{"id":620,"date":"2020-09-07T10:54:05","date_gmt":"2020-09-07T05:24:05","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=620"},"modified":"2021-07-02T15:03:45","modified_gmt":"2021-07-02T09:33:45","slug":"kseeb-solutions-class-9-maths-chapter-4-ex-4-3","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-9-maths-chapter-4-ex-4-3\/","title":{"rendered":"KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.3"},"content":{"rendered":"

KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.3<\/strong> are part of KSEEB Solutions for Class 9 Maths<\/a>. Here we have given Karnataka Board Class 9 Maths Chapter 4 Polynomials Exercise 4.3.<\/p>\n

Karnataka Board Class 9 Maths Chapter 4 Polynomials Ex 4.3<\/h2>\n

Question 1.
\nFind the remainder when x3<\/sup> + 3x2<\/sup> + 3x + 1 is divided by
\ni) x + 1
\nii) \\(x-\\frac{1}{2}\\)
\niii) x
\niv) x + \u03c0
\nv) 5 + 2x
\nAnswer:
\ni) p(x) = x3<\/sup> + 3x2<\/sup> + 3x + 1 g(x) = x – 1
\nLet x – 1 = 0, then
\nx = 1.
\nAs per Remainder theorem, r(x) = p(x) = p(a)
\np(x) = x3<\/sup> + 3x2<\/sup> + 3x + 1
\np(1) = (1)3<\/sup> + 3(1)2<\/sup> + 3(1) + 1
\n= 1 + 3(1) + 3(1) + 1
\n= 1 + 3 + 3 + 1
\nP(1) = 8
\n\u2234 r(x) = p(x) = 8
\n\u2234 Remainder is 8.<\/p>\n

ii) p(x) = x3<\/sup> + 3x2<\/sup> + 3x + 1 g(x) = \\(x-\\frac{1}{2}\\)
\nIf \\(x-\\frac{1}{2}=0\\) then \\(x=\\frac{1}{2}\\)
\np(x) = x3<\/sup> + 3x2<\/sup> + 3x + 1
\n\"KSEEB
\n\u2234 r(x) = p(x) = p(a) = \\(\\frac{15}{8}\\)
\n\u2234 Remainder is \\(\\frac{15}{8}\\)<\/p>\n

iii) p(x) = x3<\/sup> + 3x2<\/sup> + 3x + 1
\ng(x) = x
\nIf x = 0, then
\np(x) = x3<\/sup> + 3x2<\/sup> + 3x + 1
\np(0) = (0)3<\/sup> + 3(0)2<\/sup> + 3(0) + 1
\n= 0 + 3(0) + 3(0) + 1
\n= 0 + 0 + 0 + 1
\np(0) = 0
\nRemainder r(x) = 1.<\/p>\n

iv) p(x) = x3<\/sup> + 3x2<\/sup> + 3x + 1
\ng(x) = x + \u03c0
\nIf x + \u03c0 = 0, then x = -\u03c0
\np(x) = x3<\/sup> + 3x2<\/sup> + 3x + 1
\np(-\u03c0) = (-\u03c0)3<\/sup> + 3(-\u03c0)2<\/sup> + 3(-\u03c0) + 1
\np(-\u03c0t) = -\u03c023<\/sup> – 3\u03c02<\/sup> – 3\u03c0 + 1
\nr(x) = -\u03c03<\/sup> – 3v2<\/sup> – 3\u03c0+1<\/p>\n

v) p(x) = x3<\/sup> + 3x2<\/sup> + 3x + 1
\ng(x) = 5 + 2x
\nIf 5 + 2x = 0, then 2x = -5
\n\\(x=-\\frac{5}{2}\\)
\np(x) = x3<\/sup> + 3x2<\/sup> + 3x + 1
\n\"KSEEB<\/p>\n

Question 2.
\nFind the remainder when x3<\/sup> – ax2<\/sup> + 6x – a is divided by x – a.
\nAnswer:
\np(x) = x3<\/sup> – ax2<\/sup> + 6x – a
\nIf g(x) = x – a, then r(x) = ?
\nLet x – a = 0, then x = a
\np(x) = x3<\/sup> – ax2<\/sup> + 6x – a
\n\u2234 p(a) = (a)3<\/sup> – a(a)2<\/sup> + 6(a) – a
\n= a3<\/sup> – a3<\/sup> + 6a – a
\n\u2234 p(a) = 5a
\n\u2234 r(x) = p(a) = 5a.<\/p>\n

Question 3.
\nCheck whether 7 + 3x is a factor of 3x3<\/sup> + 7x.
\nAnswer:
\np(x) = 3x3<\/sup> + 7x
\nLet g(x) = 7 + 3x = 0. then
\nIf 7 + 3x = 0, then 3x = -7
\n\\(x=-\\frac{7}{3}\\)
\nIf p(x) is divided by p(z), remainder r(x) – 0, then g(x) is a factor.
\np(x) = 3x3<\/sup> + 7x
\n\"KSEEB
\nHere, \\(r(x)=-\\frac{590}{9}\\). This is not equal to Zero.
\nHence 7 + 3x is not a factor of p(x).<\/p>\n

We hope the KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.3 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 4 Polynomials Exercise 4.3, drop a comment below and we will get back to you at the earliest.<\/p>\n","protected":false},"excerpt":{"rendered":"

KSEEB Solutions for Class 9 Maths Chapter 4 Polynomials Ex 4.3 are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 4 Polynomials Exercise 4.3. Karnataka Board Class 9 Maths Chapter 4 Polynomials Ex 4.3 Question 1. Find the remainder when x3 + 3×2 + 3x …<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/620"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=620"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/620\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=620"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=620"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=620"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}