{"id":71962,"date":"2021-02-04T15:14:35","date_gmt":"2021-02-04T09:44:35","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=71962"},"modified":"2021-02-04T15:14:40","modified_gmt":"2021-02-04T09:44:40","slug":"rd-sharma-chapter-6-class-9-maths-exercise-6-5-solutions","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-6-class-9-maths-exercise-6-5-solutions\/","title":{"rendered":"RD Sharma Chapter 6 Class 9 Maths Exercise 6.5 Solutions"},"content":{"rendered":"\n\n\n<p>In the RD Sharma Chapter 6 Class 9 Maths Exercise 6.5 Solutions, we are providing the detail about the Factorization of Polynomials with Integral Coefficient using the Factor Theorem. We have given the details based on the CBSE Book of Class 9 and RD Sharma. The Integral Coefficient is the coefficient in an algebraic identity that is an integer. Here learners will get to know the complete details like- Terms, Explanation, Examples, Benefits, frequently asked questions related to the Factorization of Polynomials with Integral Coefficient using the Factor Theorem.<\/p>\n<p>In the attached PDF, there are various types of questions that help students score well in the exam. The solution to each problem is provided with tricks, tips, and shortcuts. Learners can also practice for the exam with these questions by downloading them and practice offline anytime based on their schedule.<\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-9-maths-chapter-6-factorization-of-polynomials\/\" target=\"_blank\" rel=\"noopener noreferrer\"><strong>Click Here<\/strong><\/a> to learn about Class 9 Chapter 6- Factorization of Polynomial<\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"ez-toc-toggle-icon-1\"><label for=\"item-69eba4b9378c1\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg style=\"fill: #000000;color:#000000\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #000000;color:#000000\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/label><input  type=\"checkbox\" id=\"item-69eba4b9378c1\"><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-visibility-hide-by-default' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-6-class-9-maths-exercise-6-5-solutions\/#download-rd-sharma-chapter-6-class-9-maths-exercise-65-solutions-pdf\" title=\"Download RD Sharma Chapter 6 Class 9 Maths Exercise 6.5 Solutions PDF\">Download RD Sharma Chapter 6 Class 9 Maths Exercise 6.5 Solutions PDF<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-6-class-9-maths-exercise-6-5-solutions\/#important-definitions-rd-sharma-chapter-6-class-9-maths-exercise-65-solutions\" title=\"Important Definitions RD Sharma Chapter 6 Class 9 Maths Exercise 6.5 Solutions\">Important Definitions RD Sharma Chapter 6 Class 9 Maths Exercise 6.5 Solutions<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-6-class-9-maths-exercise-6-5-solutions\/#examples-of-the-factorization-of-polynomials-with-integral-coefficient-using-factor-theorem\" title=\"Examples of the Factorization of Polynomials with Integral Coefficient using Factor Theorem\">Examples of the Factorization of Polynomials with Integral Coefficient using Factor Theorem<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-6-class-9-maths-exercise-6-5-solutions\/#frequently-asked-questions-faqs-of-rd-sharma-chapter-6-class-9-maths-exercise-65-solutions\" title=\"Frequently Asked Questions (FAQs) of RD Sharma Chapter 6 Class 9 Maths Exercise 6.5 Solutions\">Frequently Asked Questions (FAQs) of RD Sharma Chapter 6 Class 9 Maths Exercise 6.5 Solutions<\/a><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-chapter-6-class-9-maths-exercise-65-solutions-pdf\"><\/span>Download RD Sharma Chapter 6 Class 9 Maths Exercise 6.5 Solutions PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 500px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br \/><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/Solutions-for-Class-9-Maths-Chapter-6-Factorization-of-Polynomials-Exercise-6.5.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/01\/Solutions-for-Class-9-Maths-Chapter-6-Factorization-of-Polynomials-Exercise-6.5.pdf\">Solutions for Class 9 Maths Chapter 6 Factorization of Polynomials Exercise 6.5<\/a><\/p>\n<h2><span class=\"ez-toc-section\" id=\"important-definitions-rd-sharma-chapter-6-class-9-maths-exercise-65-solutions\"><\/span>Important Definitions RD Sharma Chapter 6 Class 9 Maths Exercise 6.5 Solutions<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>Suppose \u2018P\u2019 is a polynomial with integer coefficients, where P(a) \u2212 P(b) is divisible by a\u2212b for any separate integers a and b. In particular, all integer roots of P divide P(0).<\/p>\n<h3><span class=\"ez-toc-section\" id=\"examples-of-the-factorization-of-polynomials-with-integral-coefficient-using-factor-theorem\"><\/span>Examples of the Factorization of Polynomials with Integral Coefficient using Factor Theorem<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p>Here are some solved examples of the Integral Coefficient using Factor Theorem based on the RD Sharma Chapter 6 Class 9 Maths Exercise 6.5 Solutions-<\/p>\n<p><strong>Ques- Use the factor theorem to factorize the following polynomials completely.<\/strong><\/p>\n<p style=\"text-align: center;\">x<sup>3<\/sup> \u2013 13x \u2013 12<\/p>\n<p>Given,<\/p>\n<p>f(x)= x<sup>3<\/sup> \u2013 13x \u2013 12<\/p>\n<p>We need to find out at least one root by hit and trial applying the factor theorem to solve this problem.<\/p>\n<p><strong>Here is a trick:<\/strong><\/p>\n<p>Suppose a polynomial function has integer coefficients. Every rational zero will have the form qp, where &#8216;p&#8217; is a factor of the constant and &#8216;q&#8217; is a factor of the leading coefficient.<\/p>\n<p><strong>Here<\/strong><\/p>\n<p>p= \u00b11, \u00b12, \u00b13, \u00b14, \u00b16, \u00b112 and<\/p>\n<p>q= \u00b11<\/p>\n<p>Find all combinations of \u00b1qp.<\/p>\n<p>These are the probable roots of the polynomial function.<\/p>\n<p>\u00b11, \u00b12, \u00b13, \u00b14, \u00b16, \u00b112<\/p>\n<p>Substituting x=1 in f(x), we get<\/p>\n<p>f(\u22121)= (\u20131)<sup>3<\/sup> \u2013 13(\u20131) \u2013 12<\/p>\n<p>= \u22121+13\u221212<\/p>\n<p>=\u00a0 13+13<\/p>\n<p>=0<\/p>\n<p>Therefore, (x+1) is the factor of f(x)\u00a0<\/p>\n<p>Now, dividing f(x) by (x+1), we get<\/p>\n<p>x<sup>3<\/sup> \u2013 13x \u2013 12 = (x+1) (x<sup>2<\/sup> \u2013 x \u2013 12)<\/p>\n<p>=(x+1) (x<sup>2<\/sup> \u2013 4x + 3x \u2013 12)<\/p>\n<p>=(x+1) {x(x\u22124) + 3(x\u22124)}<\/p>\n<p>=(x+1) (x+3) (x\u22124)<\/p>\n<p><strong>Ques- x<sup>3<\/sup> \u2013 6x<sup>2<\/sup> + 3x + 10<\/strong><\/p>\n<p><strong>Solution-<\/strong><\/p>\n<p>\u00a0x<sup>3<\/sup> \u2013 6x<sup>2<\/sup> + 3x + 10<\/p>\n<p>Solution:<\/p>\n<p>Let p(x) = x<sup>3<\/sup> \u2013 6x<sup>2<\/sup> + 3x + 10<\/p>\n<p>Constant term = 10<\/p>\n<p>Factors of 10 are \u00b11, \u00b12, \u00b15, \u00b110<\/p>\n<p>Let x + 1 = 0 or x = -1<\/p>\n<p>p(-1) = (-1)<sup>3<\/sup> \u2013 6(-1)<sup>2<\/sup> + 3(-1) + 10 = 10 \u2013 10 = 0<\/p>\n<p>p(-1) = 0<\/p>\n<p>Let x + 2 = 0 or x = -2<\/p>\n<p>p(-2) = (-2)<sup>3<\/sup> \u2013 6(-2)<sup>2<\/sup> + 3(-2) + 10 = -8 \u2013 24 \u2013 6 + 10 = -28<\/p>\n<p>p(-2) \u2260 0<\/p>\n<p>Let x \u2013 2 = 0 or x = 2<\/p>\n<p>p(2) = (2)<sup>3<\/sup> \u2013 6(2)<sup>2<\/sup> + 3(2) + 10 = 8 \u2013 24 + 6 + 10 = 0<\/p>\n<p>p(2) = 0<\/p>\n<p>Let x \u2013 5 = 0 or x = 5<\/p>\n<p>p(5) = (5)<sup>3<\/sup> \u2013 6(5)<sup>2<\/sup> + 3(5) + 10 = 125 \u2013 150 + 15 + 10 = 0<\/p>\n<p>p(5) = 0<\/p>\n<p>Therefore, (x + 1), (x \u2013 2) and (x-5) are factors of p(x)<\/p>\n<p>Hence p(x) = (x + 1) (x \u2013 2) (x-5)<\/p>\n<p><strong>Ques- x<sup>4<\/sup> \u2013 2x<sup>3<\/sup> \u2013 7x<sup>2<\/sup> + 8x + 12<\/strong><\/p>\n<p><strong>Solution-<\/strong><\/p>\n<p>p(x) = x<sup>4<\/sup> \u2013 2x<sup>3<\/sup> \u2013 7x<sup>2<\/sup> + 8x + 12<\/p>\n<p>Constant term = 12<\/p>\n<p>Factors of 12 are \u00b11, \u00b12, \u00b13, \u00b14, \u00b16, \u00b112<\/p>\n<p>Let x \u2013 1 = 0 or x = 1<\/p>\n<p>p(1) = (1)<sup>4<\/sup> \u2013 2(1)<sup>3<\/sup> \u2013 7(1)<sup>2<\/sup> + 8(1) + 12 = 1 \u2013 2 \u2013 7 + 8 + 12 = 12<\/p>\n<p>p(1) \u2260 0<\/p>\n<p>Let x + 1 = 0 or x = -1<\/p>\n<p>p(-1) = (-1)<sup>4<\/sup> \u2013 2(-1)<sup>3<\/sup> \u2013 7(-1)<sup>2<\/sup> + 8(-1) + 12 = 1 + 2 \u2013 7 \u2013 8 + 12 = 0<\/p>\n<p>p(-1) = 0<\/p>\n<p>Let x +2 = 0 or x = -2<\/p>\n<p>p(-2) = (-2)<sup>4<\/sup> \u2013 2(-2)<sup>3<\/sup> \u2013 7(-2)<sup>2<\/sup> + 8(-2) + 12 = 16 + 16 \u2013 28 \u2013 16 + 12 = 0<\/p>\n<p>p(-2) = 0<\/p>\n<p>Let x \u2013 2 = 0 or x = 2<\/p>\n<p>p(2) = (2)<sup>4<\/sup> \u2013 2(2)<sup>3<\/sup> \u2013 7(2)<sup>2<\/sup> + 8(2) + 12 = 16 \u2013 16 \u2013 28 + 16 + 12 = 0<\/p>\n<p>p(2) = 0<\/p>\n<p>Let x \u2013 3 = 0 or x = 3<\/p>\n<p>p(3) = (3)<sup>4<\/sup> \u2013 2(3)<sup>3<\/sup> \u2013 7(3)<sup>2<\/sup> + 8(3) + 12 = 0<\/p>\n<p>p(3) = 0<\/p>\n<p>Therefore, (x + 1), (x + 2), (x \u2013 2) and (x-3) are factors of p(x)<\/p>\n<p>Hence p(x) = (x + 1)(x + 2) (x \u2013 2) (x-3)<\/p>\n<h2><span class=\"ez-toc-section\" id=\"frequently-asked-questions-faqs-of-rd-sharma-chapter-6-class-9-maths-exercise-65-solutions\"><\/span>Frequently Asked Questions (FAQs) of RD Sharma Chapter 6 Class 9 Maths Exercise 6.5 Solutions<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>Ques- How do you factor polynomials with integer coefficients?<\/p>\n<p>Ans- Suppose \u2018P\u2019 is a polynomial with integer coefficients, where P(a) \u2212 P(b) is divisible by a\u2212b for any separate integers a and b. In particular, all integer roots of P divide P(0).<\/p>\n<p><a href=\"https:\/\/cbse.gov.in\/\" target=\"_blank\" rel=\"noopener noreferrer\">Know about the CBSE Board<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"<p>In the RD Sharma Chapter 6 Class 9 Maths Exercise 6.5 Solutions, we are providing the detail about the Factorization of Polynomials with Integral Coefficient using the Factor Theorem. We have given the details based on the CBSE Book of Class 9 and RD Sharma. The Integral Coefficient is the coefficient in an algebraic identity &#8230; <a title=\"RD Sharma Chapter 6 Class 9 Maths Exercise 6.5 Solutions\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-chapter-6-class-9-maths-exercise-6-5-solutions\/\" aria-label=\"More on RD Sharma Chapter 6 Class 9 Maths Exercise 6.5 Solutions\">Read more<\/a><\/p>\n","protected":false},"author":241,"featured_media":71963,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[2985,2917,73719,73411],"tags":[3081,3086,3085],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/71962"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/241"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=71962"}],"version-history":[{"count":3,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/71962\/revisions"}],"predecessor-version":[{"id":73405,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/71962\/revisions\/73405"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/71963"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=71962"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=71962"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=71962"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}