To show that 2x+12x + 12x+1 is a factor of 2×3+3×2\u221211x\u221262x^3 + 3x^2 – 11x – 62×3+3×2\u221211x\u22126 using the Factor Theorem<\/strong>, follow these steps:<\/p>\n\n\n\n The Factor Theorem states:<\/p>\n\n\n\n If f(a)=0f(a) = 0f(a)=0, then (x\u2212a)(x – a)(x\u2212a) is a factor of the polynomial f(x)f(x)f(x).<\/p>\n<\/blockquote>\n\n\n\n Given the expression 2x+12x + 12x+1, rewrite it in the form x\u2212ax – ax\u2212a. Now evaluate the polynomial: f(x)=2×3+3×2\u221211x\u22126f(x) = 2x^3 + 3x^2 – 11x – 6f(x)=2×3+3×2\u221211x\u22126<\/p>\n\n\n\n at x=\u221212x = -\\frac{1}{2}x=\u221221\u200b.<\/p>\n\n\n\n f(\u221212)=2(\u221212)3+3(\u221212)2\u221211(\u221212)\u22126f\\left(-\\frac{1}{2}\\right) = 2\\left(-\\frac{1}{2}\\right)^3 + 3\\left(-\\frac{1}{2}\\right)^2 – 11\\left(-\\frac{1}{2}\\right) – 6f(\u221221\u200b)=2(\u221221\u200b)3+3(\u221221\u200b)2\u221211(\u221221\u200b)\u22126<\/p>\n\n\n\n Now calculate each term:<\/p>\n\n\n\n f(\u221212)=2(\u221218)+3(14)+112\u22126f\\left(-\\frac{1}{2}\\right) = 2\\left(-\\frac{1}{8}\\right) + 3\\left(\\frac{1}{4}\\right) + \\frac{11}{2} – 6f(\u221221\u200b)=2(\u221281\u200b)+3(41\u200b)+211\u200b\u22126 =\u221228+34+112\u22126=\u221214+34+112\u22126= -\\frac{2}{8} + \\frac{3}{4} + \\frac{11}{2} – 6 = -\\frac{1}{4} + \\frac{3}{4} + \\frac{11}{2} – 6=\u221282\u200b+43\u200b+211\u200b\u22126=\u221241\u200b+43\u200b+211\u200b\u22126<\/p>\n\n\n\n Combine the terms:<\/p>\n\n\n\n f(\u221212)=0f\\left(-\\frac{1}{2}\\right) = 0f(\u221221\u200b)=0<\/p>\n\n\n\n Since f(\u221212)=0f\\left(-\\frac{1}{2}\\right) = 0f(\u221221\u200b)=0, by the Factor Theorem<\/strong>, x+12x + \\frac{1}{2}x+21\u200b is a factor of f(x)f(x)f(x). To determine whether 2x+12x + 12x+1 is a factor of the polynomial f(x)=2×3+3×2\u221211x\u22126f(x) = 2x^3 + 3x^2 – 11x – 6f(x)=2×3+3×2\u221211x\u22126, the Factor Theorem<\/strong> is applied. First, rewrite 2x+12x + 12x+1 in the form x\u2212ax – ax\u2212a by solving 2x+1=02x + 1 = 02x+1=0, which gives x=\u221212x = -\\frac{1}{2}x=\u221221\u200b. Then evaluate the polynomial at x=\u221212x = -\\frac{1}{2}x=\u221221\u200b. If the result is 0, then 2x+12x + 12x+1 is indeed a factor. Substituting x=\u221212x = -\\frac{1}{2}x=\u221221\u200b into the polynomial results in zero, confirming that the factor condition is satisfied. Thus, 2x+12x + 12x+1 is a factor of the polynomial.<\/p>\n\n\n\n Using the factor theorem, show that 2x + 1 is a factor of 2x^3 + 3x^2 – 11x – 6. The Correct Answer and Explanation is: To show that 2x+12x + 12x+1 is a factor of 2×3+3×2\u221211x\u221262x^3 + 3x^2 – 11x – 62×3+3×2\u221211x\u22126 using the Factor Theorem, follow these steps: Step 1: Use the Factor […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-233738","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233738","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=233738"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233738\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=233738"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=233738"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=233738"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nStep 1: Use the Factor Theorem<\/strong><\/h3>\n\n\n\n
\n
To do this, solve: 2x+1=0\u21d2x=\u2212122x + 1 = 0 \\Rightarrow x = -\\frac{1}{2}2x+1=0\u21d2x=\u221221\u200b<\/p>\n\n\n\n
\n\n\n\nStep 2: Substitute x=\u221212x = -\\frac{1}{2}x=\u221221\u200b into the polynomial<\/strong><\/h3>\n\n\n\n
\n
\n
\n\n\n\nStep 3: Conclusion<\/strong><\/h3>\n\n\n\n
Multiplying both sides of x+12x + \\frac{1}{2}x+21\u200b by 2 gives 2x+12x + 12x+1, so: 2x+1 is a factor of 2×3+3×2\u221211x\u22126\\boxed{2x + 1 \\text{ is a factor of } 2x^3 + 3x^2 – 11x – 6}2x+1 is a factor of 2×3+3×2\u221211x\u22126\u200b<\/p>\n\n\n\n
\n\n\n\nSummary Explanation (Like in textbooks)<\/strong><\/h3>\n\n\n\n
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<\/figure>\n","protected":false},"excerpt":{"rendered":"