10:12:00","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=3984"},"modified":"2021-09-15T11:40:40","modified_gmt":"2021-09-15T06:10:40","slug":"ap-board-9th-class-maths-solutions-chapter-2-ex-2-2","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-board-9th-class-maths-solutions-chapter-2-ex-2-2\/","title":{"rendered":"AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.2"},"content":{"rendered":"

AP State Syllabus\u00a0AP Board 9th Class Maths Solutions<\/a> Chapter 2 Polynomials and Factorisation Ex 2.2 Textbook Questions and Answers.<\/p>\n

AP State Syllabus 9th Class Maths Solutions 2nd Lesson Polynomials and Factorisation Exercise 2.2<\/h2>\n

Question 1.
\nFind the value of the polynomial 4x2<\/sup> – 5x + 3, when
\n(i) x = 0
\nSolution:
\nThe value at x = 0 is
\n4(0)2<\/sup> – 5(0) + 3
\n= 3<\/p>\n

\"AP<\/p>\n

(ii) x = – 1
\nSolution:
\nThe value at x = – 1 is
\n4 (- 1)2<\/sup> – 5 (- 1) + 3
\n= 4 + 5 + 3
\n= 12<\/p>\n

iii) x = 2
\nSolution:
\nThe value at x = 2 is
\n4(2)2<\/sup> – 5(2) + 3
\n= 16 -10 + 3
\n= 9<\/p>\n

iv) x = \\(\\frac{1}{2}\\)
\nSolution:
\n\"AP<\/p>\n

Question 2.
\nFind p(0), p(1) and p(2) for each of the following polynomials.
\ni) p(x) = x2<\/sup> – x + 1
\nSolution:
\np(0) = 02<\/sup> – 0 + 1 = 1
\np(1) = 12<\/sup> – 1 + 1 = 1
\np(2) = 22<\/sup> – 2 + 1 = 3<\/p>\n

ii) P(y) = 2 + y + 2y2<\/sup> – y3<\/sup>
\nSolution:
\np(0) = 2 + 0 + 2(0)2<\/sup> – 03<\/sup> = 2
\np(1) = 2+ 1 + 2(1)2<\/sup> – 13<\/sup> = 4
\np(2) = 2 + 2 + 2(2)2<\/sup> -23<\/sup> = 4 + 8- 8 = 4<\/p>\n

\"AP<\/p>\n

iii) P(z) = z3<\/sup>
\nSolution:
\np(0) = 03<\/sup> = 0
\np(1) = 13<\/sup> = 1
\np(2) = 23<\/sup> = 8<\/p>\n

iv) p(t) = (t – 1)(t + 1) = t2<\/sup> – 1
\nSolution:
\np(0) = (0 – 1) (0 + 1) = – 1
\np(1) = t2<\/sup> – 1 = 12<\/sup> – 1 = 0
\np(2) = 22<\/sup> – 1 = 4 – 1 = 3<\/p>\n

v) p(x) = x2<\/sup> – 3x + 2
\nSolution:
\np(0) = 02<\/sup> – 3(0) + 2 = 2
\np(1) = 12<\/sup> – 3(1) + 2 = 1 – 3 + 2 = 0
\np(2) = 22<\/sup> – 3(2) + 2 = 4- 6 + 2 = 0<\/p>\n

Question 3.
\nVerify whether the values of x given in each case are the zeroes of the polynomial or not ?
\ni) p(x) = 2x + 1; x = \\(\\frac{-1}{2}\\)
\nSolution:
\nThe value of p(x) at x = \\(\\frac{-1}{2}\\) is
\n\\(\\mathrm{p}\\left(\\frac{-1}{2}\\right)=2\\left(\\frac{-1}{2}\\right)+1\\)
\n= -1 + 1 = 0
\n\u2234 x = \\(\\frac{-1}{2}\\) is a zero of p(x).<\/p>\n

\"AP<\/p>\n

(ii) p(x) = 5x – \u03c0 ; x = \\(\\frac{-3}{2}\\)
\nSolution:
\nThe value of p(x) at x = \\(\\frac{-3}{2}\\) is
\n\\(\\mathrm{p}\\left(\\frac{-3}{2}\\right)=5\\left(\\frac{-3}{2}\\right)-\\pi=\\frac{-15}{2}-\\pi \\neq 0\\)
\n\u2234 x = \\(\\frac{-3}{2}\\) is not a zero of p(x).<\/p>\n

iii) p(x) = x2<\/sup> – 1; x = \u00b11
\nSolution:
\nThe value of p(x) at x = 1 and – 1 is
\np(1) = 12<\/sup> – 1 = 0
\np(-1) = (-1)2<\/sup> -1 = 0
\n\u2234 x = \u00b11 is a zero of p(x).<\/p>\n

iv) p(x) = (x – 1) (x + 2); x = – 1, – 2
\nSolution:
\nThe value of p(x) at x = – 1 is
\np(-1) = (-1 – 1) (-1 + 2)
\n=-2 x 1 =-2 \u2260 0
\nHence x = – 1 is not a zero of p(x).
\nAnd the value of p(x) at x = – 2 is
\np (- 2) = (- 2 – 1) (- 2 + 2) = – 3 x 0 = 0
\nHence, x = – 2 is a zero of p(x).<\/p>\n

v) p(y) = y2<\/sup>; y = o
\nSolution:
\nThe value of p(y) at y = 0 is p(0) = 02<\/sup> = 0
\nHence y = 0 is a zero of p(y).<\/p>\n

\"AP<\/p>\n

vi) p(x) = ax + b ; x = \\(\\frac{-\\mathbf{b}}{\\mathbf{a}}\\)
\nSolution:
\nThe value of p(x) at x = \\(\\frac{-\\mathbf{b}}{\\mathbf{a}}\\) is
\n\\(\\mathrm{p}\\left(\\frac{-\\mathrm{b}}{\\mathrm{a}}\\right)=\\mathrm{a}\\left(\\frac{-\\mathrm{b}}{\\mathrm{a}}\\right)+\\mathrm{b}\\)
\n= -b + b = 0
\n\u2234 x = \\(\\frac{-\\mathbf{b}}{\\mathbf{a}}\\) is a zero of p(x).<\/p>\n

vii) f(x) = 3x2<\/sup> – 1; x = \\(\\frac{-1}{\\sqrt{3}}, \\frac{2}{\\sqrt{3}}\\)
\nSolution:
\n\"AP<\/p>\n

\"AP<\/p>\n

viii) f(x) = 2x – 1; x = \\(\\frac{1}{2} ;-\\frac{1}{2}\\)
\nSolution:
\n\"AP<\/p>\n

Question 4.
\nFind the zero of the polynomial in each of the following cases.
\ni) f(x) = x + 2
\nSolution:
\nx + 2 = 0
\nx = – 2<\/p>\n

ii) f(x) = x – 2
\nSolution:
\nx – 2 = 0
\nx = 2<\/p>\n

iii) f(x) = 2x + 3
\nSolution:
\n2x + 3 = 0
\n2x = – 3
\nx = \\(\\frac{-3}{2}\\)<\/p>\n

\"AP<\/p>\n

iv) f(x) = 2x – 3
\nSolution:
\n2x – 3 = 0
\n2x = 3
\nx = \\(\\frac{3}{2}\\)<\/p>\n

v) f(x) = x2<\/sup>
\nSolution:
\nx2<\/sup> = 0
\nx = 0<\/p>\n

vi) f(x) = px, p \u2260 0
\nSolutin:
\npx = 0
\nx = 0<\/p>\n

vii) f(x) = px + q; p \u2260 0; p, q are real numbers.
\nSolution:
\npx + q = 0
\npx = -q
\nx = \\(\\frac{-\\mathrm{q}}{\\mathrm{p}}\\)<\/p>\n

\"AP<\/p>\n

Question 5.
\nIf 2 is a zero of the polynomial p(x) = 2x2<\/sup> – 3x + 7a, find the value of
\na.
\nSolution:
\nGiven that 2 is a zero of p(x) = 2x2<\/sup> – 3x + 7a
\n(i.e.) p(2) = 0
\n\u21d2 2(2)2<\/sup> – 3(2) + 7a = 0
\n\u21d2 8 – 6 + 7a = 0
\n\u21d2 2 + 7a = 0
\n\u21d2 7a = – 2
\n\u21d2 a = \\(\\frac{-2}{7}\\)<\/p>\n

\"AP<\/p>\n

Question 6.
\nIf 0 and 1 are the zeroes of the polynomial f(x) = 2x3<\/sup> – 3x2<\/sup> + ax + b, find the values of a and b.
\nSolution:
\nGiven that f(0) = 0; f(1) = 0 and
\nf(x) = 2x3<\/sup> – 3x2<\/sup> + ax + b
\n\u2234 f(0) = 2(0)3<\/sup> – 3(0)2<\/sup> + a(0) + b
\n\u21d2 0 = b
\nAlso f(1) = 0
\n\u21d2 2(1)3<\/sup> – 3(1)2<\/sup> + a(1) + 0 = 0
\n\u21d2 2 – 3 + a = 0 .
\n\u21d2 a = 1
\nHence a = 1; b = 0<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Syllabus\u00a0AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.2 Textbook Questions and Answers. AP State Syllabus 9th Class Maths Solutions 2nd Lesson Polynomials and Factorisation Exercise 2.2 Question 1. Find the value of the polynomial 4×2 – 5x + 3, when (i) x = 0 Solution: The value at …<\/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\/3984"}],"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=3984"}],"version-history":[{"count":0,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/3984\/revisions"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=3984"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=3984"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=3984"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}