{"id":461,"date":"2019-07-15T22:45:18","date_gmt":"2019-07-15T22:45:18","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/chapter\/use-the-binomial-theorem\/"},"modified":"2019-07-15T22:45:18","modified_gmt":"2019-07-15T22:45:18","slug":"use-the-binomial-theorem","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ntcc-collegealgebracorequisite\/chapter\/use-the-binomial-theorem\/","title":{"raw":"Use the Binomial Theorem","rendered":"Use the Binomial Theorem"},"content":{"raw":"\n<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n \t<li>Identify binomial coefficients given the formula for a combination.<\/li>\n \t<li>Expand a binomial using the binomial theorem.<\/li>\n \t<li>Use the binomial to find a single term in a binomial.<\/li>\n<\/ul>\n<\/div>\n\n<div class=\"textbox examples\">\n<h3>expanding [latex]\\left(x+y\\right)^n[\/latex]<\/h3>\nEarlier, we learned how to use the distributive property to multiply binomials together. We found that certain binomial products resulted in a nice pattern such as [latex](a+b)^2 = a^2 + 2ab + b^2[\/latex]. But multiplying more than three binomials together was a tedious process.\n\nWe will learn now that there is a pattern we can apply to greater powers of binomial factors.\n\n<\/div>\nPreviously, we studied <strong>combinations<\/strong>. In the shortcut to finding [latex]{\\left(x+y\\right)}^{n}[\/latex], we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. In this case, we use the notation [latex]\\left(\\begin{gathered}n\\\\ r\\end{gathered}\\right)[\/latex] instead of [latex]C\\left(n,r\\right)[\/latex], but it can be calculated in the same way. So\n<p style=\"text-align: center\">[latex]\\left(\\begin{gathered}n\\\\ r\\end{gathered}\\right)=C\\left(n,r\\right)=\\dfrac{n!}{r!\\left(n-r\\right)!}[\/latex]<\/p>\nThe combination [latex]\\left(\\begin{gathered}n\\\\ r\\end{gathered}\\right)[\/latex] is called a <strong>binomial coefficient<\/strong>. An example of a binomial coefficient is [latex]\\left(\\begin{gathered}5\\\\ 2\\end{gathered}\\right)=C\\left(5,2\\right)=10[\/latex].\n<div class=\"textbox\">\n<h3>A General Note: Binomial Coefficients<\/h3>\nIf [latex]n[\/latex] and [latex]r[\/latex] are integers greater than or equal to 0 with [latex]n\\ge r[\/latex], then the <strong>binomial coefficient<\/strong> is\n<p style=\"text-align: center\">[latex]\\left(\\begin{gathered}n\\\\ r\\end{gathered}\\right)=C\\left(n,r\\right)=\\dfrac{n!}{r!\\left(n-r\\right)!}[\/latex]<\/p>\n\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<h4>Is a binomial coefficient always a whole number?<\/h4>\n<em>Yes. Just as the number of combinations must always be a whole number, a binomial coefficient will always be a whole number.<\/em>\n\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding Binomial Coefficients<\/h3>\nFind each binomial coefficient.\n<ol>\n \t<li>[latex]\\left(\\begin{gathered}5\\\\ 3\\end{gathered}\\right)[\/latex]<\/li>\n \t<li>[latex]\\left(\\begin{gathered}9\\\\ 2\\end{gathered}\\right)[\/latex]<\/li>\n \t<li>[latex]\\left(\\begin{gathered}9\\\\ 7\\end{gathered}\\right)[\/latex]<\/li>\n<\/ol>\n[reveal-answer q=\"286266\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"286266\"]\n\nUse the formula to calculate each binomial coefficient. You can also use the [latex]{n}_{}{C}_{r}[\/latex] function on your calculator.\n<p style=\"text-align: center\">[latex]\\left(\\begin{gathered}n\\\\ r\\end{gathered}\\right)=C\\left(n,r\\right)=\\dfrac{n!}{r!\\left(n-r\\right)!}[\/latex]<\/p>\n\n<ol>\n \t<li>[latex]\\left(\\begin{gathered}5\\\\ 3\\end{gathered}\\right)=\\dfrac{5!}{3!\\left(5 - 3\\right)!}=\\dfrac{5\\cdot 4\\cdot 3!}{3!2!}=10[\/latex]<\/li>\n \t<li>[latex]\\left(\\begin{gathered}9\\\\ 2\\end{gathered}\\right)=\\dfrac{9!}{2!\\left(9 - 2\\right)!}=\\dfrac{9\\cdot 8\\cdot 7!}{2!7!}=36[\/latex]<\/li>\n \t<li>[latex]\\left(\\begin{gathered}9\\\\ 7\\end{gathered}\\right)=\\dfrac{9!}{7!\\left(9 - 7\\right)!}=\\dfrac{9\\cdot 8\\cdot 7!}{7!2!}=36[\/latex]<\/li>\n<\/ol>\n<h4>Analysis of the Solution<\/h4>\nNotice that we obtained the same result for parts (b) and (c). If you look closely at the solution for these two parts, you will see that you end up with the same two factorials in the denominator, but the order is reversed, just as with combinations.\n<p style=\"text-align: center\">[latex]\\left(\\begin{gathered}n\\\\ r\\end{gathered}\\right)=\\left(\\begin{gathered}n\\\\ n-r\\end{gathered}\\right)[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\nFind each binomial coefficient.\n<ol style=\"list-style-type: lower-alpha\">\n \t<li>[latex]\\left(\\begin{gathered}7\\\\ 3\\end{gathered}\\right)[\/latex]<\/li>\n \t<li>[latex]\\left(\\begin{gathered}11\\\\ 4\\end{gathered}\\right)[\/latex]<\/li>\n<\/ol>\n[reveal-answer q=\"808348\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"808348\"]\n<ol style=\"list-style-type: lower-alpha\">\n \t<li>35<\/li>\n \t<li>330<\/li>\n<\/ol>\n[\/hidden-answer]\n\n<\/div>\n<div class=\"textbox examples\">\n<h3>tip for success<\/h3>\nThe patterns that emerge from calculating binomial coefficients and that are present in Pascal's Triangle are handy and should be memorized over time as mathematical facts much in the same way that you just \"know\" [latex]4[\/latex] and [latex]3[\/latex] make [latex]7[\/latex]. Of course, that will take a lot of time and patient practice. If you are continuing in mathematics beyond this course, it will be well worth the effort.\n\nStart by working through the next several paragraphs and the examples that follow on paper, perhaps more than once or twice, to begin to get familiar with the patterns and processes. Don't be discouraged if it takes time to become clear. It is a challenging concept.\n\n<\/div>\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! 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 style=\"text-align: center\">[latex]\\begin{align}&amp;{\\left(x+y\\right)}^{2}={x}^{2}+2xy+{y}^{2} \\\\ &amp;{\\left(x+y\\right)}^{3}={x}^{3}+3{x}^{2}y+3x{y}^{2}+{y}^{3} \\\\ &amp;{\\left(x+y\\right)}^{4}={x}^{4}+4{x}^{3}y+6{x}^{2}{y}^{2}+4x{y}^{3}+{y}^{4} \\end{align}[\/latex]<\/p>\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<p style=\"text-align: center\">[latex]\\left(\\begin{gathered}n\\\\ 0\\end{gathered}\\right),\\left(\\begin{gathered}n\\\\ 1\\end{gathered}\\right),\\left(\\begin{gathered}n\\\\ 2\\end{gathered}\\right),\\dots,\\left(\\begin{gathered}n\\\\ n\\end{gathered}\\right)[\/latex].<\/p>\nThese patterns lead us to the <strong>Binomial Theorem<\/strong>, which can be used to expand any binomial.\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}{c}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>\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<p style=\"text-align: center\">[latex]\\begin{align}&amp;{\\left(x+y\\right)}^{1}=x+y \\\\ &amp;{\\left(x+y\\right)}^{2}={x}^{2}+2xy+{y}^{2} \\\\ &amp;{\\left(x+y\\right)}^{3}={x}^{3}+3{x}^{2}y+3x{y}^{2}+{y}^{3} \\\\ &amp;{\\left(x+y\\right)}^{4}={x}^{4}+4{x}^{3}y+6{x}^{2}{y}^{2}+4x{y}^{3}+{y}^{4} \\end{align}[\/latex]<\/p>\nCan you guess the next expansion for the binomial [latex]{\\left(x+y\\right)}^{5}?[\/latex]\n\n<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=\"731\" height=\"413\">\n\nAbove is an illustration the following:\n<ul>\n \t<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>\n \t<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[latex]{x}^{5},{x}^{4}y,{x}^{3}{y}^{2},{x}^{2}{y}^{3},x{y}^{4},{y}^{5}[\/latex]<\/li>\n<\/ul>\nThe next expansion would be\n<p 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].<\/p>\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>.\n\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=\"731\" height=\"300\">\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\/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\">\n<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<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}{c}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\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a binomial, write it in expanded form.<\/h3>\n<ol>\n \t<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>\n<\/div>\n<div class=\"textbox examples\">\n<h3>tip for success<\/h3>\nRemember to try the examples and practice problems before looking at the answers!\n\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Expanding a Binomial<\/h3>\nWrite in expanded form.\n<ol>\n \t<li>[latex]{\\left(x+y\\right)}^{5}[\/latex]<\/li>\n \t<li>[latex]{\\left(3x-y\\right)}^{4}[\/latex]<\/li>\n<\/ol>\n[reveal-answer q=\"607135\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"607135\"]\n<ol>\n \t<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\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]\n\n<\/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{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>\n<\/ol>\n<h4>Analysis of the Solution<\/h4>\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[\/hidden-answer]\n\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\nWrite in expanded form.\n<ol style=\"list-style-type: lower-alpha\">\n \t<li>[latex]{\\left(x-y\\right)}^{5}[\/latex]<\/li>\n \t<li>[latex]{\\left(2x+5y\\right)}^{3}[\/latex]<\/li>\n<\/ol>\n[reveal-answer q=\"459909\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"459909\"]\n<ol style=\"list-style-type: lower-alpha\">\n \t<li>[latex]{x}^{5}-5{x}^{4}y+10{x}^{3}{y}^{2}-10{x}^{2}{y}^{3}+5x{y}^{4}-{y}^{5}[\/latex]<\/li>\n \t<li>[latex]8{x}^{3}+60{x}^{2}y+150x{y}^{2}+125{y}^{3}[\/latex]<\/li>\n<\/ol>\n[\/hidden-answer]\n\n[embed]https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=156714&amp;theme=oea&amp;iframe_resize_id=mom1[\/embed]\n\n<\/div>\n<h2>Using the Binomial Theorem to Find a Single Term<\/h2>\nExpanding a binomial with a high exponent such as [latex]{\\left(x+2y\\right)}^{16}[\/latex] can be a lengthy process.\n\nSometimes we are interested only in a certain term of a binomial expansion. We do not need to fully expand a binomial to find a single specific term.\n\nNote the pattern of coefficients in the expansion of [latex]{\\left(x+y\\right)}^{5}[\/latex].\n<p style=\"text-align: center\">[latex]{\\left(x+y\\right)}^{5}={x}^{5}+\\left(\\begin{gathered}5\\\\ 1\\end{gathered}\\right){x}^{4}y+\\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{y}^{4}+{y}^{5}[\/latex]<\/p>\nThe second term is [latex]\\left(\\begin{gathered}5\\\\ 1\\end{gathered}\\right){x}^{4}y[\/latex]. The third term is [latex]\\left(\\begin{gathered}5\\\\ 2\\end{gathered}\\right){x}^{3}{y}^{2}[\/latex]. We can generalize this result.\n<p style=\"text-align: center\">[latex]\\left(\\begin{gathered}n\\\\ r\\end{gathered}\\right){x}^{n-r}{y}^{r}[\/latex]<\/p>\n\n<div class=\"textbox\">\n<h3>A General Note: The (r+1)th Term of a Binomial Expansion<\/h3>\nThe [latex]\\left(r+1\\right)\\text{th}[\/latex] term of the binomial expansion of [latex]{\\left(x+y\\right)}^{n}[\/latex] is:\n<p style=\"text-align: center\">[latex]\\left(\\begin{gathered}n\\\\ r\\end{gathered}\\right){x}^{n-r}{y}^{r}[\/latex]<\/p>\n\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a binomial, write a specific term without fully expanding.<\/h3>\n<ol>\n \t<li>Determine the value of [latex]n[\/latex] according to the exponent.<\/li>\n \t<li>Determine [latex]\\left(r+1\\right)[\/latex].<\/li>\n \t<li>Determine [latex]r[\/latex].<\/li>\n \t<li>Replace [latex]r[\/latex] in the formula for the [latex]\\left(r+1\\right)\\text{th}[\/latex] term of the binomial expansion.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox examples\">\n<h3>tip for success<\/h3>\nNo peeking! Remember to try the examples and practice problems before looking at the answers.\n\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Writing a Given Term of a Binomial Expansion<\/h3>\nFind the tenth term of [latex]{\\left(x+2y\\right)}^{16}[\/latex] without fully expanding the binomial.\n\n[reveal-answer q=\"495201\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"495201\"]\n\nBecause we are looking for the tenth term, [latex]r+1=10[\/latex], we will use [latex]r=9[\/latex] in our calculations.\n<p style=\"text-align: center\">[latex]\\left(\\begin{gathered}n\\\\ r\\end{gathered}\\right){x}^{n-r}{y}^{r}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\left(\\begin{gathered}16\\\\ 9\\end{gathered}\\right){x}^{16 - 9}{\\left(2y\\right)}^{9}=5\\text{,}857\\text{,}280{x}^{7}{y}^{9}[\/latex]<\/p>\n[\/hidden-answer]\n\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\nFind the sixth term of [latex]{\\left(3x-y\\right)}^{9}[\/latex] without fully expanding the binomial.\n\n[reveal-answer q=\"96932\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"96932\"]\n\n[latex]-10\\text{,}206{x}^{4}{y}^{5}[\/latex]\n\n[\/hidden-answer]\n\n<\/div>\n","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Identify binomial coefficients given the formula for a combination.<\/li>\n<li>Expand a binomial using the binomial theorem.<\/li>\n<li>Use the binomial to find a single term in a binomial.<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox examples\">\n<h3>expanding [latex]\\left(x+y\\right)^n[\/latex]<\/h3>\n<p>Earlier, we learned how to use the distributive property to multiply binomials together. We found that certain binomial products resulted in a nice pattern such as [latex](a+b)^2 = a^2 + 2ab + b^2[\/latex]. But multiplying more than three binomials together was a tedious process.<\/p>\n<p>We will learn now that there is a pattern we can apply to greater powers of binomial factors.<\/p>\n<\/div>\n<p>Previously, we studied <strong>combinations<\/strong>. In the shortcut to finding [latex]{\\left(x+y\\right)}^{n}[\/latex], we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. In this case, we use the notation [latex]\\left(\\begin{gathered}n\\\\ r\\end{gathered}\\right)[\/latex] instead of [latex]C\\left(n,r\\right)[\/latex], but it can be calculated in the same way. So<\/p>\n<p style=\"text-align: center\">[latex]\\left(\\begin{gathered}n\\\\ r\\end{gathered}\\right)=C\\left(n,r\\right)=\\dfrac{n!}{r!\\left(n-r\\right)!}[\/latex]<\/p>\n<p>The combination [latex]\\left(\\begin{gathered}n\\\\ r\\end{gathered}\\right)[\/latex] is called a <strong>binomial coefficient<\/strong>. An example of a binomial coefficient is [latex]\\left(\\begin{gathered}5\\\\ 2\\end{gathered}\\right)=C\\left(5,2\\right)=10[\/latex].<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Binomial Coefficients<\/h3>\n<p>If [latex]n[\/latex] and [latex]r[\/latex] are integers greater than or equal to 0 with [latex]n\\ge r[\/latex], then the <strong>binomial coefficient<\/strong> is<\/p>\n<p style=\"text-align: center\">[latex]\\left(\\begin{gathered}n\\\\ r\\end{gathered}\\right)=C\\left(n,r\\right)=\\dfrac{n!}{r!\\left(n-r\\right)!}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<h4>Is a binomial coefficient always a whole number?<\/h4>\n<p><em>Yes. Just as the number of combinations must always be a whole number, a binomial coefficient will always be a whole number.<\/em><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding Binomial Coefficients<\/h3>\n<p>Find each binomial coefficient.<\/p>\n<ol>\n<li>[latex]\\left(\\begin{gathered}5\\\\ 3\\end{gathered}\\right)[\/latex]<\/li>\n<li>[latex]\\left(\\begin{gathered}9\\\\ 2\\end{gathered}\\right)[\/latex]<\/li>\n<li>[latex]\\left(\\begin{gathered}9\\\\ 7\\end{gathered}\\right)[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q286266\">Show Solution<\/span><\/p>\n<div id=\"q286266\" class=\"hidden-answer\" style=\"display: none\">\n<p>Use the formula to calculate each binomial coefficient. You can also use the [latex]{n}_{}{C}_{r}[\/latex] function on your calculator.<\/p>\n<p style=\"text-align: center\">[latex]\\left(\\begin{gathered}n\\\\ r\\end{gathered}\\right)=C\\left(n,r\\right)=\\dfrac{n!}{r!\\left(n-r\\right)!}[\/latex]<\/p>\n<ol>\n<li>[latex]\\left(\\begin{gathered}5\\\\ 3\\end{gathered}\\right)=\\dfrac{5!}{3!\\left(5 - 3\\right)!}=\\dfrac{5\\cdot 4\\cdot 3!}{3!2!}=10[\/latex]<\/li>\n<li>[latex]\\left(\\begin{gathered}9\\\\ 2\\end{gathered}\\right)=\\dfrac{9!}{2!\\left(9 - 2\\right)!}=\\dfrac{9\\cdot 8\\cdot 7!}{2!7!}=36[\/latex]<\/li>\n<li>[latex]\\left(\\begin{gathered}9\\\\ 7\\end{gathered}\\right)=\\dfrac{9!}{7!\\left(9 - 7\\right)!}=\\dfrac{9\\cdot 8\\cdot 7!}{7!2!}=36[\/latex]<\/li>\n<\/ol>\n<h4>Analysis of the Solution<\/h4>\n<p>Notice that we obtained the same result for parts (b) and (c). If you look closely at the solution for these two parts, you will see that you end up with the same two factorials in the denominator, but the order is reversed, just as with combinations.<\/p>\n<p style=\"text-align: center\">[latex]\\left(\\begin{gathered}n\\\\ r\\end{gathered}\\right)=\\left(\\begin{gathered}n\\\\ n-r\\end{gathered}\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find each binomial coefficient.<\/p>\n<ol style=\"list-style-type: lower-alpha\">\n<li>[latex]\\left(\\begin{gathered}7\\\\ 3\\end{gathered}\\right)[\/latex]<\/li>\n<li>[latex]\\left(\\begin{gathered}11\\\\ 4\\end{gathered}\\right)[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q808348\">Show Solution<\/span><\/p>\n<div id=\"q808348\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha\">\n<li>35<\/li>\n<li>330<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox examples\">\n<h3>tip for success<\/h3>\n<p>The patterns that emerge from calculating binomial coefficients and that are present in Pascal&#8217;s Triangle are handy and should be memorized over time as mathematical facts much in the same way that you just &#8220;know&#8221; [latex]4[\/latex] and [latex]3[\/latex] make [latex]7[\/latex]. Of course, that will take a lot of time and patient practice. If you are continuing in mathematics beyond this course, it will be well worth the effort.<\/p>\n<p>Start by working through the next several paragraphs and the examples that follow on paper, perhaps more than once or twice, to begin to get familiar with the patterns and processes. Don&#8217;t be discouraged if it takes time to become clear. It is a challenging concept.<\/p>\n<\/div>\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! If we 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\">[latex]\\begin{align}&{\\left(x+y\\right)}^{2}={x}^{2}+2xy+{y}^{2} \\\\ &{\\left(x+y\\right)}^{3}={x}^{3}+3{x}^{2}y+3x{y}^{2}+{y}^{3} \\\\ &{\\left(x+y\\right)}^{4}={x}^{4}+4{x}^{3}y+6{x}^{2}{y}^{2}+4x{y}^{3}+{y}^{4} \\end{align}[\/latex]<\/p>\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<p style=\"text-align: center\">[latex]\\left(\\begin{gathered}n\\\\ 0\\end{gathered}\\right),\\left(\\begin{gathered}n\\\\ 1\\end{gathered}\\right),\\left(\\begin{gathered}n\\\\ 2\\end{gathered}\\right),\\dots,\\left(\\begin{gathered}n\\\\ n\\end{gathered}\\right)[\/latex].<\/p>\n<p>These patterns lead us to the <strong>Binomial Theorem<\/strong>, which 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}{c}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<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<p style=\"text-align: center\">[latex]\\begin{align}&{\\left(x+y\\right)}^{1}=x+y \\\\ &{\\left(x+y\\right)}^{2}={x}^{2}+2xy+{y}^{2} \\\\ &{\\left(x+y\\right)}^{3}={x}^{3}+3{x}^{2}y+3x{y}^{2}+{y}^{3} \\\\ &{\\left(x+y\\right)}^{4}={x}^{4}+4{x}^{3}y+6{x}^{2}{y}^{2}+4x{y}^{3}+{y}^{4} \\end{align}[\/latex]<\/p>\n<p>Can you guess the next expansion for the binomial [latex]{\\left(x+y\\right)}^{5}?[\/latex]<\/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\/03234323\/CNX_Precalc_Figure_11_06_0022.jpg\" alt=\"Graph of the function f_2.\" width=\"731\" height=\"413\" \/><\/p>\n<p>Above is an illustration 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.<br \/>\n[latex]{x}^{5},{x}^{4}y,{x}^{3}{y}^{2},{x}^{2}{y}^{3},x{y}^{4},{y}^{5}[\/latex]<\/li>\n<\/ul>\n<p>The next expansion would be<\/p>\n<p 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].<\/p>\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=\"731\" height=\"300\" \/><\/p>\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\/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\" \/><\/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<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}{c}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<\/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 examples\">\n<h3>tip for success<\/h3>\n<p>Remember to try the examples and practice problems before looking at the answers!<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: 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 class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q607135\">Show Solution<\/span><\/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<h4>Analysis of the Solution<\/h4>\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<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Write in expanded form.<\/p>\n<ol style=\"list-style-type: lower-alpha\">\n<li>[latex]{\\left(x-y\\right)}^{5}[\/latex]<\/li>\n<li>[latex]{\\left(2x+5y\\right)}^{3}[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q459909\">Show Solution<\/span><\/p>\n<div id=\"q459909\" class=\"hidden-answer\" style=\"display: none\">\n<ol style=\"list-style-type: lower-alpha\">\n<li>[latex]{x}^{5}-5{x}^{4}y+10{x}^{3}{y}^{2}-10{x}^{2}{y}^{3}+5x{y}^{4}-{y}^{5}[\/latex]<\/li>\n<li>[latex]8{x}^{3}+60{x}^{2}y+150x{y}^{2}+125{y}^{3}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm156714\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=156714&#38;theme=oea&#38;iframe_resize_id=ohm156714&#38;show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Using the Binomial Theorem to Find a Single Term<\/h2>\n<p>Expanding a binomial with a high exponent such as [latex]{\\left(x+2y\\right)}^{16}[\/latex] can be a lengthy process.<\/p>\n<p>Sometimes we are interested only in a certain term of a binomial expansion. We do not need to fully expand a binomial to find a single specific term.<\/p>\n<p>Note the pattern of coefficients in the expansion of [latex]{\\left(x+y\\right)}^{5}[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]{\\left(x+y\\right)}^{5}={x}^{5}+\\left(\\begin{gathered}5\\\\ 1\\end{gathered}\\right){x}^{4}y+\\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{y}^{4}+{y}^{5}[\/latex]<\/p>\n<p>The second term is [latex]\\left(\\begin{gathered}5\\\\ 1\\end{gathered}\\right){x}^{4}y[\/latex]. The third term is [latex]\\left(\\begin{gathered}5\\\\ 2\\end{gathered}\\right){x}^{3}{y}^{2}[\/latex]. We can generalize this result.<\/p>\n<p style=\"text-align: center\">[latex]\\left(\\begin{gathered}n\\\\ r\\end{gathered}\\right){x}^{n-r}{y}^{r}[\/latex]<\/p>\n<div class=\"textbox\">\n<h3>A General Note: The (r+1)th Term of a Binomial Expansion<\/h3>\n<p>The [latex]\\left(r+1\\right)\\text{th}[\/latex] term of the binomial expansion of [latex]{\\left(x+y\\right)}^{n}[\/latex] is:<\/p>\n<p style=\"text-align: center\">[latex]\\left(\\begin{gathered}n\\\\ r\\end{gathered}\\right){x}^{n-r}{y}^{r}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given a binomial, write a specific term without fully expanding.<\/h3>\n<ol>\n<li>Determine the value of [latex]n[\/latex] according to the exponent.<\/li>\n<li>Determine [latex]\\left(r+1\\right)[\/latex].<\/li>\n<li>Determine [latex]r[\/latex].<\/li>\n<li>Replace [latex]r[\/latex] in the formula for the [latex]\\left(r+1\\right)\\text{th}[\/latex] term of the binomial expansion.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox examples\">\n<h3>tip for success<\/h3>\n<p>No peeking! Remember to try the examples and practice problems before looking at the answers.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Writing a Given Term of a Binomial Expansion<\/h3>\n<p>Find the tenth term of [latex]{\\left(x+2y\\right)}^{16}[\/latex] without fully expanding the binomial.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q495201\">Show Solution<\/span><\/p>\n<div id=\"q495201\" class=\"hidden-answer\" style=\"display: none\">\n<p>Because we are looking for the tenth term, [latex]r+1=10[\/latex], we will use [latex]r=9[\/latex] in our calculations.<\/p>\n<p style=\"text-align: center\">[latex]\\left(\\begin{gathered}n\\\\ r\\end{gathered}\\right){x}^{n-r}{y}^{r}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]\\left(\\begin{gathered}16\\\\ 9\\end{gathered}\\right){x}^{16 - 9}{\\left(2y\\right)}^{9}=5\\text{,}857\\text{,}280{x}^{7}{y}^{9}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find the sixth term of [latex]{\\left(3x-y\\right)}^{9}[\/latex] without fully expanding the binomial.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q96932\">Show Solution<\/span><\/p>\n<div id=\"q96932\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]-10\\text{,}206{x}^{4}{y}^{5}[\/latex]<\/p>\n<\/div>\n<\/div>\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-461\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <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 class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/li><li>Question ID 18745. <strong>Authored by<\/strong>: Shahbazian, Roy. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><\/ul><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 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