{"id":16325,"date":"2019-10-03T03:21:56","date_gmt":"2019-10-03T03:21:56","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/read-special-cases-squares\/"},"modified":"2024-04-30T23:20:47","modified_gmt":"2024-04-30T23:20:47","slug":"read-special-cases-squares","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/chapter\/read-special-cases-squares\/","title":{"raw":"Special Cases - Squares","rendered":"Special Cases &#8211; Squares"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Factor special products<\/li>\r\n<\/ul>\r\n<\/div>\r\nSome people find it helpful to know when they can take a shortcut to avoid doing extra work. There are some polynomials that will always factor a certain way, and for those, we offer a shortcut. Most people find it helpful to memorize the factored form of a perfect square trinomial or a difference of squares. The 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\n<div><\/div>\r\n<div style=\"text-align: left;\">We can use these equations to factor any perfect square trinomial.<\/div>\r\n<div><\/div>\r\n<div><\/div>\r\n<div class=\"textbox\">\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 style=\"text-align: center;\">[latex]{a}^{2}+2ab+{b}^{2}={\\left(a+b\\right)}^{2}[\/latex]<\/div>\r\n<div style=\"text-align: center;\">[latex]{a}^{2}-2ab+{b}^{2}={\\left(a-b\\right)}^{2}[\/latex]<\/div>\r\n<\/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>Example<\/h3>\r\nFactor [latex]25{x}^{2}+20x+4[\/latex].\r\n[reveal-answer q=\"119279\"]Show Solution[\/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\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\u00a0[latex]1[\/latex].\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 Solution[\/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] and [latex]b=1[\/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(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\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 formula.\r\n\r\nhttps:\/\/youtu.be\/UMCVGDTxxTI\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]91970[\/ohm_question]\r\n\r\n<\/div>\r\nWe can summarize our process in the following way:\r\n<div class=\"textbox\">\r\n<h3>How To: 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<h2>Factoring a Difference of Squares<\/h2>\r\nA difference of squares is a perfect square subtracted from a perfect square. This type of polynomial is unique because it can be factored into two binomials but has only two terms.\r\n<div class=\"textbox shaded\">\r\n<h3>Factor a Difference of Squares<\/h3>\r\nGiven [latex]a^2-b^2[\/latex], its factored form will be [latex]\\left(a+b\\right)\\left(a-b\\right)[\/latex].\r\n\r\n<\/div>\r\nYou will want to become familiar with the special relationship between a difference of squares and its factorization as we can use this equation to factor any differences of squares.\r\n\r\nA 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.\u00a0 Let's look at an example of difference of squares to help us understand how this works.\u00a0 We will start from the product of two binomials to see the pattern.\r\n\r\nGiven the product of two binomials: [latex]\\left(x-2\\right)\\left(x+2\\right)[\/latex], if we multiply them together, we lose the middle term that we are used to seeing as a result.\r\n<p style=\"text-align: left;\">Multiply:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\left(x-2\\right)\\left(x+2\\right)\\\\\\text{}\\\\=x^2-2x+2x-2^2\\\\\\text{}\\\\=x^2-2^2\\\\\\text{}\\\\=x^2-4\\end{array}[\/latex]<\/p>\r\n\u00a0The polynomial [latex]x^2-4[\/latex] is called a difference of squares because each term can be written as something squared. \u00a0A difference of squares will always factor in the following way:\r\n\r\nLet\u2019s factor [latex]x^{2}\u20134[\/latex]\u00a0by writing it as a trinomial, [latex]x^{2}+0x\u20134[\/latex]. \u00a0This\u00a0is similar in format to the\u00a0trinomials we have been factoring so far, so let\u2019s\u00a0use the same method.\r\n<p style=\"text-align: center;\">Find the factors of [latex]a\\cdot{c}[\/latex]\u00a0whose sum is <i>b, <\/i>in this case, 0<i>:<\/i><\/p>\r\n\r\n<table style=\"width: 20%;\">\r\n<thead>\r\n<tr>\r\n<th>Factors of [latex]\u22124[\/latex]<\/th>\r\n<th>Sum of the factors<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]1\\cdot-4=\u22124[\/latex]<\/td>\r\n<td>[latex]1-4=\u22123[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]2\\cdot\u22122=\u22124[\/latex]<\/td>\r\n<td>[latex]2-2=0[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]-1\\cdot4=\u22124[\/latex]<\/td>\r\n<td>[latex]-1+4=3[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[latex]2[\/latex], and [latex]-2[\/latex] have a sum of [latex]0[\/latex]. You can use these to factor [latex]x^{2}\u20134[\/latex].\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFactor [latex]x^{2}\u20134[\/latex].\r\n[reveal-answer q=\"23133\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"23133\"]Rewrite\u00a0[latex]0x[\/latex] as [latex]\u22122x+2x[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x^{2}+0x-4\\\\x^{2}-2x+2x-4\\end{array}[\/latex]<\/p>\r\nGroup pairs.\r\n<p style=\"text-align: center;\">[latex]\\left(x^{2}\u20132x\\right)+\\left(2x\u20134\\right)[\/latex]<\/p>\r\nFactor <i>x<\/i> out of the first group. Factor [latex]2[\/latex] out of the second group.\r\n<p style=\"text-align: center;\">[latex]x\\left(x\u20132\\right)+2\\left(x\u20132\\right)[\/latex]<\/p>\r\nFactor out [latex]\\left(x\u20132\\right)[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\left(x\u20132\\right)\\left(x+2\\right)[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]\\left(x\u20132\\right)\\left(x+2\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nSince order doesn't matter with multiplication, the answer can also be written as [latex]\\left(x+2\\right)\\left(x\u20132\\right)[\/latex].\r\n\r\nYou can check the answer by multiplying [latex]\\left(x\u20132\\right)\\left(x+2\\right)=x^{2}+2x\u20132x\u20134=x^{2}\u20134[\/latex].\r\n<div class=\"textbox\">\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 style=\"text-align: center;\">[latex]{a}^{2}-{b}^{2}=\\left(a+b\\right)\\left(a-b\\right)[\/latex]<\/div>\r\n<\/div>\r\nNow that we have seen how to factor a difference of squares with regrouping, let's try some examples using the short-cut.\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 Solution[\/reveal-answer]\r\n[hidden-answer a=\"960938\"]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].This 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\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]161674[\/ohm_question]\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 Solution[\/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\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<span style=\"color: #ff0000;\">TIP<\/span>:\u00a0<span style=\"color: #ff0000;\"> To help identify difference of squares factoring problems, make a list of perfect squares and become familiar with these values.\u00a0 You will frequently see them in these types of factoring problems.<\/span>\u00a0 In addition, these problems must have a minus sign. For example,\u00a0x^2 - 9 = (x+3)(x-3)\r\n<table style=\"border-collapse: collapse; width: 45.7525%; height: 101px;\" border=\"1\">\r\n<tbody>\r\n<tr style=\"height: 15px;\">\r\n<td style=\"width: 46.9072%; height: 15px;\">1^2<\/td>\r\n<td style=\"width: 53.0928%; height: 15px;\">1<\/td>\r\n<\/tr>\r\n<tr style=\"height: 14px;\">\r\n<td style=\"width: 46.9072%; height: 14px;\">2^2<\/td>\r\n<td style=\"width: 53.0928%; height: 14px;\">4<\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px;\">\r\n<td style=\"width: 46.9072%; height: 15px;\">3^2<\/td>\r\n<td style=\"width: 53.0928%; height: 15px;\">9<\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px;\">\r\n<td style=\"width: 46.9072%; height: 15px;\">4^2<\/td>\r\n<td style=\"width: 53.0928%; height: 15px;\">16<\/td>\r\n<\/tr>\r\n<tr style=\"height: 15px;\">\r\n<td style=\"width: 46.9072%; height: 15px;\">5^2<\/td>\r\n<td style=\"width: 53.0928%; height: 15px;\">25<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]7930[\/ohm_question]\r\n\r\n<\/div>\r\nIn the following video, we show another example of how to use the formula for factoring 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\">\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 Solution[\/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. To 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\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>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Factor special products<\/li>\n<\/ul>\n<\/div>\n<p>Some people find it helpful to know when they can take a shortcut to avoid doing extra work. There are some polynomials that will always factor a certain way, and for those, we offer a shortcut. Most people find it helpful to memorize the factored form of a perfect square trinomial or a difference of squares. The 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<div><\/div>\n<div style=\"text-align: left;\">We can use these equations to factor any perfect square trinomial.<\/div>\n<div><\/div>\n<div><\/div>\n<div class=\"textbox\">\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 style=\"text-align: center;\">[latex]{a}^{2}+2ab+{b}^{2}={\\left(a+b\\right)}^{2}[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex]{a}^{2}-2ab+{b}^{2}={\\left(a-b\\right)}^{2}[\/latex]<\/div>\n<\/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>Example<\/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 Solution<\/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<\/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\u00a0[latex]1[\/latex].<\/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 Solution<\/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] and [latex]b=1[\/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(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<\/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 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<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm91970\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=91970&theme=oea&iframe_resize_id=ohm91970&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>We can summarize our process in the following way:<\/p>\n<div class=\"textbox\">\n<h3>How To: 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<h2>Factoring a Difference of Squares<\/h2>\n<p>A difference of squares is a perfect square subtracted from a perfect square. This type of polynomial is unique because it can be factored into two binomials but has only two terms.<\/p>\n<div class=\"textbox shaded\">\n<h3>Factor a Difference of Squares<\/h3>\n<p>Given [latex]a^2-b^2[\/latex], its factored form will be [latex]\\left(a+b\\right)\\left(a-b\\right)[\/latex].<\/p>\n<\/div>\n<p>You will want to become familiar with the special relationship between a difference of squares and its factorization as we can use this equation to factor any differences of squares.<\/p>\n<p>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.\u00a0 Let&#8217;s look at an example of difference of squares to help us understand how this works.\u00a0 We will start from the product of two binomials to see the pattern.<\/p>\n<p>Given the product of two binomials: [latex]\\left(x-2\\right)\\left(x+2\\right)[\/latex], if we multiply them together, we lose the middle term that we are used to seeing as a result.<\/p>\n<p style=\"text-align: left;\">Multiply:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\left(x-2\\right)\\left(x+2\\right)\\\\\\text{}\\\\=x^2-2x+2x-2^2\\\\\\text{}\\\\=x^2-2^2\\\\\\text{}\\\\=x^2-4\\end{array}[\/latex]<\/p>\n<p>\u00a0The polynomial [latex]x^2-4[\/latex] is called a difference of squares because each term can be written as something squared. \u00a0A difference of squares will always factor in the following way:<\/p>\n<p>Let\u2019s factor [latex]x^{2}\u20134[\/latex]\u00a0by writing it as a trinomial, [latex]x^{2}+0x\u20134[\/latex]. \u00a0This\u00a0is similar in format to the\u00a0trinomials we have been factoring so far, so let\u2019s\u00a0use the same method.<\/p>\n<p style=\"text-align: center;\">Find the factors of [latex]a\\cdot{c}[\/latex]\u00a0whose sum is <i>b, <\/i>in this case, 0<i>:<\/i><\/p>\n<table style=\"width: 20%;\">\n<thead>\n<tr>\n<th>Factors of [latex]\u22124[\/latex]<\/th>\n<th>Sum of the factors<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]1\\cdot-4=\u22124[\/latex]<\/td>\n<td>[latex]1-4=\u22123[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]2\\cdot\u22122=\u22124[\/latex]<\/td>\n<td>[latex]2-2=0[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]-1\\cdot4=\u22124[\/latex]<\/td>\n<td>[latex]-1+4=3[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>[latex]2[\/latex], and [latex]-2[\/latex] have a sum of [latex]0[\/latex]. You can use these to factor [latex]x^{2}\u20134[\/latex].<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Factor [latex]x^{2}\u20134[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q23133\">Show Solution<\/span><\/p>\n<div id=\"q23133\" class=\"hidden-answer\" style=\"display: none\">Rewrite\u00a0[latex]0x[\/latex] as [latex]\u22122x+2x[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}x^{2}+0x-4\\\\x^{2}-2x+2x-4\\end{array}[\/latex]<\/p>\n<p>Group pairs.<\/p>\n<p style=\"text-align: center;\">[latex]\\left(x^{2}\u20132x\\right)+\\left(2x\u20134\\right)[\/latex]<\/p>\n<p>Factor <i>x<\/i> out of the first group. Factor [latex]2[\/latex] out of the second group.<\/p>\n<p style=\"text-align: center;\">[latex]x\\left(x\u20132\\right)+2\\left(x\u20132\\right)[\/latex]<\/p>\n<p>Factor out [latex]\\left(x\u20132\\right)[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\left(x\u20132\\right)\\left(x+2\\right)[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\left(x\u20132\\right)\\left(x+2\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Since order doesn&#8217;t matter with multiplication, the answer can also be written as [latex]\\left(x+2\\right)\\left(x\u20132\\right)[\/latex].<\/p>\n<p>You can check the answer by multiplying [latex]\\left(x\u20132\\right)\\left(x+2\\right)=x^{2}+2x\u20132x\u20134=x^{2}\u20134[\/latex].<\/p>\n<div class=\"textbox\">\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 style=\"text-align: center;\">[latex]{a}^{2}-{b}^{2}=\\left(a+b\\right)\\left(a-b\\right)[\/latex]<\/div>\n<\/div>\n<p>Now that we have seen how to factor a difference of squares with regrouping, let&#8217;s try some examples using the short-cut.<\/p>\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 Solution<\/span><\/p>\n<div id=\"q960938\" class=\"hidden-answer\" style=\"display: none\">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].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<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm161674\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=161674&theme=oea&iframe_resize_id=ohm161674&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\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 Solution<\/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<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<p><span style=\"color: #ff0000;\">TIP<\/span>:\u00a0<span style=\"color: #ff0000;\"> To help identify difference of squares factoring problems, make a list of perfect squares and become familiar with these values.\u00a0 You will frequently see them in these types of factoring problems.<\/span>\u00a0 In addition, these problems must have a minus sign. For example,\u00a0x^2 &#8211; 9 = (x+3)(x-3)<\/p>\n<table style=\"border-collapse: collapse; width: 45.7525%; height: 101px;\">\n<tbody>\n<tr style=\"height: 15px;\">\n<td style=\"width: 46.9072%; height: 15px;\">1^2<\/td>\n<td style=\"width: 53.0928%; height: 15px;\">1<\/td>\n<\/tr>\n<tr style=\"height: 14px;\">\n<td style=\"width: 46.9072%; height: 14px;\">2^2<\/td>\n<td style=\"width: 53.0928%; height: 14px;\">4<\/td>\n<\/tr>\n<tr style=\"height: 15px;\">\n<td style=\"width: 46.9072%; height: 15px;\">3^2<\/td>\n<td style=\"width: 53.0928%; height: 15px;\">9<\/td>\n<\/tr>\n<tr style=\"height: 15px;\">\n<td style=\"width: 46.9072%; height: 15px;\">4^2<\/td>\n<td style=\"width: 53.0928%; height: 15px;\">16<\/td>\n<\/tr>\n<tr style=\"height: 15px;\">\n<td style=\"width: 46.9072%; height: 15px;\">5^2<\/td>\n<td style=\"width: 53.0928%; height: 15px;\">25<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm7930\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=7930&theme=oea&iframe_resize_id=ohm7930&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>In the following video, we show another example of how to use the formula for factoring 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\">\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 Solution<\/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. To 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>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\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-16325\">\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><\/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":169554,"menu_order":14,"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\":\"\"}]","CANDELA_OUTCOMES_GUID":"933680a871454fc6912430bf6e5dd586, 4375e5a5fccc4dc98999dfe19e1c11a0","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-16325","chapter","type-chapter","status-publish","hentry"],"part":16188,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/16325","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/users\/169554"}],"version-history":[{"count":10,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/16325\/revisions"}],"predecessor-version":[{"id":19900,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/16325\/revisions\/19900"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/parts\/16188"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapters\/16325\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/media?parent=16325"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/pressbooks\/v2\/chapter-type?post=16325"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/contributor?post=16325"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/wm-developmentalemporium\/wp-json\/wp\/v2\/license?post=16325"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}