{"id":4347,"date":"2024-09-30T16:52:02","date_gmt":"2024-09-30T16:52:02","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/?post_type=chapter&#038;p=4347"},"modified":"2025-08-15T02:55:30","modified_gmt":"2025-08-15T02:55:30","slug":"mastering-polynomial-functions-theorems-zeros-and-applications-learn-it-7","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/chapter\/mastering-polynomial-functions-theorems-zeros-and-applications-learn-it-7\/","title":{"raw":"Zeros of Polynomial Functions: Learn It 7","rendered":"Zeros of Polynomial Functions: Learn It 7"},"content":{"raw":"<h2>Solving Real-world Applications of Polynomial Equations<\/h2>\r\nWe have now introduced a variety of tools for solving polynomial equations.\r\n\r\n<section class=\"textbox example\">A new bakery offers decorated sheet cakes for children\u2019s birthday parties and other special occasions. The bakery wants the volume of a small cake to be 351 cubic inches. The cake is in the shape of a rectangular solid. They want the length of the cake to be four inches longer than the width of the cake and the height of the cake to be one-third of the width. What should the dimensions of the cake pan be?[reveal-answer q=\"801673\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"801673\"]Begin by writing an equation for the volume of the cake. The volume of a rectangular solid is given by [latex]V=lwh[\/latex]. We were given that the length must be four inches longer than the width, so we can express the length of the cake as [latex]l=w+4[\/latex]. We were given that the height of the cake is one-third of the width, so we can express the height of the cake as [latex]h=\\frac{1}{3}w[\/latex]. Let\u2019s write the volume of the cake in terms of width of the cake.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}V=\\left(w+4\\right)\\left(w\\right)\\left(\\frac{1}{3}w\\right)\\\\ V=\\frac{1}{3}{w}^{3}+\\frac{4}{3}{w}^{2}\\end{array}[\/latex]<\/p>\r\nSubstitute the given volume into this equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }351=\\frac{1}{3}{w}^{3}+\\frac{4}{3}{w}^{2}\\hfill &amp; \\text{Substitute 351 for }V.\\hfill \\\\ 1053={w}^{3}+4{w}^{2}\\hfill &amp; \\text{Multiply both sides by 3}.\\hfill \\\\ \\text{ }0={w}^{3}+4{w}^{2}-1053 \\hfill &amp; \\text{Subtract 1053 from both sides}.\\hfill \\end{array}[\/latex]<\/p>\r\nDescartes' rule of signs tells us there is one positive solution. The Rational Zero Theorem tells us that the possible rational zeros are [latex]\\pm 3,\\pm 9,\\pm 13,\\pm 27,\\pm 39,\\pm 81,\\pm 117,\\pm 351[\/latex],\u00a0and [latex]\\pm 1053[\/latex].\u00a0We can use synthetic division to test these possible zeros. Only positive numbers make sense as dimensions for a cake, so we need not test any negative values. Let\u2019s begin by testing values that make the most sense as dimensions for a small sheet cake. Use synthetic division to check [latex]x=1[\/latex].\r\n\r\n[caption id=\"attachment_13122\" align=\"aligncenter\" width=\"155\"]<img class=\"wp-image-13122 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02205604\/Screen-Shot-2015-09-11-at-3.12.09-PM.png\" alt=\"Synthetic Division with divisor = 1, and quotient = {1, 4, 0, -1053}. Solution is {1, 5, 5, -1048}\" width=\"155\" height=\"100\" \/> Synthetic division[\/caption]\r\n\r\nSince 1 is not a solution, we will check [latex]x=3[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"100\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02205606\/CNX_Precalc_revised_eq12.png\" alt=\".\" width=\"100\" height=\"178\" \/> Synthetic division[\/caption]\r\n\r\nSince 3 is not a solution either, we will test [latex]x=9[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"120\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02205608\/CNX_Precalc_revised_eq22.png\" alt=\".\" width=\"120\" height=\"170\" \/> Synthetic division[\/caption]\r\n\r\nSynthetic division gives a remainder of 0, so 9 is a solution to the equation. We can use the relationships between the width and the other dimensions to determine the length and height of the sheet cake pan.\r\n<p style=\"text-align: center;\">[latex]l=w+4=9+4=13\\text{ and }h=\\frac{1}{3}w=\\frac{1}{3}\\left(9\\right)=3[\/latex]<\/p>\r\nThe sheet cake pan should have dimensions 13 inches by 9 inches by 3 inches.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]24639[\/ohm2_question]<\/section>","rendered":"<h2>Solving Real-world Applications of Polynomial Equations<\/h2>\n<p>We have now introduced a variety of tools for solving polynomial equations.<\/p>\n<section class=\"textbox example\">A new bakery offers decorated sheet cakes for children\u2019s birthday parties and other special occasions. The bakery wants the volume of a small cake to be 351 cubic inches. The cake is in the shape of a rectangular solid. They want the length of the cake to be four inches longer than the width of the cake and the height of the cake to be one-third of the width. What should the dimensions of the cake pan be?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q801673\">Show Solution<\/button><\/p>\n<div id=\"q801673\" class=\"hidden-answer\" style=\"display: none\">Begin by writing an equation for the volume of the cake. The volume of a rectangular solid is given by [latex]V=lwh[\/latex]. We were given that the length must be four inches longer than the width, so we can express the length of the cake as [latex]l=w+4[\/latex]. We were given that the height of the cake is one-third of the width, so we can express the height of the cake as [latex]h=\\frac{1}{3}w[\/latex]. Let\u2019s write the volume of the cake in terms of width of the cake.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}V=\\left(w+4\\right)\\left(w\\right)\\left(\\frac{1}{3}w\\right)\\\\ V=\\frac{1}{3}{w}^{3}+\\frac{4}{3}{w}^{2}\\end{array}[\/latex]<\/p>\n<p>Substitute the given volume into this equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\text{ }351=\\frac{1}{3}{w}^{3}+\\frac{4}{3}{w}^{2}\\hfill & \\text{Substitute 351 for }V.\\hfill \\\\ 1053={w}^{3}+4{w}^{2}\\hfill & \\text{Multiply both sides by 3}.\\hfill \\\\ \\text{ }0={w}^{3}+4{w}^{2}-1053 \\hfill & \\text{Subtract 1053 from both sides}.\\hfill \\end{array}[\/latex]<\/p>\n<p>Descartes&#8217; rule of signs tells us there is one positive solution. The Rational Zero Theorem tells us that the possible rational zeros are [latex]\\pm 3,\\pm 9,\\pm 13,\\pm 27,\\pm 39,\\pm 81,\\pm 117,\\pm 351[\/latex],\u00a0and [latex]\\pm 1053[\/latex].\u00a0We can use synthetic division to test these possible zeros. Only positive numbers make sense as dimensions for a cake, so we need not test any negative values. Let\u2019s begin by testing values that make the most sense as dimensions for a small sheet cake. Use synthetic division to check [latex]x=1[\/latex].<\/p>\n<figure id=\"attachment_13122\" aria-describedby=\"caption-attachment-13122\" style=\"width: 155px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-13122 size-full\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02205604\/Screen-Shot-2015-09-11-at-3.12.09-PM.png\" alt=\"Synthetic Division with divisor = 1, and quotient = {1, 4, 0, -1053}. Solution is {1, 5, 5, -1048}\" width=\"155\" height=\"100\" \/><figcaption id=\"caption-attachment-13122\" class=\"wp-caption-text\">Synthetic division<\/figcaption><\/figure>\n<p>Since 1 is not a solution, we will check [latex]x=3[\/latex].<\/p>\n<figure style=\"width: 100px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02205606\/CNX_Precalc_revised_eq12.png\" alt=\".\" width=\"100\" height=\"178\" \/><figcaption class=\"wp-caption-text\">Synthetic division<\/figcaption><\/figure>\n<p>Since 3 is not a solution either, we will test [latex]x=9[\/latex].<\/p>\n<figure style=\"width: 120px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02205608\/CNX_Precalc_revised_eq22.png\" alt=\".\" width=\"120\" height=\"170\" \/><figcaption class=\"wp-caption-text\">Synthetic division<\/figcaption><\/figure>\n<p>Synthetic division gives a remainder of 0, so 9 is a solution to the equation. We can use the relationships between the width and the other dimensions to determine the length and height of the sheet cake pan.<\/p>\n<p style=\"text-align: center;\">[latex]l=w+4=9+4=13\\text{ and }h=\\frac{1}{3}w=\\frac{1}{3}\\left(9\\right)=3[\/latex]<\/p>\n<p>The sheet cake pan should have dimensions 13 inches by 9 inches by 3 inches.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm24639\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=24639&theme=lumen&iframe_resize_id=ohm24639&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":15,"menu_order":32,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":206,"module-header":"learn_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/4347"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":5,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/4347\/revisions"}],"predecessor-version":[{"id":7776,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/4347\/revisions\/7776"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/parts\/206"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/4347\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/media?parent=4347"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=4347"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/contributor?post=4347"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/license?post=4347"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}