{"id":2726,"date":"2016-07-19T20:25:57","date_gmt":"2016-07-19T20:25:57","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/?post_type=chapter&#038;p=2726"},"modified":"2017-08-16T02:22:58","modified_gmt":"2017-08-16T02:22:58","slug":"read-special-cases-squares","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/chapter\/read-special-cases-squares\/","title":{"raw":"Factoring Perfect Square Trinomials and Difference of Squares Binomials","rendered":"Factoring Perfect Square Trinomials and Difference of Squares Binomials"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Special Cases - Squares\r\n<ul>\r\n \t<li>Factor a polynomial of the form:\u00a0[latex]{a}^{2}+2ab+{b}^{2}[\/latex]<\/li>\r\n \t<li>Factor a polynomial of the form:\u00a0[latex]{a}^{2}-{b}^{2}[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2><span style=\"color: #308eb0\">Why learn how to factor special cases?<\/span><\/h2>\r\n<img class=\"size-medium wp-image-2786 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/19191115\/Screen-Shot-2016-07-19-at-12.10.30-PM-300x153.png\" alt=\"Repeated pattern of interlocking plus signs, each row a different color following the rainbow spectrum.\" width=\"300\" height=\"153\" \/>\r\n\r\n&nbsp;\r\n\r\nSome people like to find patterns in the world around them, like a game. \u00a0There are some polynomials that, when factored, follow a specific pattern.\r\n\r\n&nbsp;\r\n<div class=\"textbox shaded\">\r\n<p style=\"text-align: left\"><span style=\"color: #7c5887\"><strong>Perfect square trinomials of the form<\/strong>:<\/span>\u00a0[latex]{a}^{2}+2ab+{b}^{2}[\/latex]<\/p>\r\n<p style=\"text-align: center\"><span style=\"color: #7c5887\"><strong>A difference of squares:<\/strong><\/span>\u00a0[latex]{a}^{2}-{b}^{2}[\/latex]<\/p>\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nIn this lesson you will see\u00a0you can factor each of these types of polynomials following a specific pattern.\r\n\r\nSome people find it helpful to know when they can take a shortcut to avoid doing extra work. \u00a0There are some polynomials that will always factor a certain way, and for those we offer a shortcut. \u00a0Most people find it helpful to memorize the factored form of a perfect square trinomial or a difference of squares. \u00a0The most important skill you will use in this section will be recognizing when you can use the shortcuts.\r\n<h2>Factoring a Perfect Square Trinomial<\/h2>\r\nA perfect square trinomial is a trinomial that can be written as the square of a binomial. Recall that when a binomial is squared, the result is the square of the first term added to twice the product of the two terms and the square of the last term.\r\n<div style=\"text-align: center\">[latex]\\begin{array}{ccc}\\hfill {a}^{2}+2ab+{b}^{2}&amp; =&amp; {\\left(a+b\\right)}^{2}\\hfill \\\\ &amp; \\text{and}&amp; \\\\ \\hfill {a}^{2}-2ab+{b}^{2}&amp; =&amp; {\\left(a-b\\right)}^{2}\\hfill \\end{array}[\/latex]<\/div>\r\nWe can use this equation to factor any perfect square trinomial.\r\n<div class=\"textbox shaded\">\r\n<h3>A General Note: Perfect Square Trinomials<\/h3>\r\nA perfect square trinomial can be written as the square of a binomial:\r\n<div>[latex]{a}^{2}+2ab+{b}^{2}={\\left(a+b\\right)}^{2}[\/latex]<\/div>\r\n<\/div>\r\n&nbsp;\r\n<div style=\"text-align: center\"><\/div>\r\nIn the following example we will show you how to define a, and b so you can use the shortcut.\r\n<div class=\"textbox exercises\">\r\n<h3>Exercises<\/h3>\r\nFactor [latex]25{x}^{2}+20x+4[\/latex].\r\n[reveal-answer q=\"119279\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"119279\"]\r\n\r\nFirst, notice that [latex]25{x}^{2}[\/latex] and [latex]4[\/latex] are perfect squares because [latex]25{x}^{2}={\\left(5x\\right)}^{2}[\/latex] and [latex]4={2}^{2}[\/latex].\r\n\r\nThis means that [latex]a=5x\\text{ and }b=2[\/latex]\r\n\r\nNext, check to see if the middle term is equal to [latex]2ab[\/latex], which it is:\r\n<p style=\"text-align: center\">[latex]2ab = 2\\left(5x\\right)\\left(2\\right)=20x[\/latex].<\/p>\r\n&nbsp;\r\n\r\nTherefore, the trinomial is a perfect square trinomial and can be written as [latex]{\\left(a+b\\right)}^{2}={\\left(5x+2\\right)}^{2}[\/latex].\r\n<h4>Answer<\/h4>\r\n[latex]25{x}^{2}+20x+4={\\left(5x+2\\right)}^{2}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the next example, we will show that we can use [latex]1 = 1^2[\/latex] to factor a polynomial with a term equal to 1.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor [latex]49{x}^{2}-14x+1[\/latex].\r\n[reveal-answer q=\"865849\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"865849\"]\r\n\r\nFirst, notice that [latex]49{x}^{2}[\/latex] and [latex]1[\/latex] are perfect squares because [latex]49{x}^{2}={\\left(7x\\right)}^{2}[\/latex] and [latex]1={1}^{2}[\/latex].\r\n\r\nThis means that [latex]a=7x[\/latex], we could say that [latex]b=1[\/latex], but would that give a middle term of [latex]-14x[\/latex]? We will need to choose [latex]b = -1[\/latex] to get the results we want:\r\n<p style=\"text-align: center\">[latex]2ab = 2\\left(7x\\right)\\left(-1\\right)=-14x[\/latex].<\/p>\r\nTherefore, the trinomial is a perfect square trinomial and can be written as [latex]{\\left(a-b\\right)}^{2}={\\left(7x-1\\right)}^{2}[\/latex].\r\n<h4>Answer<\/h4>\r\n[latex]49{x}^{2}-14x+1={\\left(7x-1\\right)}^{2}[\/latex]\r\n\r\n&nbsp;\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video we provide another short description of what a perfect square trinomial is, and show how to factor them using a the formula.\r\n\r\nhttps:\/\/youtu.be\/UMCVGDTxxTI\r\n\r\nWe can summarize our process in the following way:\r\n<div class=\"textbox shaded\">\r\n<h3>Given a perfect square trinomial, factor it into the square of a binomial.<strong>\r\n<\/strong><\/h3>\r\n<ol>\r\n \t<li>Confirm that the first and last term are perfect squares.<\/li>\r\n \t<li>Confirm that the middle term is twice the product of [latex]ab[\/latex].<\/li>\r\n \t<li>Write the factored form as [latex]{\\left(a+b\\right)}^{2}[\/latex], or[latex]{\\left(a-b\\right)}^{2}[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n&nbsp;\r\n<h2>Factoring a Difference of Squares<\/h2>\r\nA difference of squares is a perfect square subtracted from a perfect square. Recall that a difference of squares can be rewritten as factors containing the same terms but opposite signs because the middle terms cancel each other out when the two factors are multiplied.\r\n<div style=\"text-align: center\">[latex]{a}^{2}-{b}^{2}=\\left(a+b\\right)\\left(a-b\\right)[\/latex]<\/div>\r\nWe can use this equation to factor any differences of squares.\r\n<div class=\"textbox shaded\">\r\n<h3>A General Note: Differences of Squares<\/h3>\r\nA difference of squares can be rewritten as two factors containing the same terms but opposite signs.\r\n<div>[latex]{a}^{2}-{b}^{2}=\\left(a+b\\right)\\left(a-b\\right)[\/latex]<\/div>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor [latex]9{x}^{2}-25[\/latex].\r\n[reveal-answer q=\"960938\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"960938\"]\r\n\r\nNotice that [latex]9{x}^{2}[\/latex] and [latex]25[\/latex] are perfect squares because [latex]9{x}^{2}={\\left(3x\\right)}^{2}[\/latex] and [latex]25={5}^{2}[\/latex].\r\n\r\nThis means that [latex]a=3x,\\text{ and }b=5[\/latex]\r\n\r\nThe polynomial represents a difference of squares and can be rewritten as [latex]\\left(3x+5\\right)\\left(3x - 5\\right)[\/latex].\r\n\r\nCheck that you are correct by multiplying.\r\n\r\n[latex]\\left(3x+5\\right)\\left(3x - 5\\right)=9x^2-15x+15x-25=9x^2-25[\/latex]\r\n<h4>Answer<\/h4>\r\n[latex]9{x}^{2}-25=\\left(3x+5\\right)\\left(3x - 5\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe most helpful\u00a0thing for recognizing a difference of squares that can be factored with the shortcut is knowing which numbers are perfect squares, as you will see in the next example.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor [latex]81{y}^{2}-144[\/latex].\r\n[reveal-answer q=\"193159\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"193159\"]\r\n\r\nNotice that [latex]81{y}^{2}[\/latex] and [latex]144[\/latex] are perfect squares because [latex]81{y}^{2}={\\left(9x\\right)}^{2}[\/latex] and [latex]144={12}^{2}[\/latex].\r\n\r\nThis means that [latex]a=9x,\\text{ and }b=12[\/latex]\r\n\r\nThe polynomial represents a difference of squares and can be rewritten as [latex]\\left(9x+12\\right)\\left(9x - 12\\right)[\/latex].\r\n\r\nCheck that you are correct by multiplying.\r\n\r\n[latex]\\left(9x+12\\right)\\left(9x - 12\\right)=81x^2-108x+108x-144=81x^2-144[\/latex]\r\n<h4>Answer<\/h4>\r\n[latex]81{y}^{2}-144=\\left(9x+12\\right)\\left(9x - 12\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\nIn the following video we show another example of how to use the formula for fact a difference of squares.\r\n\r\nhttps:\/\/youtu.be\/Li9IBp5HrFA\r\n\r\nWe can summarize the process for factoring a difference of squares with the shortcut this way:\r\n<div class=\"textbox shaded\">\r\n<h3>How To: Given a difference of squares, factor it into binomials.<\/h3>\r\n<ol>\r\n \t<li>Confirm that the first and last term are perfect squares.<\/li>\r\n \t<li>Write the factored form as [latex]\\left(a+b\\right)\\left(a-b\\right)[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Think About It<\/h3>\r\nIs there a formula to factor the sum of squares, [latex]a^2+b^2[\/latex], into a product of two binomials?\r\n\r\nWrite down some ideas for how you would answer this in the box below before you look at the answer.\r\n\r\n[practice-area rows=\"1\"][\/practice-area]\r\n[reveal-answer q=\"121734\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"121734\"]\r\n\r\nThere is no way to factor a sum of squares into a product of two binomials, this is because of addition - the middle term needs to \"disappear\" and the only way to do that is with opposite signs. \u00a0to get a positive result, you must multiply two numbers with the same signs.\r\n\r\nThe only time a sum of squares can be factored is if they share any common factors, as in the following case:\r\n\r\n[latex]9x^2+36[\/latex]\r\n\r\n[latex]9x^2={(3x)}^2, \\text{ and }36 = 6^2[\/latex]\r\n\r\nThe only way to factor this expression is by pulling out the GCF which is 9.\r\n\r\n[latex]9x^2+36=9(x^2+4)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n<h2>Factor Completely<\/h2>\r\nSometimes you may encounter a polynomial that takes an extra step to factor. In our next example we will first find the GCF of a trinomial, and after factoring it out we will be able to factor again so that we end up with a product of a monomial, and two binomials.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFactor completely [latex]6m^2k-3mk-3k[\/latex].\r\n[reveal-answer q=\"698742\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"698742\"]Whenever you factor, first try the easy route and ask yourself if there is a GCF. In this case, there is one, and it is 3k.\r\n\r\nFactor 3k from the trinomial:\r\n\r\n[latex]6m^2k-3mk-3k=3k\\left(2m^2-m-1\\right)[\/latex]\r\n\r\nWe are left with a trinomial that can be factored using your choice of factoring method. We will create a table to find the factors of [latex]2\\cdot{-1}=-2[\/latex] that sum to [latex]-1[\/latex]\r\n<table class=\" aligncenter\" style=\"width: 20%\" summary=\"A table with five rows and two columns. The first row has columns labeled: Factors of 2 times 3=6 and Sum of Factors. The entries in the first column are: 1, 6; -1, -6; 2,3; and -2,-3. The entries in the second column are: 7, -7, 5, and -5.\">\r\n<thead>\r\n<tr>\r\n<th>Factors of [latex]2\\cdot-1=-2[\/latex]<\/th>\r\n<th>Sum of Factors<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]2,-1[\/latex]<\/td>\r\n<td>[latex]1[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-2,1[\/latex]<\/td>\r\n<td>[latex]-1[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nOur factors are [latex]-2,1[\/latex], so we can factor by grouping:\r\n\r\nRewrite the middle term with the\u00a0factors we found with the table:\r\n<p style=\"text-align: center\">[latex]\\left(2m^2-m-1\\right)=2m^2-2m+m-1[\/latex]<\/p>\r\nRegroup and find the GCF of each group:\r\n<p style=\"text-align: center\">[latex](2m^2-2m)+(m-1)=2m(m-1)+1(m-1)[\/latex]<\/p>\r\nNow\u00a0factor [latex](m-1)[\/latex] from each term:\r\n<p style=\"text-align: center\">[latex]2m^2-m-1=(m-1)(2m+1)[\/latex]<\/p>\r\n<p style=\"text-align: left\">Don't forget the original GCF that we factored out! Our final factored form is:<\/p>\r\n<p style=\"text-align: center\">[latex]6m^2k-3mk-3k=3k (m-1)(2m+1)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn our last example we show that it is important to factor out a GCF if there is one before you being using the techniques shown in this module.\r\n\r\nhttps:\/\/youtu.be\/hMAImz2BuPc\r\n<h2>Summary<\/h2>\r\nIn this section we used factoring with special cases. The last topic we covered was what it means to factor completely.","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Special Cases &#8211; Squares\n<ul>\n<li>Factor a polynomial of the form:\u00a0[latex]{a}^{2}+2ab+{b}^{2}[\/latex]<\/li>\n<li>Factor a polynomial of the form:\u00a0[latex]{a}^{2}-{b}^{2}[\/latex]<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<h2><span style=\"color: #308eb0\">Why learn how to factor special cases?<\/span><\/h2>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-2786 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/121\/2016\/07\/19191115\/Screen-Shot-2016-07-19-at-12.10.30-PM-300x153.png\" alt=\"Repeated pattern of interlocking plus signs, each row a different color following the rainbow spectrum.\" width=\"300\" height=\"153\" \/><\/p>\n<p>&nbsp;<\/p>\n<p>Some people like to find patterns in the world around them, like a game. \u00a0There are some polynomials that, when factored, follow a specific pattern.<\/p>\n<p>&nbsp;<\/p>\n<div class=\"textbox shaded\">\n<p style=\"text-align: left\"><span style=\"color: #7c5887\"><strong>Perfect square trinomials of the form<\/strong>:<\/span>\u00a0[latex]{a}^{2}+2ab+{b}^{2}[\/latex]<\/p>\n<p style=\"text-align: center\"><span style=\"color: #7c5887\"><strong>A difference of squares:<\/strong><\/span>\u00a0[latex]{a}^{2}-{b}^{2}[\/latex]<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>In this lesson you will see\u00a0you can factor each of these types of polynomials following a specific pattern.<\/p>\n<p>Some people find it helpful to know when they can take a shortcut to avoid doing extra work. \u00a0There are some polynomials that will always factor a certain way, and for those we offer a shortcut. \u00a0Most people find it helpful to memorize the factored form of a perfect square trinomial or a difference of squares. \u00a0The most important skill you will use in this section will be recognizing when you can use the shortcuts.<\/p>\n<h2>Factoring a Perfect Square Trinomial<\/h2>\n<p>A perfect square trinomial is a trinomial that can be written as the square of a binomial. Recall that when a binomial is squared, the result is the square of the first term added to twice the product of the two terms and the square of the last term.<\/p>\n<div style=\"text-align: center\">[latex]\\begin{array}{ccc}\\hfill {a}^{2}+2ab+{b}^{2}& =& {\\left(a+b\\right)}^{2}\\hfill \\\\ & \\text{and}& \\\\ \\hfill {a}^{2}-2ab+{b}^{2}& =& {\\left(a-b\\right)}^{2}\\hfill \\end{array}[\/latex]<\/div>\n<p>We can use this equation to factor any perfect square trinomial.<\/p>\n<div class=\"textbox shaded\">\n<h3>A General Note: Perfect Square Trinomials<\/h3>\n<p>A perfect square trinomial can be written as the square of a binomial:<\/p>\n<div>[latex]{a}^{2}+2ab+{b}^{2}={\\left(a+b\\right)}^{2}[\/latex]<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<div style=\"text-align: center\"><\/div>\n<p>In the following example we will show you how to define a, and b so you can use the shortcut.<\/p>\n<div class=\"textbox exercises\">\n<h3>Exercises<\/h3>\n<p>Factor [latex]25{x}^{2}+20x+4[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q119279\">Show Answer<\/span><\/p>\n<div id=\"q119279\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, notice that [latex]25{x}^{2}[\/latex] and [latex]4[\/latex] are perfect squares because [latex]25{x}^{2}={\\left(5x\\right)}^{2}[\/latex] and [latex]4={2}^{2}[\/latex].<\/p>\n<p>This means that [latex]a=5x\\text{ and }b=2[\/latex]<\/p>\n<p>Next, check to see if the middle term is equal to [latex]2ab[\/latex], which it is:<\/p>\n<p style=\"text-align: center\">[latex]2ab = 2\\left(5x\\right)\\left(2\\right)=20x[\/latex].<\/p>\n<p>&nbsp;<\/p>\n<p>Therefore, the trinomial is a perfect square trinomial and can be written as [latex]{\\left(a+b\\right)}^{2}={\\left(5x+2\\right)}^{2}[\/latex].<\/p>\n<h4>Answer<\/h4>\n<p>[latex]25{x}^{2}+20x+4={\\left(5x+2\\right)}^{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the next example, we will show that we can use [latex]1 = 1^2[\/latex] to factor a polynomial with a term equal to 1.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor [latex]49{x}^{2}-14x+1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q865849\">Show Answer<\/span><\/p>\n<div id=\"q865849\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, notice that [latex]49{x}^{2}[\/latex] and [latex]1[\/latex] are perfect squares because [latex]49{x}^{2}={\\left(7x\\right)}^{2}[\/latex] and [latex]1={1}^{2}[\/latex].<\/p>\n<p>This means that [latex]a=7x[\/latex], we could say that [latex]b=1[\/latex], but would that give a middle term of [latex]-14x[\/latex]? We will need to choose [latex]b = -1[\/latex] to get the results we want:<\/p>\n<p style=\"text-align: center\">[latex]2ab = 2\\left(7x\\right)\\left(-1\\right)=-14x[\/latex].<\/p>\n<p>Therefore, the trinomial is a perfect square trinomial and can be written as [latex]{\\left(a-b\\right)}^{2}={\\left(7x-1\\right)}^{2}[\/latex].<\/p>\n<h4>Answer<\/h4>\n<p>[latex]49{x}^{2}-14x+1={\\left(7x-1\\right)}^{2}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video we provide another short description of what a perfect square trinomial is, and show how to factor them using a the formula.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Factor Perfect Square Trinomials Using a Formula\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/UMCVGDTxxTI?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>We can summarize our process in the following way:<\/p>\n<div class=\"textbox shaded\">\n<h3>Given a perfect square trinomial, factor it into the square of a binomial.<strong><br \/>\n<\/strong><\/h3>\n<ol>\n<li>Confirm that the first and last term are perfect squares.<\/li>\n<li>Confirm that the middle term is twice the product of [latex]ab[\/latex].<\/li>\n<li>Write the factored form as [latex]{\\left(a+b\\right)}^{2}[\/latex], or[latex]{\\left(a-b\\right)}^{2}[\/latex].<\/li>\n<\/ol>\n<\/div>\n<p>&nbsp;<\/p>\n<h2>Factoring a Difference of Squares<\/h2>\n<p>A difference of squares is a perfect square subtracted from a perfect square. Recall that a difference of squares can be rewritten as factors containing the same terms but opposite signs because the middle terms cancel each other out when the two factors are multiplied.<\/p>\n<div style=\"text-align: center\">[latex]{a}^{2}-{b}^{2}=\\left(a+b\\right)\\left(a-b\\right)[\/latex]<\/div>\n<p>We can use this equation to factor any differences of squares.<\/p>\n<div class=\"textbox shaded\">\n<h3>A General Note: Differences of Squares<\/h3>\n<p>A difference of squares can be rewritten as two factors containing the same terms but opposite signs.<\/p>\n<div>[latex]{a}^{2}-{b}^{2}=\\left(a+b\\right)\\left(a-b\\right)[\/latex]<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor [latex]9{x}^{2}-25[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q960938\">Show Answer<\/span><\/p>\n<div id=\"q960938\" class=\"hidden-answer\" style=\"display: none\">\n<p>Notice that [latex]9{x}^{2}[\/latex] and [latex]25[\/latex] are perfect squares because [latex]9{x}^{2}={\\left(3x\\right)}^{2}[\/latex] and [latex]25={5}^{2}[\/latex].<\/p>\n<p>This means that [latex]a=3x,\\text{ and }b=5[\/latex]<\/p>\n<p>The polynomial represents a difference of squares and can be rewritten as [latex]\\left(3x+5\\right)\\left(3x - 5\\right)[\/latex].<\/p>\n<p>Check that you are correct by multiplying.<\/p>\n<p>[latex]\\left(3x+5\\right)\\left(3x - 5\\right)=9x^2-15x+15x-25=9x^2-25[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]9{x}^{2}-25=\\left(3x+5\\right)\\left(3x - 5\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The most helpful\u00a0thing for recognizing a difference of squares that can be factored with the shortcut is knowing which numbers are perfect squares, as you will see in the next example.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor [latex]81{y}^{2}-144[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q193159\">Show Answer<\/span><\/p>\n<div id=\"q193159\" class=\"hidden-answer\" style=\"display: none\">\n<p>Notice that [latex]81{y}^{2}[\/latex] and [latex]144[\/latex] are perfect squares because [latex]81{y}^{2}={\\left(9x\\right)}^{2}[\/latex] and [latex]144={12}^{2}[\/latex].<\/p>\n<p>This means that [latex]a=9x,\\text{ and }b=12[\/latex]<\/p>\n<p>The polynomial represents a difference of squares and can be rewritten as [latex]\\left(9x+12\\right)\\left(9x - 12\\right)[\/latex].<\/p>\n<p>Check that you are correct by multiplying.<\/p>\n<p>[latex]\\left(9x+12\\right)\\left(9x - 12\\right)=81x^2-108x+108x-144=81x^2-144[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]81{y}^{2}-144=\\left(9x+12\\right)\\left(9x - 12\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<p>In the following video we show another example of how to use the formula for fact a difference of squares.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex:  Factor a Difference of Squares\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/Li9IBp5HrFA?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>We can summarize the process for factoring a difference of squares with the shortcut this way:<\/p>\n<div class=\"textbox shaded\">\n<h3>How To: Given a difference of squares, factor it into binomials.<\/h3>\n<ol>\n<li>Confirm that the first and last term are perfect squares.<\/li>\n<li>Write the factored form as [latex]\\left(a+b\\right)\\left(a-b\\right)[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Think About It<\/h3>\n<p>Is there a formula to factor the sum of squares, [latex]a^2+b^2[\/latex], into a product of two binomials?<\/p>\n<p>Write down some ideas for how you would answer this in the box below before you look at the answer.<\/p>\n<p><textarea aria-label=\"Your Answer\" rows=\"1\"><\/textarea><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q121734\">Show Answer<\/span><\/p>\n<div id=\"q121734\" class=\"hidden-answer\" style=\"display: none\">\n<p>There is no way to factor a sum of squares into a product of two binomials, this is because of addition &#8211; the middle term needs to &#8220;disappear&#8221; and the only way to do that is with opposite signs. \u00a0to get a positive result, you must multiply two numbers with the same signs.<\/p>\n<p>The only time a sum of squares can be factored is if they share any common factors, as in the following case:<\/p>\n<p>[latex]9x^2+36[\/latex]<\/p>\n<p>[latex]9x^2={(3x)}^2, \\text{ and }36 = 6^2[\/latex]<\/p>\n<p>The only way to factor this expression is by pulling out the GCF which is 9.<\/p>\n<p>[latex]9x^2+36=9(x^2+4)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<h2>Factor Completely<\/h2>\n<p>Sometimes you may encounter a polynomial that takes an extra step to factor. In our next example we will first find the GCF of a trinomial, and after factoring it out we will be able to factor again so that we end up with a product of a monomial, and two binomials.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Factor completely [latex]6m^2k-3mk-3k[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q698742\">Show Answer<\/span><\/p>\n<div id=\"q698742\" class=\"hidden-answer\" style=\"display: none\">Whenever you factor, first try the easy route and ask yourself if there is a GCF. In this case, there is one, and it is 3k.<\/p>\n<p>Factor 3k from the trinomial:<\/p>\n<p>[latex]6m^2k-3mk-3k=3k\\left(2m^2-m-1\\right)[\/latex]<\/p>\n<p>We are left with a trinomial that can be factored using your choice of factoring method. We will create a table to find the factors of [latex]2\\cdot{-1}=-2[\/latex] that sum to [latex]-1[\/latex]<\/p>\n<table class=\"aligncenter\" style=\"width: 20%\" summary=\"A table with five rows and two columns. The first row has columns labeled: Factors of 2 times 3=6 and Sum of Factors. The entries in the first column are: 1, 6; -1, -6; 2,3; and -2,-3. The entries in the second column are: 7, -7, 5, and -5.\">\n<thead>\n<tr>\n<th>Factors of [latex]2\\cdot-1=-2[\/latex]<\/th>\n<th>Sum of Factors<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]2,-1[\/latex]<\/td>\n<td>[latex]1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-2,1[\/latex]<\/td>\n<td>[latex]-1[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Our factors are [latex]-2,1[\/latex], so we can factor by grouping:<\/p>\n<p>Rewrite the middle term with the\u00a0factors we found with the table:<\/p>\n<p style=\"text-align: center\">[latex]\\left(2m^2-m-1\\right)=2m^2-2m+m-1[\/latex]<\/p>\n<p>Regroup and find the GCF of each group:<\/p>\n<p style=\"text-align: center\">[latex](2m^2-2m)+(m-1)=2m(m-1)+1(m-1)[\/latex]<\/p>\n<p>Now\u00a0factor [latex](m-1)[\/latex] from each term:<\/p>\n<p style=\"text-align: center\">[latex]2m^2-m-1=(m-1)(2m+1)[\/latex]<\/p>\n<p style=\"text-align: left\">Don&#8217;t forget the original GCF that we factored out! Our final factored form is:<\/p>\n<p style=\"text-align: center\">[latex]6m^2k-3mk-3k=3k (m-1)(2m+1)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In our last example we show that it is important to factor out a GCF if there is one before you being using the techniques shown in this module.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Ex:  Factoring Polynomials with Common Factors\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/hMAImz2BuPc?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Summary<\/h2>\n<p>In this section we used factoring with special cases. The last topic we covered was what it means to factor completely.<\/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-2726\">\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>Screenshot: Method to the Madness. <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><li>Image: Shortcut This Way. <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><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>Factor Perfect Square Trinomials Using a Formula. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/UMCVGDTxxTI\">https:\/\/youtu.be\/UMCVGDTxxTI<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex: Factor a Difference of Squares. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com). <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/Li9IBp5HrFA\">https:\/\/youtu.be\/Li9IBp5HrFA<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Unit 12: Factoring, from Developmental Math: An Open Program. <strong>Provided by<\/strong>: Monterey Institute of Technology and Education. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/nrocnetwork.org\/dm-opentext\">http:\/\/nrocnetwork.org\/dm-opentext<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li><strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/tFSEpOB262M\">https:\/\/youtu.be\/tFSEpOB262M<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex 3: Factor a Sum or Difference of Cubes. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/J_0ctMrl5_0\">https:\/\/youtu.be\/J_0ctMrl5_0<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Factor Expressions with Negative Exponents. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/4w99g0GZOCk\">https:\/\/youtu.be\/4w99g0GZOCk<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Factor Expressions with Fractional Exponents. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/R6BzjR2O4z8\">https:\/\/youtu.be\/R6BzjR2O4z8<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Factor Expressions Using Substitution. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/QUznZt6yrgI\">https:\/\/youtu.be\/QUznZt6yrgI<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex: Factoring Polynomials with Common Factors. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/hMAImz2BuPc\">https:\/\/youtu.be\/hMAImz2BuPc<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":21,"menu_order":22,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Factor Perfect Square Trinomials Using a Formula\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/UMCVGDTxxTI\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex: Factor a Difference of Squares\",\"author\":\"James Sousa (Mathispower4u.com)\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/Li9IBp5HrFA\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Screenshot: Method to the Madness\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Image: Shortcut This Way\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Unit 12: Factoring, from Developmental Math: An Open Program\",\"author\":\"\",\"organization\":\"Monterey Institute of Technology and Education\",\"url\":\"http:\/\/nrocnetwork.org\/dm-opentext\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen 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