To simplify the given expression:x+5x\u22122\u22c5(x2+3x\u22129)\u2212(x2\u22122x)\u22125x\u22129\u203e\u2212(5x\u221210)\\frac{x+5}{x-2} \\cdot \\left( x^2 + 3x – 9 \\right) – \\left( x^2 – 2x \\right) – \\overline{5x – 9} – \\left( 5x – 10 \\right)x\u22122x+5\u200b\u22c5(x2+3x\u22129)\u2212(x2\u22122x)\u22125x\u22129\u200b\u2212(5x\u221210)<\/p>\n\n\n\n
Step 1: Start by simplifying each term<\/h3>\n\n\n\n
We will deal with the expression one part at a time.<\/p>\n\n\n\n
- \n
- The fraction term:<\/strong> x+5x\u22122\u22c5(x2+3x\u22129)\\frac{x+5}{x-2} \\cdot \\left( x^2 + 3x – 9 \\right)x\u22122x+5\u200b\u22c5(x2+3x\u22129) This is a multiplication of a rational function with a polynomial. So, multiply each term in the polynomial by the fraction: (x+5)\u22c5(x2+3x\u22129)x\u22122\\frac{(x+5) \\cdot (x^2 + 3x – 9)}{x – 2}x\u22122(x+5)\u22c5(x2+3x\u22129)\u200b Expanding (x+5)(x2+3x\u22129)(x+5)(x^2 + 3x – 9)(x+5)(x2+3x\u22129): (x+5)(x2+3x\u22129)=x(x2+3x\u22129)+5(x2+3x\u22129)(x+5)(x^2 + 3x – 9) = x(x^2 + 3x – 9) + 5(x^2 + 3x – 9)(x+5)(x2+3x\u22129)=x(x2+3x\u22129)+5(x2+3x\u22129) =x3+3×2\u22129x+5×2+15x\u221245= x^3 + 3x^2 – 9x + 5x^2 + 15x – 45=x3+3×2\u22129x+5×2+15x\u221245 =x3+8×2+6x\u221245= x^3 + 8x^2 + 6x – 45=x3+8×2+6x\u221245 So, the first term becomes: x3+8×2+6x\u221245x\u22122\\frac{x^3 + 8x^2 + 6x – 45}{x – 2}x\u22122×3+8×2+6x\u221245\u200b<\/li>\n\n\n\n
- The second and third terms:<\/strong> We have \u2212(x2\u22122x)-(x^2 – 2x)\u2212(x2\u22122x) and \u22125x\u22129\u203e-\\overline{5x – 9}\u22125x\u22129\u200b. These are simpler, so they are just: \u2212x2+2xand\u2212(5x\u22129)-x^2 + 2x \\quad \\text{and} \\quad -(5x – 9)\u2212x2+2xand\u2212(5x\u22129) which becomes: \u22125x+9-5x + 9\u22125x+9<\/li>\n\n\n\n
- The fourth term:<\/strong> We also have \u2212(5x\u221210)-(5x – 10)\u2212(5x\u221210), which simplifies to: \u22125x+10-5x + 10\u22125x+10<\/li>\n<\/ol>\n\n\n\n
Step 2: Combine all terms<\/h3>\n\n\n\n
Now, put everything together:x3+8×2+6x\u221245x\u22122\u2212x2+2x\u22125x+9\u22125x+10\\frac{x^3 + 8x^2 + 6x – 45}{x – 2} – x^2 + 2x – 5x + 9 – 5x + 10x\u22122×3+8×2+6x\u221245\u200b\u2212x2+2x\u22125x+9\u22125x+10<\/p>\n\n\n\n
Simplify the linear terms:=x3+8×2+6x\u221245x\u22122\u2212x2\u22128x+19= \\frac{x^3 + 8x^2 + 6x – 45}{x – 2} – x^2 – 8x + 19=x\u22122×3+8×2+6x\u221245\u200b\u2212x2\u22128x+19<\/p>\n\n\n\n
This is the simplified expression. The rational term remains as a quotient, while the polynomial part has been combined.<\/p>\n\n\n\n
Explanation:<\/h3>\n\n\n\n
This type of expression involves working with polynomials and rational functions. The primary steps include distributing terms, collecting like terms, and keeping the rational function intact. The simplification of each individual term is necessary before combining everything into one final expression. The result is a rational expression with a polynomial subtraction.<\/p>\n\n\n\n
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<\/figure>\n","protected":false},"excerpt":{"rendered":"\\frac{x+5}{x-2} \\Big) x^2+3x-9 \\newline -(x^2-2x) \\newline \\overline{5x-9} \\newline -(5x-10) The Correct Answer and Explanation is: To simplify the given expression:x+5x\u22122\u22c5(x2+3x\u22129)\u2212(x2\u22122x)\u22125x\u22129\u203e\u2212(5x\u221210)\\frac{x+5}{x-2} \\cdot \\left( x^2 + 3x – 9 \\right) – \\left( x^2 – 2x \\right) – \\overline{5x – 9} – \\left( 5x – 10 \\right)x\u22122x+5\u200b\u22c5(x2+3x\u22129)\u2212(x2\u22122x)\u22125x\u22129\u200b\u2212(5x\u221210) Step 1: Start by simplifying each term We will deal with […]<\/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-266026","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/266026","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=266026"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/266026\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=266026"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=266026"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=266026"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
- The second and third terms:<\/strong> We have \u2212(x2\u22122x)-(x^2 – 2x)\u2212(x2\u22122x) and \u22125x\u22129\u203e-\\overline{5x – 9}\u22125x\u22129\u200b. These are simpler, so they are just: \u2212x2+2xand\u2212(5x\u22129)-x^2 + 2x \\quad \\text{and} \\quad -(5x – 9)\u2212x2+2xand\u2212(5x\u22129) which becomes: \u22125x+9-5x + 9\u22125x+9<\/li>\n\n\n\n