The polynomial function y = x3 + 8×2 – 12x – 144 has a root located at (-6,0).<\/p>\n\n\n\n
a) Use polynomial division AND factoring to write the polynomial in fully factored form. Show all work that supports your answer. Fully factored form: y<\/p>\n\n\n\n
b) Identify the degree and all roots of the original polynomial function. *Write roots as (x, y) points! Degree Crossing Root(s) Bouncing Root(s)<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n Let’s solve the given polynomial function step by step.<\/p>\n\n\n\n [ We divide ( x^3 + 8x^2 – 12x – 144 ) by ( x + 6 ) using synthetic division.<\/p>\n\n\n\n Set up the division with ( x+6 ):<\/p>\n\n\n\n The quotient is ( x^2 + 2x – 24 ), and the remainder is 0, confirming that ( x+6 ) is a factor.<\/p>\n\n\n\n Now, we factor ( x^2 + 2x – 24 ).<\/p>\n\n\n\n We need two numbers that multiply to -24 and add to 2. These numbers are 6 and -4.<\/p>\n\n\n\n [ Since we already divided by ( x+6 ), the fully factored form of the polynomial is:<\/p>\n\n\n\n [ or written with exponents:<\/p>\n\n\n\n [ The given polynomial function is ( y = x^3 + 8x^2 – 12x – 144 ), and we are given that one of its roots is ( (-6,0) ). This means ( x+6 ) is a factor of the polynomial.<\/p>\n\n\n\n To find the fully factored form, we perform polynomial division. Using synthetic division, we divide ( x^3 + 8x^2 – 12x – 144 ) by ( x+6 ), which gives the quotient ( x^2 + 2x – 24 ) with a remainder of 0, confirming ( x+6 ) as a factor.<\/p>\n\n\n\n Next, we factor ( x^2 + 2x – 24 ). Finding two numbers that multiply to -24 and add to 2, we get 6 and -4. So, it factors as ( (x+6)(x-4) ). Since we already had an ( x+6 ) factor from the division, the fully factored form of the polynomial is:<\/p>\n\n\n\n [ This form helps us identify the roots. The exponent 2 on ( (x+6) ) means the root ( x = -6 ) has even multiplicity, so the graph of the function bounces<\/strong> at ( x = -6 ). The root ( x = 4 ) has an exponent of 1, meaning it crosses<\/strong> the x-axis at ( x = 4 ).<\/p>\n\n\n\n Thus, the function has a degree of 3<\/strong> (highest power of x), a bouncing root at (-6,0)<\/strong>, and a crossing root at (4,0)<\/strong>.<\/p>\n\n\n\n This solution shows how polynomial division and factoring help break down the function into its simplest form and understand its behavior graphically.<\/p>\n","protected":false},"excerpt":{"rendered":" The polynomial function y = x3 + 8×2 – 12x – 144 has a root located at (-6,0). a) Use polynomial division AND factoring to write the polynomial in fully factored form. Show all work that supports your answer. Fully factored form: y b) Identify the degree and all roots of the original polynomial function. […]<\/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-201998","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201998","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=201998"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201998\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=201998"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=201998"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=201998"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Given Polynomial:<\/h3>\n\n\n\n
y = x^3 + 8x^2 – 12x – 144
]
We are given that ( (-6,0) ) is a root, which means ( x+6 ) is a factor of the polynomial.<\/p>\n\n\n\nStep 1: Polynomial Division<\/strong><\/h4>\n\n\n\n
Synthetic Division:<\/strong><\/h5>\n\n\n\n
-6 | 1 8 -12 -144\n | -6 -12 144\n -----------------\n 1 2 -24 0<\/code><\/pre>\n\n\n\nStep 2: Factoring the Quotient<\/strong><\/h4>\n\n\n\n
x^2 + 2x – 24 = (x + 6)(x – 4)
]<\/p>\n\n\n\nStep 3: Fully Factored Form<\/strong><\/h4>\n\n\n\n
y = (x + 6)(x + 6)(x – 4)
]<\/p>\n\n\n\n
y = (x + 6)^2 (x – 4)
]<\/p>\n\n\n\n
\n\n\n\nPart B: Degree and Roots<\/strong><\/h3>\n\n\n\n
\n
\n
Final Answer:<\/strong><\/h3>\n\n\n\n
\n
\n\n\n\nExplanation (300 Words)<\/strong><\/h3>\n\n\n\n
y = (x+6)^2 (x-4)
]<\/p>\n\n\n\n