{"id":1753,"date":"2025-07-28T18:59:39","date_gmt":"2025-07-28T18:59:39","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&#038;p=1753"},"modified":"2026-03-26T19:20:10","modified_gmt":"2026-03-26T19:20:10","slug":"binomial-theorem-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/binomial-theorem-learn-it-2\/","title":{"raw":"Binomial Theorem: Learn It 2","rendered":"Binomial Theorem: Learn It 2"},"content":{"raw":"<h2>The Binomial Theorem<\/h2>\r\nWhen we expand [latex]{\\left(x+y\\right)}^{n}[\/latex] by multiplying, the result is called a <strong>binomial expansion<\/strong>, and it includes binomial coefficients. If we wanted to expand [latex]{\\left(x+y\\right)}^{52}[\/latex], we might multiply [latex]\\left(x+y\\right)[\/latex] by itself fifty-two times. This could take hours!\r\n\r\nLet's examine some simple binomial expansions, we can find patterns that will lead us to a shortcut for finding more complicated binomial expansions.\r\n<p style=\"text-align: center;\"><img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03234323\/CNX_Precalc_Figure_11_06_0022.jpg\" alt=\"Graph of the function f_2.\" width=\"550\" height=\"311\" \/><\/p>\r\nNotice that:\r\n<ul>\r\n \t<li>There are [latex]n+1[\/latex] terms in the expansion of [latex]{\\left(x+y\\right)}^{n}[\/latex].<\/li>\r\n \t<li>The degree (or sum of the exponents) for each term is [latex]n[\/latex].<\/li>\r\n \t<li>The powers on [latex]x[\/latex] begin with [latex]n[\/latex] and decrease to [latex]0[\/latex].<\/li>\r\n \t<li>The powers on [latex]y[\/latex] begin with [latex]0[\/latex] and increase to [latex]n[\/latex].<\/li>\r\n \t<li>The coefficients are symmetric.<\/li>\r\n<\/ul>\r\nBut where do those coefficients come from? The binomial coefficients are symmetric. We can see these coefficients in an array known as <strong>Pascal's Triangle<\/strong>.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03234326\/CNX_Precalc_Figure_11_06_0012.jpg\" alt=\"Pascal's Triangle\" width=\"549\" height=\"225\" \/>\r\n\r\n[reveal-answer q=\"47861\"]How to generate the Pascal's Triangle?[\/reveal-answer]\r\n[hidden-answer a=\"47861\"]To generate Pascal\u2019s Triangle, we start by writing a 1. In the row below, row 2, we write two 1\u2019s. In the 3rd row, flank the ends of the rows with 1\u2019s, and add [latex]1+1[\/latex] to find the middle number, 2. In the [latex]n\\text{th}[\/latex] row, flank the ends of the row with 1\u2019s. Each element in the triangle is the sum of the two elements immediately above it.[\/hidden-answer]\r\n\r\nLet's put it side-by-side to see the connection between Pascal\u2019s Triangle and binomial expansion. Pay attention to the coefficients!\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03234328\/CNX_Precalc_Figure_11_06_0032.jpg\" alt=\"Pascal's Triangle expanded to show the values of the triangle as x and y terms with exponents\" width=\"553\" height=\"121\" \/>\r\n\r\nThese patterns lead us to the <strong>Binomial Theorem<\/strong>, which can be used to expand any binomial.\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>Binomial Theorem<\/h3>\r\nThe <strong>Binomial Theorem<\/strong> is a formula that can be used to expand any binomial.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}{\\left(x+y\\right)}^{n}&amp; =\\sum\\limits _{k=0}^{n}\\left(\\begin{gathered}n\\\\ k\\end{gathered}\\right){x}^{n-k}{y}^{k} \\\\ &amp; ={x}^{n}+\\left(\\begin{gathered}n\\\\ 1\\end{gathered}\\right){x}^{n - 1}y+\\left(\\begin{gathered}n\\\\ 2\\end{gathered}\\right){x}^{n - 2}{y}^{2}+\\dots+\\left(\\begin{gathered}n\\\\ n - 1\\end{gathered}\\right)x{y}^{n - 1}+{y}^{n} \\end{align}[\/latex]<\/p>\r\n\r\n<\/section><section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a binomial, write it in expanded form.<\/strong>\r\n<ol>\r\n \t<li>Determine the value of [latex]n[\/latex] according to the exponent.<\/li>\r\n \t<li>Evaluate the [latex]k=0[\/latex] through [latex]k=n[\/latex] using the Binomial Theorem formula.<\/li>\r\n \t<li>Simplify.<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Write in expanded form.\r\n<ol>\r\n \t<li>[latex]{\\left(x+y\\right)}^{5}[\/latex]<\/li>\r\n \t<li>[latex]{\\left(3x-y\\right)}^{4}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"607135\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"607135\"]\r\n<ol>\r\n \t<li>Substitute [latex]n=5[\/latex] into the formula. Evaluate the [latex]k=0[\/latex] through [latex]k=5[\/latex] terms. Simplify.\r\n<div style=\"text-align: center;\">\r\n\r\n[latex]\\begin{align}{\\left(x+y\\right)}^{5} &amp; =\\left(\\begin{gathered}5\\\\ 0\\end{gathered}\\right){x}^{5}{y}^{0}+\\left(\\begin{gathered}5\\\\ 1\\end{gathered}\\right){x}^{4}{y}^{1}+\\left(\\begin{gathered}5\\\\ 2\\end{gathered}\\right){x}^{3}{y}^{2}+\\left(\\begin{gathered}5\\\\ 3\\end{gathered}\\right){x}^{2}{y}^{3}+\\left(\\begin{gathered}5\\\\ 4\\end{gathered}\\right){x}^{1}{y}^{4}+\\left(\\begin{gathered}5\\\\ 5\\end{gathered}\\right){x}^{0}{y}^{5} \\\\ {\\left(x+y\\right)}^{5} &amp; ={x}^{5}+5{x}^{4}y+10{x}^{3}{y}^{2}+10{x}^{2}{y}^{3}+5x{y}^{4}+{y}^{5} \\end{align}[\/latex]\r\n\r\n<\/div><\/li>\r\n \t<li>Substitute [latex]n=4[\/latex] into the formula. Evaluate the [latex]k=0[\/latex] through [latex]k=4[\/latex] terms. Notice that [latex]3x[\/latex] is in the place that was occupied by [latex]x[\/latex] and that [latex]-y[\/latex] is in the place that was occupied by [latex]y[\/latex]. So we substitute them. Simplify.\r\n<div style=\"text-align: center;\">[latex]\\begin{align}{\\left(3x-y\\right)}^{4} &amp; =\\left(\\begin{gathered}4\\\\ 0\\end{gathered}\\right){\\left(3x\\right)}^{4}{\\left(-y\\right)}^{0}+\\left(\\begin{gathered}4\\\\ 1\\end{gathered}\\right){\\left(3x\\right)}^{3}{\\left(-y\\right)}^{1}+\\left(\\begin{gathered}4\\\\ 2\\end{gathered}\\right){\\left(3x\\right)}^{2}{\\left(-y\\right)}^{2}+\\left(\\begin{gathered}4\\\\ 3\\end{gathered}\\right){\\left(3x\\right)}^{1}{\\left(-y\\right)}^{3}+\\left(\\begin{gathered}4\\\\ 4\\end{gathered}\\right){\\left(3x\\right)}^{0}{\\left(-y\\right)}^{4} \\\\ {\\left(3x-y\\right)}^{4} &amp; =81{x}^{4}-108{x}^{3}y+54{x}^{2}{y}^{2}-12x{y}^{3}+{y}^{4} \\end{align}[\/latex]<\/div><\/li>\r\n<\/ol>\r\n<strong>Analysis of the Solution<\/strong>\r\n\r\nNotice the alternating signs in part b. This happens because [latex]\\left(-y\\right)[\/latex] raised to odd powers is negative, but [latex]\\left(-y\\right)[\/latex] raised to even powers is positive. This will occur whenever the binomial contains a subtraction sign.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]322005[\/ohm_question]<\/section>","rendered":"<h2>The Binomial Theorem<\/h2>\n<p>When we expand [latex]{\\left(x+y\\right)}^{n}[\/latex] by multiplying, the result is called a <strong>binomial expansion<\/strong>, and it includes binomial coefficients. If we wanted to expand [latex]{\\left(x+y\\right)}^{52}[\/latex], we might multiply [latex]\\left(x+y\\right)[\/latex] by itself fifty-two times. This could take hours!<\/p>\n<p>Let&#8217;s examine some simple binomial expansions, we can find patterns that will lead us to a shortcut for finding more complicated binomial expansions.<\/p>\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03234323\/CNX_Precalc_Figure_11_06_0022.jpg\" alt=\"Graph of the function f_2.\" width=\"550\" height=\"311\" \/><\/p>\n<p>Notice that:<\/p>\n<ul>\n<li>There are [latex]n+1[\/latex] terms in the expansion of [latex]{\\left(x+y\\right)}^{n}[\/latex].<\/li>\n<li>The degree (or sum of the exponents) for each term is [latex]n[\/latex].<\/li>\n<li>The powers on [latex]x[\/latex] begin with [latex]n[\/latex] and decrease to [latex]0[\/latex].<\/li>\n<li>The powers on [latex]y[\/latex] begin with [latex]0[\/latex] and increase to [latex]n[\/latex].<\/li>\n<li>The coefficients are symmetric.<\/li>\n<\/ul>\n<p>But where do those coefficients come from? The binomial coefficients are symmetric. We can see these coefficients in an array known as <strong>Pascal&#8217;s Triangle<\/strong>.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03234326\/CNX_Precalc_Figure_11_06_0012.jpg\" alt=\"Pascal's Triangle\" width=\"549\" height=\"225\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q47861\">How to generate the Pascal&#8217;s Triangle?<\/button><\/p>\n<div id=\"q47861\" class=\"hidden-answer\" style=\"display: none\">To generate Pascal\u2019s Triangle, we start by writing a 1. In the row below, row 2, we write two 1\u2019s. In the 3rd row, flank the ends of the rows with 1\u2019s, and add [latex]1+1[\/latex] to find the middle number, 2. In the [latex]n\\text{th}[\/latex] row, flank the ends of the row with 1\u2019s. Each element in the triangle is the sum of the two elements immediately above it.<\/div>\n<\/div>\n<p>Let&#8217;s put it side-by-side to see the connection between Pascal\u2019s Triangle and binomial expansion. Pay attention to the coefficients!<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/03234328\/CNX_Precalc_Figure_11_06_0032.jpg\" alt=\"Pascal's Triangle expanded to show the values of the triangle as x and y terms with exponents\" width=\"553\" height=\"121\" \/><\/p>\n<p>These patterns lead us to the <strong>Binomial Theorem<\/strong>, which can be used to expand any binomial.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>Binomial Theorem<\/h3>\n<p>The <strong>Binomial Theorem<\/strong> is a formula that can be used to expand any binomial.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}{\\left(x+y\\right)}^{n}& =\\sum\\limits _{k=0}^{n}\\left(\\begin{gathered}n\\\\ k\\end{gathered}\\right){x}^{n-k}{y}^{k} \\\\ & ={x}^{n}+\\left(\\begin{gathered}n\\\\ 1\\end{gathered}\\right){x}^{n - 1}y+\\left(\\begin{gathered}n\\\\ 2\\end{gathered}\\right){x}^{n - 2}{y}^{2}+\\dots+\\left(\\begin{gathered}n\\\\ n - 1\\end{gathered}\\right)x{y}^{n - 1}+{y}^{n} \\end{align}[\/latex]<\/p>\n<\/section>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\"><strong>How To: Given a binomial, write it in expanded form.<\/strong><\/p>\n<ol>\n<li>Determine the value of [latex]n[\/latex] according to the exponent.<\/li>\n<li>Evaluate the [latex]k=0[\/latex] through [latex]k=n[\/latex] using the Binomial Theorem formula.<\/li>\n<li>Simplify.<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Write in expanded form.<\/p>\n<ol>\n<li>[latex]{\\left(x+y\\right)}^{5}[\/latex]<\/li>\n<li>[latex]{\\left(3x-y\\right)}^{4}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q607135\">Show Solution<\/button><\/p>\n<div id=\"q607135\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>Substitute [latex]n=5[\/latex] into the formula. Evaluate the [latex]k=0[\/latex] through [latex]k=5[\/latex] terms. Simplify.\n<div style=\"text-align: center;\">\n<p>[latex]\\begin{align}{\\left(x+y\\right)}^{5} & =\\left(\\begin{gathered}5\\\\ 0\\end{gathered}\\right){x}^{5}{y}^{0}+\\left(\\begin{gathered}5\\\\ 1\\end{gathered}\\right){x}^{4}{y}^{1}+\\left(\\begin{gathered}5\\\\ 2\\end{gathered}\\right){x}^{3}{y}^{2}+\\left(\\begin{gathered}5\\\\ 3\\end{gathered}\\right){x}^{2}{y}^{3}+\\left(\\begin{gathered}5\\\\ 4\\end{gathered}\\right){x}^{1}{y}^{4}+\\left(\\begin{gathered}5\\\\ 5\\end{gathered}\\right){x}^{0}{y}^{5} \\\\ {\\left(x+y\\right)}^{5} & ={x}^{5}+5{x}^{4}y+10{x}^{3}{y}^{2}+10{x}^{2}{y}^{3}+5x{y}^{4}+{y}^{5} \\end{align}[\/latex]<\/p>\n<\/div>\n<\/li>\n<li>Substitute [latex]n=4[\/latex] into the formula. Evaluate the [latex]k=0[\/latex] through [latex]k=4[\/latex] terms. Notice that [latex]3x[\/latex] is in the place that was occupied by [latex]x[\/latex] and that [latex]-y[\/latex] is in the place that was occupied by [latex]y[\/latex]. So we substitute them. Simplify.\n<div style=\"text-align: center;\">[latex]\\begin{align}{\\left(3x-y\\right)}^{4} & =\\left(\\begin{gathered}4\\\\ 0\\end{gathered}\\right){\\left(3x\\right)}^{4}{\\left(-y\\right)}^{0}+\\left(\\begin{gathered}4\\\\ 1\\end{gathered}\\right){\\left(3x\\right)}^{3}{\\left(-y\\right)}^{1}+\\left(\\begin{gathered}4\\\\ 2\\end{gathered}\\right){\\left(3x\\right)}^{2}{\\left(-y\\right)}^{2}+\\left(\\begin{gathered}4\\\\ 3\\end{gathered}\\right){\\left(3x\\right)}^{1}{\\left(-y\\right)}^{3}+\\left(\\begin{gathered}4\\\\ 4\\end{gathered}\\right){\\left(3x\\right)}^{0}{\\left(-y\\right)}^{4} \\\\ {\\left(3x-y\\right)}^{4} & =81{x}^{4}-108{x}^{3}y+54{x}^{2}{y}^{2}-12x{y}^{3}+{y}^{4} \\end{align}[\/latex]<\/div>\n<\/li>\n<\/ol>\n<p><strong>Analysis of the Solution<\/strong><\/p>\n<p>Notice the alternating signs in part b. This happens because [latex]\\left(-y\\right)[\/latex] raised to odd powers is negative, but [latex]\\left(-y\\right)[\/latex] raised to even powers is positive. This will occur whenever the binomial contains a subtraction sign.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm322005\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=322005&theme=lumen&iframe_resize_id=ohm322005&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":13,"menu_order":14,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":513,"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\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1753"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/users\/13"}],"version-history":[{"count":3,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1753\/revisions"}],"predecessor-version":[{"id":6054,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1753\/revisions\/6054"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/513"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/1753\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=1753"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=1753"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=1753"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=1753"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}