{"id":2109,"date":"2015-11-12T18:30:41","date_gmt":"2015-11-12T18:30:41","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&#038;p=2109"},"modified":"2015-11-12T18:30:41","modified_gmt":"2015-11-12T18:30:41","slug":"using-the-binomial-theorem","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/chapter\/using-the-binomial-theorem\/","title":{"raw":"Using the Binomial Theorem","rendered":"Using the Binomial Theorem"},"content":{"raw":"<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! If we examine some simple binomial expansions, we can find patterns that will lead us to a shortcut for finding more complicated binomial expansions.\n<\/p><div style=\"text-align: center;\">[latex]\\begin{array}{l}{\\left(x+y\\right)}^{2}={x}^{2}+2xy+{y}^{2}\\hfill \\\\ {\\left(x+y\\right)}^{3}={x}^{3}+3{x}^{2}y+3x{y}^{2}+{y}^{3}\\hfill \\\\ {\\left(x+y\\right)}^{4}={x}^{4}+4{x}^{3}y+6{x}^{2}{y}^{2}+4x{y}^{3}+{y}^{4}\\hfill \\end{array}[\/latex]<\/div>\nFirst, let\u2019s examine the exponents. With each successive term, the exponent for [latex]x[\/latex] decreases and the exponent for [latex]y[\/latex] increases. The sum of the two exponents is [latex]n[\/latex] for each term.\n\nNext, let\u2019s examine the coefficients. Notice that the coefficients increase and then decrease in a symmetrical pattern. The coefficients follow a pattern:\n<div style=\"text-align: center;\">[latex]\\left(\\begin{array}{c}n\\\\ 0\\end{array}\\right),\\left(\\begin{array}{c}n\\\\ 1\\end{array}\\right),\\left(\\begin{array}{c}n\\\\ 2\\end{array}\\right),...,\\left(\\begin{array}{c}n\\\\ n\\end{array}\\right)[\/latex].<\/div>\nThese patterns lead us to the <strong>Binomial Theorem<\/strong>, which can be used to expand any binomial.\n<div style=\"text-align: center;\">[latex]\\begin{array}{ll}{\\left(x+y\\right)}^{n}\\hfill &amp; =\\sum _{k=0}^{n}\\left(\\begin{array}{c}n\\\\ k\\end{array}\\right){x}^{n-k}{y}^{k}\\hfill \\\\ \\hfill &amp; ={x}^{n}+\\left(\\begin{array}{c}n\\\\ 1\\end{array}\\right){x}^{n - 1}y+\\left(\\begin{array}{c}n\\\\ 2\\end{array}\\right){x}^{n - 2}{y}^{2}+...+\\left(\\begin{array}{c}n\\\\ n - 1\\end{array}\\right)x{y}^{n - 1}+{y}^{n}\\hfill \\end{array}[\/latex]<\/div>\nAnother way to see the coefficients is to examine the expansion of a binomial in general form, [latex]x+y[\/latex], to successive powers 1, 2, 3, and 4.\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}{\\left(x+y\\right)}^{1}=x+y\\hfill \\\\ {\\left(x+y\\right)}^{2}={x}^{2}+2xy+{y}^{2}\\hfill \\\\ {\\left(x+y\\right)}^{3}={x}^{3}+3{x}^{2}y+3x{y}^{2}+{y}^{3}\\hfill \\\\ {\\left(x+y\\right)}^{4}={x}^{4}+4{x}^{3}y+6{x}^{2}{y}^{2}+4x{y}^{3}+{y}^{4}\\hfill \\end{array}[\/latex]<\/div>\nCan you guess the next expansion for the binomial [latex]{\\left(x+y\\right)}^{5}?[\/latex]\n\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202615\/CNX_Precalc_Figure_11_06_0022.jpg\" alt=\"Graph of the function f_2.\" width=\"731\" height=\"413\" data-media-type=\"image\/jpg\"\/><b>Figure 1<\/b>[\/caption]\n\nSee Figure 1, which illustrates the following:\n<ul><li>There are [latex]n+1[\/latex] terms in the expansion of [latex]{\\left(x+y\\right)}^{n}[\/latex].<\/li>\n\t<li>The degree (or sum of the exponents) for each term is [latex]n[\/latex].<\/li>\n\t<li>The powers on [latex]x[\/latex] begin with [latex]n[\/latex] and decrease to 0.<\/li>\n\t<li>The powers on [latex]y[\/latex] begin with 0 and increase to [latex]n[\/latex].<\/li>\n\t<li>The coefficients are symmetric.<\/li>\n<\/ul>\nTo determine the expansion on [latex]{\\left(x+y\\right)}^{5}[\/latex], we see [latex]n=5[\/latex], thus, there will be 5+1 = 6 terms. Each term has a combined degree of 5. In descending order for powers of [latex]x[\/latex], the pattern is as follows:\n<ul><li>Introduce [latex]{x}^{5}[\/latex], and then for each successive term reduce the exponent on [latex]x[\/latex] by 1 until [latex]{x}^{0}=1[\/latex] is reached.<\/li>\n\t<li>Introduce [latex]{y}^{0}=1[\/latex], and then increase the exponent on [latex]y[\/latex] by 1 until [latex]{y}^{5}[\/latex] is reached.\n<div>[latex]{x}^{5},{x}^{4}y,{x}^{3}{y}^{2},{x}^{2}{y}^{3},x{y}^{4},{y}^{5}[\/latex]<\/div><\/li>\n<\/ul>\nThe next expansion would be\n<div style=\"text-align: center;\">[latex]{\\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}[\/latex].<\/div>\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>, shown in Figure 2.\n\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202617\/CNX_Precalc_Figure_11_06_0012.jpg\" alt=\"Pascal's Triangle\" width=\"731\" height=\"300\" data-media-type=\"image\/jpg\"\/><b>Figure 2<\/b>[\/caption]\n\nTo generate Pascal\u2019s Triangle, we start by writing a 1. In the row below, row 2, we write two 1\u2019s. In the 3<sup>rd<\/sup> 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.\n\nTo see the connection between Pascal\u2019s Triangle and binomial coefficients, let us revisit the expansion of the binomials in general form.\n\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202618\/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\" data-media-type=\"image\/jpg\"\/><div class=\"textbox\">\n<h3>A General Note: The Binomial Theorem<\/h3>\nThe <strong>Binomial Theorem<\/strong> is a formula that can be used to expand any binomial.\n<div style=\"text-align: center;\">[latex]\\begin{array}{ll}{\\left(x+y\\right)}^{n}\\hfill &amp; =\\sum _{k=0}^{n}\\left(\\begin{array}{c}n\\\\ k\\end{array}\\right){x}^{n-k}{y}^{k}\\hfill \\\\ \\hfill &amp; ={x}^{n}+\\left(\\begin{array}{c}n\\\\ 1\\end{array}\\right){x}^{n - 1}y+\\left(\\begin{array}{c}n\\\\ 2\\end{array}\\right){x}^{n - 2}{y}^{2}+...+\\left(\\begin{array}{c}n\\\\ n - 1\\end{array}\\right)x{y}^{n - 1}+{y}^{n}\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a binomial, write it in expanded form.<\/h3>\n<ol><li>Determine the value of [latex]n[\/latex] according to the exponent.<\/li>\n\t<li>Evaluate the [latex]k=0[\/latex] through [latex]k=n[\/latex] using the Binomial Theorem formula.<\/li>\n\t<li>Simplify.<\/li>\n<\/ol><\/div>\n<div class=\"textbox shaded\">\n<h3>Example 2: Expanding a Binomial<\/h3>\nWrite in expanded form.\n<ol><li>[latex]{\\left(x+y\\right)}^{5}[\/latex]<\/li>\n\t<li>[latex]{\\left(3x-y\\right)}^{4}[\/latex]<\/li>\n<\/ol><\/div>\n<div class=\"textbox shaded\">\n<h3>Solution<\/h3>\n<ol><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;\">[latex]\\begin{array}{ll}{\\left(x+y\\right)}^{5}\\hfill &amp; =\\left(\\begin{array}{c}5\\\\ 0\\end{array}\\right){x}^{5}{y}^{0}+\\left(\\begin{array}{c}5\\\\ 1\\end{array}\\right){x}^{4}{y}^{1}+\\left(\\begin{array}{c}5\\\\ 2\\end{array}\\right){x}^{3}{y}^{2}+\\left(\\begin{array}{c}5\\\\ 3\\end{array}\\right){x}^{2}{y}^{3}+\\left(\\begin{array}{c}5\\\\ 4\\end{array}\\right){x}^{1}{y}^{4}+\\left(\\begin{array}{c}5\\\\ 5\\end{array}\\right){x}^{0}{y}^{5}\\hfill \\\\ {\\left(x+y\\right)}^{5}\\hfill &amp; ={x}^{5}+5{x}^{4}y+10{x}^{3}{y}^{2}+10{x}^{2}{y}^{3}+5x{y}^{4}+{y}^{5}\\hfill \\end{array}[\/latex]<\/div><\/li>\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.\n<div style=\"text-align: center;\">[latex]\\begin{array}{ll}{\\left(3x-y\\right)}^{4}\\hfill &amp; =\\left(\\begin{array}{c}4\\\\ 0\\end{array}\\right){\\left(3x\\right)}^{4}{\\left(-y\\right)}^{0}+\\left(\\begin{array}{c}4\\\\ 1\\end{array}\\right){\\left(3x\\right)}^{3}{\\left(-y\\right)}^{1}+\\left(\\begin{array}{c}4\\\\ 2\\end{array}\\right){\\left(3x\\right)}^{2}{\\left(-y\\right)}^{2}+\\left(\\begin{array}{c}4\\\\ 3\\end{array}\\right){\\left(3x\\right)}^{1}{\\left(-y\\right)}^{3}+\\left(\\begin{array}{c}4\\\\ 4\\end{array}\\right){\\left(3x\\right)}^{0}{\\left(-y\\right)}^{4}\\hfill \\\\ {\\left(3x-y\\right)}^{4}\\hfill &amp; =81{x}^{4}-108{x}^{3}y+54{x}^{2}{y}^{2}-12x{y}^{3}+{y}^{4}\\hfill \\end{array}[\/latex]<\/div><\/li>\n<\/ol><\/div>\n<div>\n<h3>Analysis of the Solution<\/h3>\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.\n\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It 2<\/h3>\nWrite in expanded form.\n<p style=\"padding-left: 60px;\">a. [latex]{\\left(x-y\\right)}^{5}[\/latex]<\/p>\n<p style=\"padding-left: 60px;\">b. [latex]{\\left(2x+5y\\right)}^{3}[\/latex]<\/p>\n\n<a href=\"https:\/\/courses.candelalearning.com\/precalctwo1xmaster\/chapter\/solutions-33\/\" target=\"_blank\">Solution<\/a>\n<\/div>","rendered":"<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! If we examine some simple binomial expansions, we can find patterns that will lead us to a shortcut for finding more complicated binomial expansions.\n<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}{\\left(x+y\\right)}^{2}={x}^{2}+2xy+{y}^{2}\\hfill \\\\ {\\left(x+y\\right)}^{3}={x}^{3}+3{x}^{2}y+3x{y}^{2}+{y}^{3}\\hfill \\\\ {\\left(x+y\\right)}^{4}={x}^{4}+4{x}^{3}y+6{x}^{2}{y}^{2}+4x{y}^{3}+{y}^{4}\\hfill \\end{array}[\/latex]<\/div>\n<p>First, let\u2019s examine the exponents. With each successive term, the exponent for [latex]x[\/latex] decreases and the exponent for [latex]y[\/latex] increases. The sum of the two exponents is [latex]n[\/latex] for each term.<\/p>\n<p>Next, let\u2019s examine the coefficients. Notice that the coefficients increase and then decrease in a symmetrical pattern. The coefficients follow a pattern:<\/p>\n<div style=\"text-align: center;\">[latex]\\left(\\begin{array}{c}n\\\\ 0\\end{array}\\right),\\left(\\begin{array}{c}n\\\\ 1\\end{array}\\right),\\left(\\begin{array}{c}n\\\\ 2\\end{array}\\right),...,\\left(\\begin{array}{c}n\\\\ n\\end{array}\\right)[\/latex].<\/div>\n<p>These patterns lead us to the <strong>Binomial Theorem<\/strong>, which can be used to expand any binomial.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ll}{\\left(x+y\\right)}^{n}\\hfill & =\\sum _{k=0}^{n}\\left(\\begin{array}{c}n\\\\ k\\end{array}\\right){x}^{n-k}{y}^{k}\\hfill \\\\ \\hfill & ={x}^{n}+\\left(\\begin{array}{c}n\\\\ 1\\end{array}\\right){x}^{n - 1}y+\\left(\\begin{array}{c}n\\\\ 2\\end{array}\\right){x}^{n - 2}{y}^{2}+...+\\left(\\begin{array}{c}n\\\\ n - 1\\end{array}\\right)x{y}^{n - 1}+{y}^{n}\\hfill \\end{array}[\/latex]<\/div>\n<p>Another way to see the coefficients is to examine the expansion of a binomial in general form, [latex]x+y[\/latex], to successive powers 1, 2, 3, and 4.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{l}{\\left(x+y\\right)}^{1}=x+y\\hfill \\\\ {\\left(x+y\\right)}^{2}={x}^{2}+2xy+{y}^{2}\\hfill \\\\ {\\left(x+y\\right)}^{3}={x}^{3}+3{x}^{2}y+3x{y}^{2}+{y}^{3}\\hfill \\\\ {\\left(x+y\\right)}^{4}={x}^{4}+4{x}^{3}y+6{x}^{2}{y}^{2}+4x{y}^{3}+{y}^{4}\\hfill \\end{array}[\/latex]<\/div>\n<p>Can you guess the next expansion for the binomial [latex]{\\left(x+y\\right)}^{5}?[\/latex]<\/p>\n<div style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202615\/CNX_Precalc_Figure_11_06_0022.jpg\" alt=\"Graph of the function f_2.\" width=\"731\" height=\"413\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 1<\/b><\/p>\n<\/div>\n<p>See Figure 1, which illustrates the following:<\/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 0.<\/li>\n<li>The powers on [latex]y[\/latex] begin with 0 and increase to [latex]n[\/latex].<\/li>\n<li>The coefficients are symmetric.<\/li>\n<\/ul>\n<p>To determine the expansion on [latex]{\\left(x+y\\right)}^{5}[\/latex], we see [latex]n=5[\/latex], thus, there will be 5+1 = 6 terms. Each term has a combined degree of 5. In descending order for powers of [latex]x[\/latex], the pattern is as follows:<\/p>\n<ul>\n<li>Introduce [latex]{x}^{5}[\/latex], and then for each successive term reduce the exponent on [latex]x[\/latex] by 1 until [latex]{x}^{0}=1[\/latex] is reached.<\/li>\n<li>Introduce [latex]{y}^{0}=1[\/latex], and then increase the exponent on [latex]y[\/latex] by 1 until [latex]{y}^{5}[\/latex] is reached.\n<div>[latex]{x}^{5},{x}^{4}y,{x}^{3}{y}^{2},{x}^{2}{y}^{3},x{y}^{4},{y}^{5}[\/latex]<\/div>\n<\/li>\n<\/ul>\n<p>The next expansion would be<\/p>\n<div style=\"text-align: center;\">[latex]{\\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}[\/latex].<\/div>\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>, shown in Figure 2.<\/p>\n<div style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202617\/CNX_Precalc_Figure_11_06_0012.jpg\" alt=\"Pascal's Triangle\" width=\"731\" height=\"300\" data-media-type=\"image\/jpg\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 2<\/b><\/p>\n<\/div>\n<p>To generate Pascal\u2019s Triangle, we start by writing a 1. In the row below, row 2, we write two 1\u2019s. In the 3<sup>rd<\/sup> 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.<\/p>\n<p>To see the connection between Pascal\u2019s Triangle and binomial coefficients, let us revisit the expansion of the binomials in general form.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25202618\/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\" data-media-type=\"image\/jpg\" \/><\/p>\n<div class=\"textbox\">\n<h3>A General Note: The Binomial Theorem<\/h3>\n<p>The <strong>Binomial Theorem<\/strong> is a formula that can be used to expand any binomial.<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{array}{ll}{\\left(x+y\\right)}^{n}\\hfill & =\\sum _{k=0}^{n}\\left(\\begin{array}{c}n\\\\ k\\end{array}\\right){x}^{n-k}{y}^{k}\\hfill \\\\ \\hfill & ={x}^{n}+\\left(\\begin{array}{c}n\\\\ 1\\end{array}\\right){x}^{n - 1}y+\\left(\\begin{array}{c}n\\\\ 2\\end{array}\\right){x}^{n - 2}{y}^{2}+...+\\left(\\begin{array}{c}n\\\\ n - 1\\end{array}\\right)x{y}^{n - 1}+{y}^{n}\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a binomial, write it in expanded form.<\/h3>\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<\/div>\n<div class=\"textbox shaded\">\n<h3>Example 2: Expanding a Binomial<\/h3>\n<p>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>\n<div class=\"textbox shaded\">\n<h3>Solution<\/h3>\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;\">[latex]\\begin{array}{ll}{\\left(x+y\\right)}^{5}\\hfill & =\\left(\\begin{array}{c}5\\\\ 0\\end{array}\\right){x}^{5}{y}^{0}+\\left(\\begin{array}{c}5\\\\ 1\\end{array}\\right){x}^{4}{y}^{1}+\\left(\\begin{array}{c}5\\\\ 2\\end{array}\\right){x}^{3}{y}^{2}+\\left(\\begin{array}{c}5\\\\ 3\\end{array}\\right){x}^{2}{y}^{3}+\\left(\\begin{array}{c}5\\\\ 4\\end{array}\\right){x}^{1}{y}^{4}+\\left(\\begin{array}{c}5\\\\ 5\\end{array}\\right){x}^{0}{y}^{5}\\hfill \\\\ {\\left(x+y\\right)}^{5}\\hfill & ={x}^{5}+5{x}^{4}y+10{x}^{3}{y}^{2}+10{x}^{2}{y}^{3}+5x{y}^{4}+{y}^{5}\\hfill \\end{array}[\/latex]<\/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{array}{ll}{\\left(3x-y\\right)}^{4}\\hfill & =\\left(\\begin{array}{c}4\\\\ 0\\end{array}\\right){\\left(3x\\right)}^{4}{\\left(-y\\right)}^{0}+\\left(\\begin{array}{c}4\\\\ 1\\end{array}\\right){\\left(3x\\right)}^{3}{\\left(-y\\right)}^{1}+\\left(\\begin{array}{c}4\\\\ 2\\end{array}\\right){\\left(3x\\right)}^{2}{\\left(-y\\right)}^{2}+\\left(\\begin{array}{c}4\\\\ 3\\end{array}\\right){\\left(3x\\right)}^{1}{\\left(-y\\right)}^{3}+\\left(\\begin{array}{c}4\\\\ 4\\end{array}\\right){\\left(3x\\right)}^{0}{\\left(-y\\right)}^{4}\\hfill \\\\ {\\left(3x-y\\right)}^{4}\\hfill & =81{x}^{4}-108{x}^{3}y+54{x}^{2}{y}^{2}-12x{y}^{3}+{y}^{4}\\hfill \\end{array}[\/latex]<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<div>\n<h3>Analysis of the Solution<\/h3>\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 class=\"bcc-box bcc-success\">\n<h3>Try It 2<\/h3>\n<p>Write in expanded form.<\/p>\n<p style=\"padding-left: 60px;\">a. [latex]{\\left(x-y\\right)}^{5}[\/latex]<\/p>\n<p style=\"padding-left: 60px;\">b. [latex]{\\left(2x+5y\\right)}^{3}[\/latex]<\/p>\n<p><a href=\"https:\/\/courses.candelalearning.com\/precalctwo1xmaster\/chapter\/solutions-33\/\" target=\"_blank\">Solution<\/a>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-2109\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: OpenStax College. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":276,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-2109","chapter","type-chapter","status-publish","hentry"],"part":2103,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2109","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/users\/276"}],"version-history":[{"count":1,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2109\/revisions"}],"predecessor-version":[{"id":2138,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2109\/revisions\/2138"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/2103"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/2109\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/media?parent=2109"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=2109"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/contributor?post=2109"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/wp-json\/wp\/v2\/license?post=2109"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}