{"id":341,"date":"2021-06-04T00:05:46","date_gmt":"2021-06-04T00:05:46","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/chapter\/multiplying-two-binomials-using-the-distributive-property\/"},"modified":"2022-01-05T16:42:50","modified_gmt":"2022-01-05T16:42:50","slug":"multiplying-two-binomials-using-the-distributive-property","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/chapter\/multiplying-two-binomials-using-the-distributive-property\/","title":{"raw":"8.4.3: Multiplying Polynomials","rendered":"8.4.3: Multiplying Polynomials"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Multiply Polynomials<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Key words<\/h3>\r\n<ul>\r\n \t<li><strong>Monomial<\/strong>: a polynomial with one term<\/li>\r\n \t<li><strong>Binomial<\/strong>: a polynomial with two terms<\/li>\r\n \t<li><strong>Trinomial<\/strong>: a polynomial with three terms<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h3>The Distributive Property<\/h3>\r\nIn the last section we multiplied polynomials by <em><strong>monomials<\/strong><\/em>. In this section we will expand our use of the distributive property. Consider the following example.\r\n<table id=\"eip-id1168466233256\" class=\"unnumbered unstyled\" summary=\"The top line shows parentheses x plus 3, times p, with red arrows from the p to the x and to the 3. The next line says, \">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224428\/CNX_BMath_Figure_10_03_049_img-01.png\" alt=\".\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>We distribute the [latex]p[\/latex] to get<\/td>\r\n<td>[latex]x\\color{red}{p}+3\\color{red}{p}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>What if we have [latex]\\left(x+7\\right)[\/latex] instead of [latex]p[\/latex] ?\r\n\r\nThink of the [latex]x+7[\/latex] as the [latex]\\color{red}{p}[\/latex] above.<\/td>\r\n<td><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224430\/CNX_BMath_Figure_10_03_049_img-03.png\" alt=\".\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Distribute [latex]\\left(x+7\\right)[\/latex] .<\/td>\r\n<td><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224432\/CNX_BMath_Figure_10_03_049_img-04.png\" alt=\".\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Distribute again.<\/td>\r\n<td>[latex]{x}^{2}+7x+3x+21[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Combine like terms.<\/td>\r\n<td>[latex]{x}^{2}+10x+21[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNotice that before combining like terms, we had four terms. We multiplied the two terms of the first <em><strong>binomial<\/strong><\/em> by the two terms of the second binomial\u2014four multiplications.\r\n\r\nBe careful to distinguish between a sum and a product.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{cccc}\\hfill \\mathbf{\\text{Sum}}\\hfill &amp; &amp; &amp; \\hfill \\mathbf{\\text{Product}}\\hfill \\\\ \\hfill x+x\\hfill &amp; &amp; &amp; \\hfill x\\cdot x\\hfill \\\\ \\hfill 2x\\hfill &amp; &amp; &amp; \\hfill {x}^{2}\\hfill \\\\ \\hfill \\text{combine like terms}\\hfill &amp; &amp; &amp; \\hfill \\text{add exponents of like bases}\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nMultiply: [latex]\\left(x+6\\right)\\left(x+8\\right)[\/latex]\r\n<h4>Solution<\/h4>\r\n<table id=\"eip-id1168468281692\" class=\"unnumbered unstyled\" summary=\"The top line shows parentheses x plus 6 times parentheses x plus 8. The next line shows parentheses x plus 6 times red parentheses x plus 8, with red arrows from x plus 8 to x and to 6. The next line says, \">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]\\left(x+6\\right)\\left(x+8\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><\/td>\r\n<td><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224433\/CNX_BMath_Figure_10_03_050_img-01.png\" alt=\".\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Distribute [latex]\\left(x+8\\right)[\/latex] .<\/td>\r\n<td>[latex]x\\color{red}{(x+8)}+6\\color{red}{(x+8)}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Distribute again.<\/td>\r\n<td>[latex]{x}^{2}+8x+6x+48[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify.<\/td>\r\n<td>[latex]{x}^{2}+14x+48[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n[ohm_question]146207[\/ohm_question]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nNow we'll see how to multiply binomials where the variable has a coefficient.\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nMultiply: [latex]\\left(2x+9\\right)\\left(3x+4\\right)[\/latex]\r\n<h4>Solution<\/h4>\r\n<table id=\"eip-id1168467332174\" class=\"unnumbered unstyled\" summary=\"The top line shows parentheses 2x plus 9 times parentheses 3x plus 4. The next line says, \">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]\\left(2x+9\\right)\\left(3x+4\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Distribute. [latex]\\left(3x+4\\right)[\/latex]<\/td>\r\n<td>[latex]2x\\color{red}{(3x+4)}+9\\color{red}{(3x+4)}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Distribute again.<\/td>\r\n<td>[latex]6{x}^{2}+8x+27x+36[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify.<\/td>\r\n<td>[latex]6{x}^{2}+35x+36[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n[ohm_question]146208[\/ohm_question]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nIn the previous examples, the binomials were sums. When there are differences, we pay special attention to make sure the signs of the product are correct.\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nMultiply: [latex]\\left(4y+3\\right)\\left(6y - 5\\right)[\/latex]\r\n<h4>Solution<\/h4>\r\n<table id=\"eip-id1168469625351\" class=\"unnumbered unstyled\" summary=\"The top line shows parentheses 4y plus 3 times parentheses 6y minus 5. The next line says, \">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>[latex]\\left(4y+3\\right)\\left(6y - 5\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Distribute.<\/td>\r\n<td>[latex]4y\\color{red}{(6y-5)}+3\\color{red}{(6y-5)}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Distribute again.<\/td>\r\n<td>[latex]24{y}^{2}-20y+18y - 15[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Simplify.<\/td>\r\n<td>[latex]24{y}^{2}-2y - 15[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n[ohm_question]146209[\/ohm_question]\r\n\r\n<\/div>\r\n&nbsp;\r\n<h3>Area Model for Multiplying Binomials<\/h3>\r\nNow let's explore multiplying two binomials visually. For those that use pictures to learn, we can draw an area model to help make sense of the process. We'll use each binomial as one of the dimensions of a rectangle, and their product as the area.\r\n\r\nThe model below shows [latex]\\left(x+4\\right)\\left(x+2\\right)[\/latex]:\r\n\r\n[caption id=\"attachment_2062\" align=\"aligncenter\" width=\"538\"]<img class=\"wp-image-2062\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5676\/2021\/06\/05164011\/Visual-binomial-multiplication-1024x419.png\" alt=\"Rectangles showing area method of multiplying\" width=\"538\" height=\"220\" \/> Figure 1. Area Method[\/caption]\r\n\r\nEach binomial is expanded into variable terms and constants, [latex]x+4[\/latex], along the top of the model and [latex]x+2[\/latex] along the left side. The product of each pair of terms is a colored rectangle. Like terms have the same color. The total area is the sum of all of these small rectangles, [latex]x^{2}+2x+4x+8[\/latex], If we combine all the like terms, we can write the product, or area, as [latex]x^{2}+6x+8[\/latex].\r\n\r\nHere is the same example using the distributive property:\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSimplify. [latex]\\left(x+4\\right)\\left(x+2\\right)[\/latex]\r\n<h4>Solution<\/h4>\r\n<p style=\"text-align: center;\">[latex]\\left(x+4\\right)\\left(x+2\\right)[\/latex]<\/p>\r\nDistribute the [latex]x[\/latex] onto [latex]x+2[\/latex], then distribute 4 onto [latex]x+2[\/latex].\r\n<p style=\"text-align: center;\">[latex]x\\left(x+2\\right)+4\\left(x+2\\right)[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]x\\left(x\\right)+x\\left(2\\right)+4\\left(x\\right)+4\\left(2\\right)[\/latex]<\/p>\r\nMultiply.\r\n<p style=\"text-align: center;\">[latex]x^{2}+2x+4x+8[\/latex]<\/p>\r\nCombine like terms [latex]\\left(2x+4x\\right)[\/latex].\r\n<p style=\"text-align: center;\">[latex]x^{2}+6x+8[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]\\left(x+4\\right)\\left(2x+2\\right)=x^{2}+6x+8[\/latex]\r\n\r\n<\/div>\r\nLook back at the model above to see where each piece of [latex]x^{2}+2x+4x+8[\/latex] comes from. Can you see where you multiply [latex]x[\/latex] by [latex]x + 2[\/latex], and where you get [latex]x^{2}[\/latex]\u00a0from [latex]x\\left(x\\right)[\/latex]?\r\n\r\nWhen multiplying binomials each term in one binomial is multiplied by each term in the other binomial. Look at the example above: the [latex]x[\/latex] in [latex]x+4[\/latex] gets multiplied by both the [latex]x[\/latex] and the [latex]2[\/latex] from [latex]x+2[\/latex], and the [latex]4[\/latex] gets multiplied by both the [latex]x[\/latex] and the [latex]2[\/latex].\r\n\r\nThe following video provides an example of multiplying two binomials using an area model as well as repeated distribution.\r\n\r\nhttps:\/\/youtu.be\/u4Hgl0BrUlo\r\n<h2><\/h2>\r\n<h2>The Table Method<\/h2>\r\nThe table method is just a simplified form of the area model, where we use the rows and columns of a table rather than rectangles.\r\n\r\nWe may see a binomial multiplied by itself written as\u00a0[latex]{\\left(x+3\\right)}^{2}[\/latex] instead of\u00a0[latex]\\left(x+3\\right)\\left(x+3\\right)[\/latex]. \u00a0To find this product, let's use the table method. We will place the terms of each binomial along the top row and first column of a table, like this:\r\n<table style=\"width: 50%;\">\r\n<thead>\r\n<tr>\r\n<td style=\"width: 23.2394%;\"><\/td>\r\n<td style=\"width: 38.2629%;\">[latex]x[\/latex]<\/td>\r\n<td style=\"width: 38.2629%;\">[latex]+3[\/latex]<\/td>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 23.2394%;\">[latex]x[\/latex]<\/td>\r\n<td style=\"width: 38.2629%;\"><\/td>\r\n<td style=\"width: 38.2629%;\"><\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 23.2394%;\">[latex]+3[\/latex]<\/td>\r\n<td style=\"width: 38.2629%;\"><\/td>\r\n<td style=\"width: 38.2629%;\"><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNow multiply the term in each column by the term in each row to get the terms of the resulting polynomial. Note how we keep the signs on the terms, even when they are positive, this will help us write the new polynomial.\r\n<table style=\"width: 50%;\">\r\n<thead>\r\n<tr>\r\n<td style=\"width: 23.2394%;\"><\/td>\r\n<td style=\"width: 38.2629%;\">[latex]x[\/latex]<\/td>\r\n<td style=\"width: 38.2629%;\">[latex]+3[\/latex]<\/td>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 23.2394%;\">[latex]x[\/latex]<\/td>\r\n<td style=\"width: 38.2629%;\">[latex]x\\cdot{x}=x^2[\/latex]<\/td>\r\n<td style=\"width: 38.2629%;\">[latex]3\\cdot{x}=+3x[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 23.2394%;\">[latex]+3[\/latex]<\/td>\r\n<td style=\"width: 38.2629%;\">[latex]x\\cdot{3}=+3x[\/latex]<\/td>\r\n<td style=\"width: 38.2629%;\">\u00a0[latex]3\\cdot{3}=+9[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNow we can write the terms of the polynomial from the entries in the table:\r\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}&amp;\\;\\;\\;\\;\\left(x+3\\right)^{2}\\\\&amp;= x^2 + 3x + 3x + 9 \\\\ &amp;= x^{2}+6x +9\\end{aligned}\\end{equation}[\/latex]<\/p>\r\nDo you see how similar this method is to the area model?\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFind the product.[latex]\\left(3\u2013s\\right)\\left(1-s\\right)[\/latex]\r\n<h4>Solution<\/h4>\r\nNotice how the binomials are not written in standard form. \u00a0There is nothing different in the way we find the product. \u00a0At the end we will reorganize terms so they are in descending order as a matter of convention.\r\n\r\n[latex]\\left(3\u2013s\\right)\\left(1\u2013s\\right)[\/latex]\r\n\r\nUse a table this time.\r\n<table style=\"width: 40%;\">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<th>[latex]3[\/latex]<\/th>\r\n<th>[latex]-s[\/latex]<\/th>\r\n<\/tr>\r\n<tr>\r\n<th>[latex]1[\/latex]<\/th>\r\n<td>[latex]3[\/latex]<\/td>\r\n<td>[latex]-s[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<th>[latex]-s[\/latex]<\/th>\r\n<td>[latex]-3s[\/latex]<\/td>\r\n<td>[latex]s^2[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNotice how the <em>s<\/em> term is now positive. Collect the terms and simplify.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\left(3\u2013s\\right)\\left(1\u2013s\\right)\\\\\\text{ }\\\\=3-3s-s+s^2\\\\\\text{ }\\\\=3-4s+s^2\\end{array}[\/latex]<\/p>\r\nAs a matter of convention, we will organize the terms so the one with greatest degree comes first. Pay close attention to the signs on the terms when you reorganize them. The [latex]3[\/latex] is positive, so we will use a plus in front of it, and the [latex]4[\/latex] is negative so we use a minus in front of it.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\left(3\u2013s\\right)\\left(1\u2013s\\right)\\\\\\text{ }\\\\=s^{2}-4s+3\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]\\left(3\u2013s\\right)\\left(1\u2013s\\right)=s^2-4s+3[\/latex]\r\n\r\n<\/div>\r\n<h2>Multiplying Two Binomials Using the Vertical Method<\/h2>\r\nThe table method and the area model are just different representations of the Distributive Property. We\u00a0can use the Distributive Property to find the product of any two polynomials. The Vertical Method is another way to complete the Distributive Property. It is very much like the method we use to multiply whole numbers. Look carefully at this example of multiplying two-digit numbers.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224522\/CNX_BMath_Figure_10_03_058_img.png\" alt=\"A vertical multiplication problem is shown. 23 times 46 is written with a line underneath. Beneath the line is 138. Beside 138 is written \" \/>\r\nWe start by multiplying [latex]23[\/latex] by [latex]6[\/latex] to get [latex]138[\/latex].\r\n\r\nThen we multiply [latex]23[\/latex] by [latex]4[\/latex], lining up the partial product in the correct columns so that we're really multiplying by 40.\r\n\r\nLast, we add the partial products.\r\n\r\nHorizontally, this would be written, [latex]23\\times 46=23(40+6)=23(40)+23(6)=920+138=10-58[\/latex]. So, multiplying vertically is just another representation of the distributive property.\r\n\r\nNow we'll apply the vertical method to multiply two binomials.\r\n<div class=\"textbox exercises\">\r\n<h3>example<\/h3>\r\nMultiply using the vertical method: [latex]\\left(5x - 1\\right)\\left(2x - 7\\right)[\/latex]\r\n<h4>Solution<\/h4>\r\nIt does not matter which binomial goes on the top, thanks to the commutative property of multiplication.\r\n\r\nWe line up the columns when we multiply:\r\n<table id=\"eip-id1168469481295\" class=\"unnumbered unstyled\" summary=\"A vertical multiplication problem is shown. 2x minus 7 times 5x minus 1 is written with a line underneath. The next line says, \">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224523\/CNX_BMath_Figure_10_03_059_img-01.png\" alt=\".\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Multiply [latex]2x - 7[\/latex] by [latex]-1[\/latex] .<\/td>\r\n<td><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224524\/CNX_BMath_Figure_10_03_059_img-02.png\" alt=\".\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Multiply [latex]2x - 7[\/latex] by [latex]5x[\/latex] .<\/td>\r\n<td><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224526\/CNX_BMath_Figure_10_03_059_img-03.png\" alt=\".\" \/><\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Add like terms.<\/td>\r\n<td><img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224527\/CNX_BMath_Figure_10_03_059_img-04.png\" alt=\".\" \/><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNotice the partial products are the same as the terms in using the Distributive Property.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224528\/CNX_BMath_Figure_10_03_060_img.png\" alt=\"On the left, 5x minus 1 times 2x minus 7 is shown. Below that is 10 x squared minus 35x minus 2x plus 7. The first two terms are in blue, the second two in red. Beneath that is 10 x squared minus 37x plus 7. On the right, a vertical multiplication problem is shown. 2xx minus 7 times 5x minus 1 is written with a line underneath. Beneath the line is a red negative 2x plus 7. Beneath that is 10 x squared minus 35 x in blue. Beneath that, there is another line. Beneath that line is 10 x squared minus 37x plus 7.\" \/>\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\n[ohm_question]146216[\/ohm_question]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nThe <em><strong>Commutative Property of Multiplication<\/strong><\/em> tells us that\u00a0terms can be multiplied in either order. The expression [latex]\\left(x+2\\right)\\left(x+4\\right)[\/latex] has the same product as [latex]\\left(x+4\\right)\\left(x+2\\right)[\/latex], [latex]x^{2}+6x+8[\/latex]. The order in which we multiply binomials does not matter. What matters is that we multiply each term in one binomial by each term in the other binomial.\r\n\r\nLet's look at another example using the Distributive Property.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSimplify [latex]\\left(4x\u201310\\right)\\left(2x+3\\right)[\/latex] using the Distributive Property.\r\n<h4>Solution<\/h4>\r\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}&amp;\\;\\;\\;\\;(4x\u201310)(2x+3)\\\\&amp;=4x(2x)+4x(3)-10(2x)-10(3)\\\\&amp;=8x^2+12x-20x-30\\\\&amp;=8x^2-8x-30\\end{aligned}\\end{equation}[\/latex]<\/p>\r\nBe careful about including the negative sign on the [latex]\u201110[\/latex], since 10 is subtracted.\r\n<h4>Answer<\/h4>\r\n[latex]\\left(4x\u201310\\right)\\left(2x+3\\right)=8x^{2}\u20138x\u201330[\/latex]\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSimplify [latex]\\left(x\u20137\\right)^2[\/latex]\r\n<h4>Solution<\/h4>\r\nWrite the product of the binomial:\r\n<p style=\"text-align: center;\">[latex]{\\left(x-7\\right)}^2=\\left(x\u20137\\right)\\left(x\u20137\\right)[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Let's use the table method, just because. Note how we carry the negative sign with the\u00a0[latex]7[\/latex].<\/p>\r\n\r\n<table style=\"width: 20%;\">\r\n<tbody>\r\n<tr>\r\n<td><\/td>\r\n<td>\u00a0[latex]x[\/latex]<\/td>\r\n<td>\u00a0[latex]-7[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>\u00a0[latex]x[\/latex]<\/td>\r\n<td>\u00a0\u00a0[latex]x^2[\/latex]<\/td>\r\n<td>\u00a0\u00a0[latex]-7x[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>\u00a0\u00a0[latex]-7[\/latex]<\/td>\r\n<td>\u00a0\u00a0[latex]-7x[\/latex]<\/td>\r\n<td>\u00a0\u00a0[latex]49[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nCollect the terms, and simplify. Note how we keep the sign on each term.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}x^2-7x-7x+49\\\\\\text{ }\\\\=x^2-14x+49\\end{array}[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]x^2-14x+49[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n\r\n<img class=\"wp-image-2132 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/03\/22011815\/traffic-sign-160659-300x265.png\" alt=\"Caution\" width=\"51\" height=\"45\" \/>Caution! It is VERY important to remember that we<span style=\"font-size: 1rem; text-align: center;\">\u00a0can't move the exponent into a grouped sum because of the order of operations!!!!!<\/span>\r\n<p style=\"text-align: center;\"><strong>INCORRECT:<\/strong> [latex]\\left(2+x\\right)^{2}\\neq2^{2}+x^{2}[\/latex]<\/p>\r\n<p style=\"text-align: center;\"><strong>\u00a0CORRECT:<\/strong> [latex]\\left(2+x\\right)^{2}=\\left(2+x\\right)\\left(2+x\\right)[\/latex]<\/p>\r\n\r\n<\/div>\r\nThe video that follows shows another example of using a table to multiply two binomials.\r\n\r\nhttps:\/\/youtu.be\/tWsLJ_pn5mQ\r\n\r\n&nbsp;\r\n\r\nThe method you choose to use doesn't matter as the answers will always be the same.\r\n\r\nA common form for products occurs when the binomial is squared.\u00a0 All we have to remember is that squaring a binomial is equivalent to multiplying it by itself.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nSimplify:\u00a0 [latex]\\left(2x+6\\right)^{2}[\/latex]\r\n<h4>Solution<\/h4>\r\n<p style=\"text-align: left;\">We will use the Distributive Property. First write the square as the product of two binomials.<\/p>\r\n[latex]\\begin{equation}\\begin{aligned}&amp;\\;\\;\\;\\;\\left(2x+6\\right)^{2}\\\\&amp;=\\left(2x+6\\right)\\left(2x+6\\right)\\\\&amp;=2x(2x)+2x(6)+6(2x)+6(6)\\\\&amp;=4x^2+12x+12x+36\\\\&amp;=4x^2+12x+36\\end{aligned}\\end{equation}[\/latex]\r\n\r\nNotice that the two [latex]x[\/latex]-terms are identical. This will always be the case when squaring a binomial.\r\n<h4>Answer<\/h4>\r\n[latex](2x+6)^{2}=4x^{2}+24x+36[\/latex]\r\n\r\n<\/div>\r\nThe next example shows another common form the product of binomials can take, where each of the terms in the two binomials is the same, but the signs in the middle are different.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nMultiply the binomials. [latex]\\left(x+8\\right)\\left(x\u20138\\right)[\/latex]\r\n<h4>Solution<\/h4>\r\n[latex]\\begin{equation}\\begin{aligned}&amp;\\;\\;\\;\\;(x+8)(x-8)\\\\&amp;=x(x)+x(-8)+8(x)+8(-8)\\\\&amp;=x^2+8x-8x-64\\\\&amp;=x^2-64\\end{aligned}\\end{equation}[\/latex]\r\n\r\nNotice that\u00a0the [latex]x[\/latex]-term in the answer disappears since the two [latex]x[\/latex]-terms are opposites.\r\n<h4>Answer<\/h4>\r\n[latex]\\left(x+8\\right)\\left(x-8\\right)=x^{2}-64[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Think About It<\/h3>\r\n<p style=\"text-align: left;\">There are predictable outcomes when you square a binomial sum or difference. In general terms, for a binomial difference,<\/p>\r\n<p style=\"text-align: center;\">[latex]\\left(a-b\\right)^{2}=\\left(a-b\\right)\\left(a-b\\right)[\/latex],<\/p>\r\n<p style=\"text-align: left;\">the resulting product, after being simplified, will look like this:<\/p>\r\n<p style=\"text-align: center;\">[latex]a^2-2ab+b^2[\/latex].<\/p>\r\n<p style=\"text-align: left;\">The product of a binomial sum will have the following predictable outcome:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\left(a+b\\right)^{2}=\\left(a+b\\right)\\left(a+b\\right)=a^2+2ab+b^2[\/latex].<\/p>\r\nThe\u00a0product of a binomial sum and binomial difference of the same two monomial will have the following predictable outcome:\r\n<p style=\"text-align: center;\">[latex]\\left(a+b\\right)\\left(a-b\\right)=a^2-b^2[\/latex].<\/p>\r\n<p style=\"text-align: left;\">Note that a and b in these generalizations could be integers, fractions, or variables with any kind of constant. \u00a0You will learn more about predictable patterns from products of binomials in later math classes.<\/p>\r\n\r\n<\/div>\r\nIn this section we showed how to multiply two binomials using the distributive property, an area model, by using a table, and the vertical method.\u00a0 All of these methods are variations of the Distributive Property. Practice each method, and decide which one you prefer.\r\n\r\nThe Distributive Property can be expanded to the product of any two polynomials; each term in the first polynomial must be multiplied into each term in the second polynomial.\r\n<div class=\"textbox examples\">\r\n<h3>Example<\/h3>\r\nFind the product:\u00a0 [latex](3x+4)(2x^2-3x-8)[\/latex]\r\n<h4>Solution<\/h4>\r\nDistributive Property:\r\n\r\n[latex]\\begin{equation}\\begin{aligned}&amp;\\;\\;\\;\\;(\\color{red}{3x}\\color{blue}{+4})(2x^2-3x-8)\\\\&amp;=\\color{red}{3x}(2x^2)+\\color{red}{3x}(-3x)+\\color{red}{3x}(-8)\\color{blue}{+4}(2x^2)\\color{blue}{+4}(-3x)\\color{blue}{+4}(-8)\\\\&amp;=6x^3-9x^2-24x+8x^2-12x-32\\\\&amp;=6x^3-x^2-36x-32\\end{aligned}\\end{equation}[\/latex]\r\n\r\nAnswer\r\n\r\n[latex](3x+4)(2x^2-3x-8)=6x^3-x^2-36x-32[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example<\/h3>\r\nFind the product:\u00a0 [latex](5x^2-x+3)(2x^2+4x-7)[\/latex]\r\n<h4>Solution<\/h4>\r\nDistributive Property:\r\n\r\n[latex]\\begin{equation}\\begin{aligned}&amp;\\;\\;\\;\\;(\\color{red}{5x^2}\\color{blue}{-x}+\\color{green}{3})(2x^2+4x-7)\\\\&amp;=\\color{red}{5x^2}(2x^2)+\\color{red}{5x^2}(4x)+\\color{red}{5x^2}(-7)\\color{blue}{-x}(2x^2)\\color{blue}{-x}(4x)\\color{blue}{-x}(-7)+\\color{green}{3}(2x^2)+\\color{green}{3}(4x)+\\color{green}{3}(-7)\\\\&amp;=10x^4+20x^3-35x^2-2x^3-4x^2+7x+6x^2+12x-21\\\\&amp;=10x^4+20x^3-2x^3-35x^2-4x^2+6x^2+7x+12x-21\\\\&amp;=10x^4+18x^3-33x^2+19x-21\\end{aligned}\\end{equation}[\/latex]\r\n\r\nAnswer\r\n\r\n[latex](5x^2-x+3)(2x^2+4x-7)=10x^4+18x^3-33x^2+19x-21[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\nFind the product:\u00a0 [latex](x+5)(x^2-3x+4)[\/latex]\r\n\r\n[reveal-answer q=\"hjm611\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm611\"][latex]x^3+2x^2-11x+20[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\nFind the product:\u00a0 [latex](2x^2-3x+1)(x^2-4x+3)[\/latex]\r\n\r\n[reveal-answer q=\"hjm931\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm931\"][latex]2x^4-11x^3+19x^2-13x+3[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n<h2>Summary<\/h2>\r\nMultiplication of binomials and polynomials requires an understanding of the distributive property, rules for exponents, and a keen eye for collecting like terms. Whether the polynomials are monomials, binomials, or trinomials, carefully multiply each term in one polynomial by each term in the other polynomial. Be careful to watch the addition and subtraction signs and negative coefficients. A product is written in simplified form if all of its like terms have been combined.","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Multiply Polynomials<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Key words<\/h3>\n<ul>\n<li><strong>Monomial<\/strong>: a polynomial with one term<\/li>\n<li><strong>Binomial<\/strong>: a polynomial with two terms<\/li>\n<li><strong>Trinomial<\/strong>: a polynomial with three terms<\/li>\n<\/ul>\n<\/div>\n<h3>The Distributive Property<\/h3>\n<p>In the last section we multiplied polynomials by <em><strong>monomials<\/strong><\/em>. In this section we will expand our use of the distributive property. Consider the following example.<\/p>\n<table id=\"eip-id1168466233256\" class=\"unnumbered unstyled\" summary=\"The top line shows parentheses x plus 3, times p, with red arrows from the p to the x and to the 3. The next line says,\">\n<tbody>\n<tr>\n<td><\/td>\n<td><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224428\/CNX_BMath_Figure_10_03_049_img-01.png\" alt=\".\" \/><\/td>\n<\/tr>\n<tr>\n<td>We distribute the [latex]p[\/latex] to get<\/td>\n<td>[latex]x\\color{red}{p}+3\\color{red}{p}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>What if we have [latex]\\left(x+7\\right)[\/latex] instead of [latex]p[\/latex] ?<\/p>\n<p>Think of the [latex]x+7[\/latex] as the [latex]\\color{red}{p}[\/latex] above.<\/td>\n<td><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224430\/CNX_BMath_Figure_10_03_049_img-03.png\" alt=\".\" \/><\/td>\n<\/tr>\n<tr>\n<td>Distribute [latex]\\left(x+7\\right)[\/latex] .<\/td>\n<td><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224432\/CNX_BMath_Figure_10_03_049_img-04.png\" alt=\".\" \/><\/td>\n<\/tr>\n<tr>\n<td>Distribute again.<\/td>\n<td>[latex]{x}^{2}+7x+3x+21[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Combine like terms.<\/td>\n<td>[latex]{x}^{2}+10x+21[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Notice that before combining like terms, we had four terms. We multiplied the two terms of the first <em><strong>binomial<\/strong><\/em> by the two terms of the second binomial\u2014four multiplications.<\/p>\n<p>Be careful to distinguish between a sum and a product.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{cccc}\\hfill \\mathbf{\\text{Sum}}\\hfill & & & \\hfill \\mathbf{\\text{Product}}\\hfill \\\\ \\hfill x+x\\hfill & & & \\hfill x\\cdot x\\hfill \\\\ \\hfill 2x\\hfill & & & \\hfill {x}^{2}\\hfill \\\\ \\hfill \\text{combine like terms}\\hfill & & & \\hfill \\text{add exponents of like bases}\\hfill \\end{array}[\/latex]<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Multiply: [latex]\\left(x+6\\right)\\left(x+8\\right)[\/latex]<\/p>\n<h4>Solution<\/h4>\n<table id=\"eip-id1168468281692\" class=\"unnumbered unstyled\" summary=\"The top line shows parentheses x plus 6 times parentheses x plus 8. The next line shows parentheses x plus 6 times red parentheses x plus 8, with red arrows from x plus 8 to x and to 6. The next line says,\">\n<tbody>\n<tr>\n<td><\/td>\n<td>[latex]\\left(x+6\\right)\\left(x+8\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224433\/CNX_BMath_Figure_10_03_050_img-01.png\" alt=\".\" \/><\/td>\n<\/tr>\n<tr>\n<td>Distribute [latex]\\left(x+8\\right)[\/latex] .<\/td>\n<td>[latex]x\\color{red}{(x+8)}+6\\color{red}{(x+8)}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Distribute again.<\/td>\n<td>[latex]{x}^{2}+8x+6x+48[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify.<\/td>\n<td>[latex]{x}^{2}+14x+48[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm146207\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=146207&theme=oea&iframe_resize_id=ohm146207&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>Now we&#8217;ll see how to multiply binomials where the variable has a coefficient.<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Multiply: [latex]\\left(2x+9\\right)\\left(3x+4\\right)[\/latex]<\/p>\n<h4>Solution<\/h4>\n<table id=\"eip-id1168467332174\" class=\"unnumbered unstyled\" summary=\"The top line shows parentheses 2x plus 9 times parentheses 3x plus 4. The next line says,\">\n<tbody>\n<tr>\n<td><\/td>\n<td>[latex]\\left(2x+9\\right)\\left(3x+4\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Distribute. [latex]\\left(3x+4\\right)[\/latex]<\/td>\n<td>[latex]2x\\color{red}{(3x+4)}+9\\color{red}{(3x+4)}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Distribute again.<\/td>\n<td>[latex]6{x}^{2}+8x+27x+36[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify.<\/td>\n<td>[latex]6{x}^{2}+35x+36[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm146208\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=146208&theme=oea&iframe_resize_id=ohm146208&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>In the previous examples, the binomials were sums. When there are differences, we pay special attention to make sure the signs of the product are correct.<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Multiply: [latex]\\left(4y+3\\right)\\left(6y - 5\\right)[\/latex]<\/p>\n<h4>Solution<\/h4>\n<table id=\"eip-id1168469625351\" class=\"unnumbered unstyled\" summary=\"The top line shows parentheses 4y plus 3 times parentheses 6y minus 5. The next line says,\">\n<tbody>\n<tr>\n<td><\/td>\n<td>[latex]\\left(4y+3\\right)\\left(6y - 5\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Distribute.<\/td>\n<td>[latex]4y\\color{red}{(6y-5)}+3\\color{red}{(6y-5)}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Distribute again.<\/td>\n<td>[latex]24{y}^{2}-20y+18y - 15[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>Simplify.<\/td>\n<td>[latex]24{y}^{2}-2y - 15[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm146209\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=146209&theme=oea&iframe_resize_id=ohm146209&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<h3>Area Model for Multiplying Binomials<\/h3>\n<p>Now let&#8217;s explore multiplying two binomials visually. For those that use pictures to learn, we can draw an area model to help make sense of the process. We&#8217;ll use each binomial as one of the dimensions of a rectangle, and their product as the area.<\/p>\n<p>The model below shows [latex]\\left(x+4\\right)\\left(x+2\\right)[\/latex]:<\/p>\n<div id=\"attachment_2062\" style=\"width: 548px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-2062\" class=\"wp-image-2062\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5676\/2021\/06\/05164011\/Visual-binomial-multiplication-1024x419.png\" alt=\"Rectangles showing area method of multiplying\" width=\"538\" height=\"220\" \/><\/p>\n<p id=\"caption-attachment-2062\" class=\"wp-caption-text\">Figure 1. Area Method<\/p>\n<\/div>\n<p>Each binomial is expanded into variable terms and constants, [latex]x+4[\/latex], along the top of the model and [latex]x+2[\/latex] along the left side. The product of each pair of terms is a colored rectangle. Like terms have the same color. The total area is the sum of all of these small rectangles, [latex]x^{2}+2x+4x+8[\/latex], If we combine all the like terms, we can write the product, or area, as [latex]x^{2}+6x+8[\/latex].<\/p>\n<p>Here is the same example using the distributive property:<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Simplify. [latex]\\left(x+4\\right)\\left(x+2\\right)[\/latex]<\/p>\n<h4>Solution<\/h4>\n<p style=\"text-align: center;\">[latex]\\left(x+4\\right)\\left(x+2\\right)[\/latex]<\/p>\n<p>Distribute the [latex]x[\/latex] onto [latex]x+2[\/latex], then distribute 4 onto [latex]x+2[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]x\\left(x+2\\right)+4\\left(x+2\\right)[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]x\\left(x\\right)+x\\left(2\\right)+4\\left(x\\right)+4\\left(2\\right)[\/latex]<\/p>\n<p>Multiply.<\/p>\n<p style=\"text-align: center;\">[latex]x^{2}+2x+4x+8[\/latex]<\/p>\n<p>Combine like terms [latex]\\left(2x+4x\\right)[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]x^{2}+6x+8[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\left(x+4\\right)\\left(2x+2\\right)=x^{2}+6x+8[\/latex]<\/p>\n<\/div>\n<p>Look back at the model above to see where each piece of [latex]x^{2}+2x+4x+8[\/latex] comes from. Can you see where you multiply [latex]x[\/latex] by [latex]x + 2[\/latex], and where you get [latex]x^{2}[\/latex]\u00a0from [latex]x\\left(x\\right)[\/latex]?<\/p>\n<p>When multiplying binomials each term in one binomial is multiplied by each term in the other binomial. Look at the example above: the [latex]x[\/latex] in [latex]x+4[\/latex] gets multiplied by both the [latex]x[\/latex] and the [latex]2[\/latex] from [latex]x+2[\/latex], and the [latex]4[\/latex] gets multiplied by both the [latex]x[\/latex] and the [latex]2[\/latex].<\/p>\n<p>The following video provides an example of multiplying two binomials using an area model as well as repeated distribution.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Multiply Binomials Using An Area Model and Using Repeated Distribution\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/u4Hgl0BrUlo?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2><\/h2>\n<h2>The Table Method<\/h2>\n<p>The table method is just a simplified form of the area model, where we use the rows and columns of a table rather than rectangles.<\/p>\n<p>We may see a binomial multiplied by itself written as\u00a0[latex]{\\left(x+3\\right)}^{2}[\/latex] instead of\u00a0[latex]\\left(x+3\\right)\\left(x+3\\right)[\/latex]. \u00a0To find this product, let&#8217;s use the table method. We will place the terms of each binomial along the top row and first column of a table, like this:<\/p>\n<table style=\"width: 50%;\">\n<thead>\n<tr>\n<td style=\"width: 23.2394%;\"><\/td>\n<td style=\"width: 38.2629%;\">[latex]x[\/latex]<\/td>\n<td style=\"width: 38.2629%;\">[latex]+3[\/latex]<\/td>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td style=\"width: 23.2394%;\">[latex]x[\/latex]<\/td>\n<td style=\"width: 38.2629%;\"><\/td>\n<td style=\"width: 38.2629%;\"><\/td>\n<\/tr>\n<tr>\n<td style=\"width: 23.2394%;\">[latex]+3[\/latex]<\/td>\n<td style=\"width: 38.2629%;\"><\/td>\n<td style=\"width: 38.2629%;\"><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Now multiply the term in each column by the term in each row to get the terms of the resulting polynomial. Note how we keep the signs on the terms, even when they are positive, this will help us write the new polynomial.<\/p>\n<table style=\"width: 50%;\">\n<thead>\n<tr>\n<td style=\"width: 23.2394%;\"><\/td>\n<td style=\"width: 38.2629%;\">[latex]x[\/latex]<\/td>\n<td style=\"width: 38.2629%;\">[latex]+3[\/latex]<\/td>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td style=\"width: 23.2394%;\">[latex]x[\/latex]<\/td>\n<td style=\"width: 38.2629%;\">[latex]x\\cdot{x}=x^2[\/latex]<\/td>\n<td style=\"width: 38.2629%;\">[latex]3\\cdot{x}=+3x[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 23.2394%;\">[latex]+3[\/latex]<\/td>\n<td style=\"width: 38.2629%;\">[latex]x\\cdot{3}=+3x[\/latex]<\/td>\n<td style=\"width: 38.2629%;\">\u00a0[latex]3\\cdot{3}=+9[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Now we can write the terms of the polynomial from the entries in the table:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}&\\;\\;\\;\\;\\left(x+3\\right)^{2}\\\\&= x^2 + 3x + 3x + 9 \\\\ &= x^{2}+6x +9\\end{aligned}\\end{equation}[\/latex]<\/p>\n<p>Do you see how similar this method is to the area model?<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Find the product.[latex]\\left(3\u2013s\\right)\\left(1-s\\right)[\/latex]<\/p>\n<h4>Solution<\/h4>\n<p>Notice how the binomials are not written in standard form. \u00a0There is nothing different in the way we find the product. \u00a0At the end we will reorganize terms so they are in descending order as a matter of convention.<\/p>\n<p>[latex]\\left(3\u2013s\\right)\\left(1\u2013s\\right)[\/latex]<\/p>\n<p>Use a table this time.<\/p>\n<table style=\"width: 40%;\">\n<tbody>\n<tr>\n<td><\/td>\n<th>[latex]3[\/latex]<\/th>\n<th>[latex]-s[\/latex]<\/th>\n<\/tr>\n<tr>\n<th>[latex]1[\/latex]<\/th>\n<td>[latex]3[\/latex]<\/td>\n<td>[latex]-s[\/latex]<\/td>\n<\/tr>\n<tr>\n<th>[latex]-s[\/latex]<\/th>\n<td>[latex]-3s[\/latex]<\/td>\n<td>[latex]s^2[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Notice how the <em>s<\/em> term is now positive. Collect the terms and simplify.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\left(3\u2013s\\right)\\left(1\u2013s\\right)\\\\\\text{ }\\\\=3-3s-s+s^2\\\\\\text{ }\\\\=3-4s+s^2\\end{array}[\/latex]<\/p>\n<p>As a matter of convention, we will organize the terms so the one with greatest degree comes first. Pay close attention to the signs on the terms when you reorganize them. The [latex]3[\/latex] is positive, so we will use a plus in front of it, and the [latex]4[\/latex] is negative so we use a minus in front of it.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}\\left(3\u2013s\\right)\\left(1\u2013s\\right)\\\\\\text{ }\\\\=s^{2}-4s+3\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\left(3\u2013s\\right)\\left(1\u2013s\\right)=s^2-4s+3[\/latex]<\/p>\n<\/div>\n<h2>Multiplying Two Binomials Using the Vertical Method<\/h2>\n<p>The table method and the area model are just different representations of the Distributive Property. We\u00a0can use the Distributive Property to find the product of any two polynomials. The Vertical Method is another way to complete the Distributive Property. It is very much like the method we use to multiply whole numbers. Look carefully at this example of multiplying two-digit numbers.<\/p>\n<p><img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224522\/CNX_BMath_Figure_10_03_058_img.png\" alt=\"A vertical multiplication problem is shown. 23 times 46 is written with a line underneath. Beneath the line is 138. Beside 138 is written\" \/><br \/>\nWe start by multiplying [latex]23[\/latex] by [latex]6[\/latex] to get [latex]138[\/latex].<\/p>\n<p>Then we multiply [latex]23[\/latex] by [latex]4[\/latex], lining up the partial product in the correct columns so that we&#8217;re really multiplying by 40.<\/p>\n<p>Last, we add the partial products.<\/p>\n<p>Horizontally, this would be written, [latex]23\\times 46=23(40+6)=23(40)+23(6)=920+138=10-58[\/latex]. So, multiplying vertically is just another representation of the distributive property.<\/p>\n<p>Now we&#8217;ll apply the vertical method to multiply two binomials.<\/p>\n<div class=\"textbox exercises\">\n<h3>example<\/h3>\n<p>Multiply using the vertical method: [latex]\\left(5x - 1\\right)\\left(2x - 7\\right)[\/latex]<\/p>\n<h4>Solution<\/h4>\n<p>It does not matter which binomial goes on the top, thanks to the commutative property of multiplication.<\/p>\n<p>We line up the columns when we multiply:<\/p>\n<table id=\"eip-id1168469481295\" class=\"unnumbered unstyled\" summary=\"A vertical multiplication problem is shown. 2x minus 7 times 5x minus 1 is written with a line underneath. The next line says,\">\n<tbody>\n<tr>\n<td><\/td>\n<td><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224523\/CNX_BMath_Figure_10_03_059_img-01.png\" alt=\".\" \/><\/td>\n<\/tr>\n<tr>\n<td>Multiply [latex]2x - 7[\/latex] by [latex]-1[\/latex] .<\/td>\n<td><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224524\/CNX_BMath_Figure_10_03_059_img-02.png\" alt=\".\" \/><\/td>\n<\/tr>\n<tr>\n<td>Multiply [latex]2x - 7[\/latex] by [latex]5x[\/latex] .<\/td>\n<td><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224526\/CNX_BMath_Figure_10_03_059_img-03.png\" alt=\".\" \/><\/td>\n<\/tr>\n<tr>\n<td>Add like terms.<\/td>\n<td><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224527\/CNX_BMath_Figure_10_03_059_img-04.png\" alt=\".\" \/><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Notice the partial products are the same as the terms in using the Distributive Property.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/277\/2017\/04\/24224528\/CNX_BMath_Figure_10_03_060_img.png\" alt=\"On the left, 5x minus 1 times 2x minus 7 is shown. Below that is 10 x squared minus 35x minus 2x plus 7. The first two terms are in blue, the second two in red. Beneath that is 10 x squared minus 37x plus 7. On the right, a vertical multiplication problem is shown. 2xx minus 7 times 5x minus 1 is written with a line underneath. Beneath the line is a red negative 2x plus 7. Beneath that is 10 x squared minus 35 x in blue. Beneath that, there is another line. Beneath that line is 10 x squared minus 37x plus 7.\" \/><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm146216\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=146216&theme=oea&iframe_resize_id=ohm146216&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>The <em><strong>Commutative Property of Multiplication<\/strong><\/em> tells us that\u00a0terms can be multiplied in either order. The expression [latex]\\left(x+2\\right)\\left(x+4\\right)[\/latex] has the same product as [latex]\\left(x+4\\right)\\left(x+2\\right)[\/latex], [latex]x^{2}+6x+8[\/latex]. The order in which we multiply binomials does not matter. What matters is that we multiply each term in one binomial by each term in the other binomial.<\/p>\n<p>Let&#8217;s look at another example using the Distributive Property.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Simplify [latex]\\left(4x\u201310\\right)\\left(2x+3\\right)[\/latex] using the Distributive Property.<\/p>\n<h4>Solution<\/h4>\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}&\\;\\;\\;\\;(4x\u201310)(2x+3)\\\\&=4x(2x)+4x(3)-10(2x)-10(3)\\\\&=8x^2+12x-20x-30\\\\&=8x^2-8x-30\\end{aligned}\\end{equation}[\/latex]<\/p>\n<p>Be careful about including the negative sign on the [latex]\u201110[\/latex], since 10 is subtracted.<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\left(4x\u201310\\right)\\left(2x+3\\right)=8x^{2}\u20138x\u201330[\/latex]<\/p>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Simplify [latex]\\left(x\u20137\\right)^2[\/latex]<\/p>\n<h4>Solution<\/h4>\n<p>Write the product of the binomial:<\/p>\n<p style=\"text-align: center;\">[latex]{\\left(x-7\\right)}^2=\\left(x\u20137\\right)\\left(x\u20137\\right)[\/latex]<\/p>\n<p style=\"text-align: left;\">Let&#8217;s use the table method, just because. Note how we carry the negative sign with the\u00a0[latex]7[\/latex].<\/p>\n<table style=\"width: 20%;\">\n<tbody>\n<tr>\n<td><\/td>\n<td>\u00a0[latex]x[\/latex]<\/td>\n<td>\u00a0[latex]-7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>\u00a0[latex]x[\/latex]<\/td>\n<td>\u00a0\u00a0[latex]x^2[\/latex]<\/td>\n<td>\u00a0\u00a0[latex]-7x[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>\u00a0\u00a0[latex]-7[\/latex]<\/td>\n<td>\u00a0\u00a0[latex]-7x[\/latex]<\/td>\n<td>\u00a0\u00a0[latex]49[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Collect the terms, and simplify. Note how we keep the sign on each term.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{c}x^2-7x-7x+49\\\\\\text{ }\\\\=x^2-14x+49\\end{array}[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]x^2-14x+49[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-2132 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1468\/2016\/03\/22011815\/traffic-sign-160659-300x265.png\" alt=\"Caution\" width=\"51\" height=\"45\" \/>Caution! It is VERY important to remember that we<span style=\"font-size: 1rem; text-align: center;\">\u00a0can&#8217;t move the exponent into a grouped sum because of the order of operations!!!!!<\/span><\/p>\n<p style=\"text-align: center;\"><strong>INCORRECT:<\/strong> [latex]\\left(2+x\\right)^{2}\\neq2^{2}+x^{2}[\/latex]<\/p>\n<p style=\"text-align: center;\"><strong>\u00a0CORRECT:<\/strong> [latex]\\left(2+x\\right)^{2}=\\left(2+x\\right)\\left(2+x\\right)[\/latex]<\/p>\n<\/div>\n<p>The video that follows shows another example of using a table to multiply two binomials.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Multiply Binomials Using a Table\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/tWsLJ_pn5mQ?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>&nbsp;<\/p>\n<p>The method you choose to use doesn&#8217;t matter as the answers will always be the same.<\/p>\n<p>A common form for products occurs when the binomial is squared.\u00a0 All we have to remember is that squaring a binomial is equivalent to multiplying it by itself.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Simplify:\u00a0 [latex]\\left(2x+6\\right)^{2}[\/latex]<\/p>\n<h4>Solution<\/h4>\n<p style=\"text-align: left;\">We will use the Distributive Property. First write the square as the product of two binomials.<\/p>\n<p>[latex]\\begin{equation}\\begin{aligned}&\\;\\;\\;\\;\\left(2x+6\\right)^{2}\\\\&=\\left(2x+6\\right)\\left(2x+6\\right)\\\\&=2x(2x)+2x(6)+6(2x)+6(6)\\\\&=4x^2+12x+12x+36\\\\&=4x^2+12x+36\\end{aligned}\\end{equation}[\/latex]<\/p>\n<p>Notice that the two [latex]x[\/latex]-terms are identical. This will always be the case when squaring a binomial.<\/p>\n<h4>Answer<\/h4>\n<p>[latex](2x+6)^{2}=4x^{2}+24x+36[\/latex]<\/p>\n<\/div>\n<p>The next example shows another common form the product of binomials can take, where each of the terms in the two binomials is the same, but the signs in the middle are different.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Multiply the binomials. [latex]\\left(x+8\\right)\\left(x\u20138\\right)[\/latex]<\/p>\n<h4>Solution<\/h4>\n<p>[latex]\\begin{equation}\\begin{aligned}&\\;\\;\\;\\;(x+8)(x-8)\\\\&=x(x)+x(-8)+8(x)+8(-8)\\\\&=x^2+8x-8x-64\\\\&=x^2-64\\end{aligned}\\end{equation}[\/latex]<\/p>\n<p>Notice that\u00a0the [latex]x[\/latex]-term in the answer disappears since the two [latex]x[\/latex]-terms are opposites.<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\left(x+8\\right)\\left(x-8\\right)=x^{2}-64[\/latex]<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Think About It<\/h3>\n<p style=\"text-align: left;\">There are predictable outcomes when you square a binomial sum or difference. In general terms, for a binomial difference,<\/p>\n<p style=\"text-align: center;\">[latex]\\left(a-b\\right)^{2}=\\left(a-b\\right)\\left(a-b\\right)[\/latex],<\/p>\n<p style=\"text-align: left;\">the resulting product, after being simplified, will look like this:<\/p>\n<p style=\"text-align: center;\">[latex]a^2-2ab+b^2[\/latex].<\/p>\n<p style=\"text-align: left;\">The product of a binomial sum will have the following predictable outcome:<\/p>\n<p style=\"text-align: center;\">[latex]\\left(a+b\\right)^{2}=\\left(a+b\\right)\\left(a+b\\right)=a^2+2ab+b^2[\/latex].<\/p>\n<p>The\u00a0product of a binomial sum and binomial difference of the same two monomial will have the following predictable outcome:<\/p>\n<p style=\"text-align: center;\">[latex]\\left(a+b\\right)\\left(a-b\\right)=a^2-b^2[\/latex].<\/p>\n<p style=\"text-align: left;\">Note that a and b in these generalizations could be integers, fractions, or variables with any kind of constant. \u00a0You will learn more about predictable patterns from products of binomials in later math classes.<\/p>\n<\/div>\n<p>In this section we showed how to multiply two binomials using the distributive property, an area model, by using a table, and the vertical method.\u00a0 All of these methods are variations of the Distributive Property. Practice each method, and decide which one you prefer.<\/p>\n<p>The Distributive Property can be expanded to the product of any two polynomials; each term in the first polynomial must be multiplied into each term in the second polynomial.<\/p>\n<div class=\"textbox examples\">\n<h3>Example<\/h3>\n<p>Find the product:\u00a0 [latex](3x+4)(2x^2-3x-8)[\/latex]<\/p>\n<h4>Solution<\/h4>\n<p>Distributive Property:<\/p>\n<p>[latex]\\begin{equation}\\begin{aligned}&\\;\\;\\;\\;(\\color{red}{3x}\\color{blue}{+4})(2x^2-3x-8)\\\\&=\\color{red}{3x}(2x^2)+\\color{red}{3x}(-3x)+\\color{red}{3x}(-8)\\color{blue}{+4}(2x^2)\\color{blue}{+4}(-3x)\\color{blue}{+4}(-8)\\\\&=6x^3-9x^2-24x+8x^2-12x-32\\\\&=6x^3-x^2-36x-32\\end{aligned}\\end{equation}[\/latex]<\/p>\n<p>Answer<\/p>\n<p>[latex](3x+4)(2x^2-3x-8)=6x^3-x^2-36x-32[\/latex]<\/p>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example<\/h3>\n<p>Find the product:\u00a0 [latex](5x^2-x+3)(2x^2+4x-7)[\/latex]<\/p>\n<h4>Solution<\/h4>\n<p>Distributive Property:<\/p>\n<p>[latex]\\begin{equation}\\begin{aligned}&\\;\\;\\;\\;(\\color{red}{5x^2}\\color{blue}{-x}+\\color{green}{3})(2x^2+4x-7)\\\\&=\\color{red}{5x^2}(2x^2)+\\color{red}{5x^2}(4x)+\\color{red}{5x^2}(-7)\\color{blue}{-x}(2x^2)\\color{blue}{-x}(4x)\\color{blue}{-x}(-7)+\\color{green}{3}(2x^2)+\\color{green}{3}(4x)+\\color{green}{3}(-7)\\\\&=10x^4+20x^3-35x^2-2x^3-4x^2+7x+6x^2+12x-21\\\\&=10x^4+20x^3-2x^3-35x^2-4x^2+6x^2+7x+12x-21\\\\&=10x^4+18x^3-33x^2+19x-21\\end{aligned}\\end{equation}[\/latex]<\/p>\n<p>Answer<\/p>\n<p>[latex](5x^2-x+3)(2x^2+4x-7)=10x^4+18x^3-33x^2+19x-21[\/latex]<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p>Find the product:\u00a0 [latex](x+5)(x^2-3x+4)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm611\">Show Answer<\/span><\/p>\n<div id=\"qhjm611\" class=\"hidden-answer\" style=\"display: none\">[latex]x^3+2x^2-11x+20[\/latex]<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p>Find the product:\u00a0 [latex](2x^2-3x+1)(x^2-4x+3)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm931\">Show Answer<\/span><\/p>\n<div id=\"qhjm931\" class=\"hidden-answer\" style=\"display: none\">[latex]2x^4-11x^3+19x^2-13x+3[\/latex]<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<h2>Summary<\/h2>\n<p>Multiplication of binomials and polynomials requires an understanding of the distributive property, rules for exponents, and a keen eye for collecting like terms. Whether the polynomials are monomials, binomials, or trinomials, carefully multiply each term in one polynomial by each term in the other polynomial. Be careful to watch the addition and subtraction signs and negative coefficients. A product is written in simplified form if all of its like terms have been combined.<\/p>\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-341\">\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>Question ID 146210, 146209,  146208, 146207. <strong>Authored 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><li>Trinomial examples; Try It hjm931; hjm611; Figure 1. Area Method. <strong>Authored by<\/strong>: Hazel McKenna. <strong>Provided by<\/strong>: Utah Valley University. <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>Polynomial Multiplication Involving Binomials and Trinomials. <strong>Authored by<\/strong>: James Sousa (mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/www.youtube.com\/watch?v=bBKbldmlbqI&#038;t=7s\">https:\/\/www.youtube.com\/watch?v=bBKbldmlbqI&#038;t=7s<\/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 class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Revised and adapted: Prealgebra. <strong>Provided by<\/strong>: OpenStax. <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\/caa57dab-41c7-455e-bd6f-f443cda5519c@9.757<\/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":23485,"menu_order":7,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Revised and adapted: Prealgebra\",\"author\":\"\",\"organization\":\"OpenStax\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at http:\/\/cnx.org\/contents\/caa57dab-41c7-455e-bd6f-f443cda5519c@9.757\"},{\"type\":\"original\",\"description\":\"Question ID 146210, 146209,  146208, 146207\",\"author\":\"Lumen Learning\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Polynomial Multiplication Involving Binomials and Trinomials\",\"author\":\"James Sousa (mathispower4u.com)\",\"organization\":\"\",\"url\":\"https:\/\/www.youtube.com\/watch?v=bBKbldmlbqI&t=7s\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Trinomial examples; Try It hjm931; hjm611; Figure 1. 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