{"id":1319,"date":"2020-09-15T09:42:30","date_gmt":"2020-09-15T04:12:30","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=1319"},"modified":"2021-07-02T15:01:51","modified_gmt":"2021-07-02T09:31:51","slug":"kseeb-sslc-class-10-maths-solutions-chapter-9-ex-9-3","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-sslc-class-10-maths-solutions-chapter-9-ex-9-3\/","title":{"rendered":"KSEEB SSLC Class 10 Maths Solutions Chapter 9 Polynomials Ex 9.3"},"content":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 9 Polynomials Ex 9.3<\/strong> are part of KSEEB SSLC Class 10 Maths Solutions<\/a>. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 9 Polynomials Exercise 9.3.<\/p>\n

Karnataka SSLC Class 10 Maths Solutions Chapter 9 Polynomials Exercise 9.3<\/h2>\n

Question 1.
\nDivide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the : following.
\n(i) p(x) = x3<\/sup> – 3x2<\/sup> + 5x – 3; g(x) = x2<\/sup> – 2
\n(ii) p(x) = x4<\/sup> – 3x2<\/sup> + 4x + 5; g(x) = x2<\/sup> + 1 – x
\n(iii) p(x) = x4<\/sup> – 5x + 6; g(x) = 2 – x2<\/sup>.
\nSolution:
\n(i) p(x) = x3<\/sup> – 3x2<\/sup> + 5x – 3
\ng(x) =x2<\/sup> – 2
\nq(x) = ?
\nr(x) = ?
\n\"KSEEB
\n\u2234 Quotient, q(x) = (x – 3)
\nRemainder, r(x) = (7x – 9)<\/p>\n

(ii) p(x) = x4<\/sup> – 3x2<\/sup> + 4x + 5
\ng(x) = x2<\/sup> + 1 – x
\nq(x) = ?
\nr(x) = ?
\n\"KSEEB
\n\u2234 Quotient, q(x) = x2<\/sup> + x – 3
\nRemainder, r(x) = 8<\/p>\n

(iii) p(x) = x4<\/sup> – 5x + 6
\ng(x) = 2 – x2<\/sup>
\nq(x) = ?
\nr(x) = ?
\n\"KSEEB
\n\u2234 Quotient, q(x) = -x2<\/sup> – 2
\nRemainder, r(x) = -5x + 10<\/p>\n

Question 2.
\nCheck whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial :
\n(i) t2<\/sup> – 3; 2t4<\/sup> + 3t3<\/sup> – 2t2<\/sup> – 9t – 12
\n(ii) x2<\/sup> + 3x + 1; 3x4<\/sup> + 5x3<\/sup> – 7x2<\/sup> + 2x + 2
\n(iii) x3<\/sup> – 3x + 1; x5<\/sup> – 4x3<\/sup> + x2<\/sup> + 3x + 1
\nSolution:
\n(i) t2<\/sup> – 3; 2t4<\/sup> + 3t3<\/sup> – 2t2<\/sup> – 9t – 12
\n\"KSEEB
\n\u2234 t2<\/sup> – 3 is the factor of 2t4<\/sup> + 3t3<\/sup> – 2t2<\/sup> – 9t – 12 because remainder is zeroes<\/p>\n

(ii) x2<\/sup> + 3x + 1; 3x4<\/sup> + 5x3<\/sup> – 7x2<\/sup> + 2x + 2
\n\"KSEEB
\nHere remainder is zero
\n\u2234 x2<\/sup> + 3x + 1 is the factor of 3x4<\/sup> + 5x3<\/sup> – 7x2<\/sup> + 2x + 2<\/p>\n

(iii) x3<\/sup> – 3x + 1; x5<\/sup> – 4x3<\/sup> + x2<\/sup> + 3x + 1
\n\"KSEEB
\nHere remainder r(x) \u2260 0 \u2235 2 \u2260 0
\n\u2234 x3<\/sup> – 3x + 1 is not the factor of x5<\/sup> – 4x3<\/sup> + x2<\/sup> + 3x + 1<\/p>\n

Question 3.
\nObtain all other zeroes of 3x4<\/sup> + 6x3<\/sup> – 2x2<\/sup> – 10x – 5, if two of its zeroes are \\(\\sqrt{\\frac{5}{3}}\\) and \\(-\\sqrt{\\frac{5}{3}}\\)
\nSolution:
\nRoots of expressions are \\(\\sqrt{\\frac{5}{3}}\\) and \\(-\\sqrt{\\frac{5}{3}}\\)
\n\"KSEEB
\n\u2234 In q(x) = x2<\/sup> + 2x + 1, there are two more roots.
\nx2<\/sup> + 2x + 1 = (x + 1)2<\/sup>
\nother roots are: x = -1, -1<\/p>\n

Question 4.
\nOn dividing x3<\/sup> – 3x2<\/sup> + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and -2x + 4, respectively. Find g(x).
\nSolution:
\nx3<\/sup> – 3x2<\/sup> + x + 2 = g(x) \u00d7 (x – 2) + (-2x + 4)
\ng(x) \u00d7 (x – 2) = x3<\/sup> – 3x2<\/sup> + x + 2 – (-2x + 4)
\n= x3<\/sup> – 3x2<\/sup> + x + 2
\n\"KSEEB<\/p>\n

Question 5.
\nGive examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
\n(i) deg p(x) = deg q(x)
\n(ii) deg q(x) = deg r(x)
\n(iii) deg r(x) = 0
\nSolution:
\ni) P(x) = 2x2<\/sup> – 2x + 14, g(x) = 2,
\nq(x) = x2<\/sup> – x + 7 and r(x) = 0
\nVerification
\nP(x) = g(x) q(x) + r(x)
\n2x2<\/sup> – 2x + 14 = 2(x2<\/sup> – x + 7) + 0
\n= 2x2<\/sup> – 2x + 14
\n\u2234 LHS = RHS<\/p>\n

ii) P(x) = x3<\/sup> + x2<\/sup> + x + 1, g(x) = x2<\/sup> – 1, q(x) = x + 1 and r(x) = 2x + 2
\nVerification
\nP(x) = q(x) g(x) + r(x)
\nx3<\/sup> + x2<\/sup> + x + 1 =(x+ 1) (x2<\/sup> – 1) + 2x + 2
\n= x3<\/sup> – x + x2<\/sup> – 1 + 2x + 2
\n= x3<\/sup> + x2<\/sup> + x + 1
\nLHS = RHS<\/p>\n

iii) P(x) = x3<\/sup> + 2x2<\/sup> – x + 2, g(x) = x2<\/sup> – 1, q(x) = x + 2 and r(x) = 4
\np(x) = g(x) q(x) + r(x)
\nx3<\/sup> + 2x2<\/sup> – x + 2 = (x2<\/sup> – 1) (x + 2) + 4
\n= x3<\/sup> + 2x2<\/sup> – x – 2 + 4
\n= x3<\/sup> + 2x2<\/sup> – x + 2
\nLHS = RHS<\/p>\n

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 9 Polynomials Ex 9.3 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 9 Polynomials Exercise 9.3, drop a comment below and we will get back to you at the earliest.<\/p>\n","protected":false},"excerpt":{"rendered":"

KSEEB SSLC Class 10 Maths Solutions Chapter 9 Polynomials Ex 9.3 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 9 Polynomials Exercise 9.3. Karnataka SSLC Class 10 Maths Solutions Chapter 9 Polynomials Exercise 9.3 Question 1. Divide the polynomial p(x) by the polynomial …<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/1319"}],"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\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=1319"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/1319\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=1319"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=1319"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=1319"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}