{"id":4779,"date":"2017-06-07T18:57:02","date_gmt":"2017-06-07T18:57:02","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/chapter\/read-factor-trinomials-part-ii\/"},"modified":"2017-08-16T02:02:18","modified_gmt":"2017-08-16T02:02:18","slug":"read-factor-trinomials-part-ii","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/aacc-collegealgebrafoundations\/chapter\/read-factor-trinomials-part-ii\/","title":{"raw":"Factor Trinomials With Leading Coefficient Greater Than 1","rendered":"Factor Trinomials With Leading Coefficient Greater Than 1"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Factor trinomials of the form [latex]ax^2+bx+c[\/latex]<\/li>\r\n \t<li>Recognize where to place negative signs when factoring a trinomial<\/li>\r\n \t<li>Recognize when a polynomial is a difference of squares, and how it would factor as the product of two binomials<\/li>\r\n<\/ul>\r\n<\/div>\r\nBecause we couldn't get enough factoring, we thought we would give an encore.\r\n\r\n[caption id=\"attachment_4874\" align=\"aligncenter\" width=\"436\"]<img class=\"wp-image-4874\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2011\/2017\/06\/07185702\/Screen-Shot-2016-06-13-at-5.22.39-PM-300x205.png\" alt=\"The word &quot;Encore&quot; in lights\" width=\"436\" height=\"298\" \/> Encore[\/caption]\r\n\r\nIn this section we will continue to factor trinomials so you can become proficient with using the techniques presented in the last section. \u00a0The goal is for you to be comfortable with recognizing where to place negative signs, and whether a trinomial can even be factored. \u00a0Additionally, we will introduce one special case to look out for at the end of this page.\r\n\r\nNot all trinomials look like [latex]x^{2}+5x+6[\/latex], where the coefficient in front of the [latex]x^{2}[\/latex]\u00a0term is 1. In these cases, your first step should be to look for common factors for the three terms.\r\n<table>\r\n<thead>\r\n<tr>\r\n<th>Trinomial<\/th>\r\n<th>Factor out Common Factor<\/th>\r\n<th>Factored<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>[latex]2x^{2}+10x+12[\/latex]<\/td>\r\n<td>[latex]2(x^{2}+5x+6)[\/latex]<\/td>\r\n<td>[latex]2\\left(x+2\\right)\\left(x+3\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\u22125a^{2}\u221215a\u221210[\/latex]<\/td>\r\n<td>[latex]\u22125(a^{2}+3a+2)[\/latex]<\/td>\r\n<td>[latex]\u22125\\left(a+2\\right)\\left(a+1\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]c^{3}\u20138c^{2}+15c[\/latex]<\/td>\r\n<td>[latex]c\\left(c^{2}\u20138c+15\\right)[\/latex]<\/td>\r\n<td>[latex]c\\left(c\u20135\\right)\\left(c\u20133\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]y^{4}\u20139y^{3}\u201310y^{2}[\/latex]<\/td>\r\n<td>[latex]y^{2}\\left(y^{2}\u20139y\u201310\\right)[\/latex]<\/td>\r\n<td>[latex]y^{2}\\left(y\u201310\\right)\\left(y+1\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNotice that once you have identified and pulled out the common factor, you can factor the remaining trinomial as usual. This process is shown below.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFactor\u00a0[latex]3x^{3}\u20133x^{2}\u201390x[\/latex].\r\n\r\n[reveal-answer q=\"298928\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"298928\"]Since 3 is a common factor for the three terms, factor out the 3.\r\n<p style=\"text-align: center\">[latex]3\\left(x^{3}\u2013x^{2}\u201330x\\right)[\/latex]<\/p>\r\n<i>x<\/i> is also a common factor, so factor out <i>x<\/i>.\r\n<p style=\"text-align: center\">[latex]3x\\left(x^{2}\u2013x\u201330\\right)[\/latex]<\/p>\r\nNow you can factor the trinomial\u00a0[latex]x^{2}\u2013x\u201330[\/latex]. To find <i>r<\/i> and <i>s<\/i>, identify two numbers whose product is [latex]\u221230[\/latex] and whose sum is [latex]\u22121[\/latex].\r\n\r\nThe pair of factors is [latex]\u22126[\/latex] and [latex]5[\/latex]. So replace [latex]\u2013x[\/latex] with [latex]\u22126x+5x[\/latex].\r\n<p style=\"text-align: center\">[latex]3x\\left(x^{2}\u20136x+5x\u201330\\right)[\/latex]<\/p>\r\nUse grouping to consider the terms in pairs.\r\n<p style=\"text-align: center\">[latex]3x\\left[\\left(x^{2}\u20136x\\right)+\\left(5x\u201330\\right)\\right][\/latex]<\/p>\r\nFactor <i>x<\/i> out of the first group and factor 5 out of the second group.\r\n<p style=\"text-align: center\">[latex]3x\\left[\\left(x\\left(x\u20136\\right)\\right)+5\\left(x\u20136\\right)\\right][\/latex]<\/p>\r\nThen factor out [latex]x\u20136[\/latex].\r\n<p style=\"text-align: center\">[latex]3x\\left(x\u20136\\right)\\left(x+5\\right)[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]3x\\left(x\u20136\\right)\\left(x+5\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe following video contains two more examples of factoring a quadratic trinomial where the first step is to factor out a GCF. We use the shortcut method instead of factoring by grouping.\r\n\r\nhttps:\/\/youtu.be\/pgH77rAtsbs\r\n\r\nThe general form of trinomials with a leading coefficient of <i>a<\/i> is [latex]ax^{2}+bx+c[\/latex]. Sometimes the factor of <i>a<\/i> can be factored as you saw above; this happens when <i>a<\/i> can be factored out of all three terms. The remaining trinomial that still needs factoring will then be simpler, with the leading term only being an [latex]x^{2}[\/latex]\u00a0term, instead of an [latex]ax^{2}[\/latex]\u00a0term.\r\n\r\nHowever, if the coefficients of all three terms of a trinomial don\u2019t have a common factor, then you will need to factor the trinomial with a coefficient of something other than 1.\r\n<div class=\"textbox shaded\">\r\n<h3>Factoring Trinomials in the form [latex]ax^{2}+bx+c[\/latex]<\/h3>\r\nTo factor a trinomial in the form [latex]ax^{2}+bx+c[\/latex], find two integers, <i>r<\/i> and <i>s<\/i>, whose sum is <i>b<\/i> and whose product is <i>ac.<\/i>\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}r\\cdot{s}=a\\cdot{c}\\\\r+s=b\\end{array}[\/latex]<\/p>\r\nRewrite the trinomial as [latex]ax^{2}+rx+sx+c[\/latex] and then use grouping and the distributive property to factor the polynomial.\r\n\r\n<\/div>\r\nThis is almost the same as factoring trinomials in the form [latex]x^{2}+bx+c[\/latex],<i> <\/i>as in this form\u00a0[latex]a=1[\/latex].<i> <\/i>Now you are looking for two factors whose product is [latex]a\\cdot{c}[\/latex],\u00a0and whose sum is <i>b<\/i>.\r\n\r\nLet\u2019s see how this strategy works by factoring [latex]6z^{2}+11z+4[\/latex].\r\n\r\nIn this trinomial, [latex]a=6[\/latex], [latex]b=11[\/latex], and [latex]c=4[\/latex]. According to the strategy, you need to find two factors, <i>r<\/i> and <i>s<\/i>, whose sum is [latex]b=11[\/latex]\u00a0and whose product is [latex]a\\cdot{c}=6\\cdot4=24[\/latex]. You can make a chart to organize the possible factor combinations. (Notice that this chart only has positive numbers. Since <i>ac<\/i> is positive and <i>b<\/i> is positive, you can be certain that the two factors you're looking for are also positive numbers.)\r\n<table style=\"width: 20%\">\r\n<thead>\r\n<tr>\r\n<th>Factors whose product is 24<\/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\\cdot24=24[\/latex]<\/td>\r\n<td>[latex]1+24=25[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]2\\cdot12=24[\/latex]<\/td>\r\n<td>[latex]2+12=14[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]3\\cdot8=24[\/latex]<\/td>\r\n<td>[latex]3+8=11[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]4\\cdot6=24[\/latex]<\/td>\r\n<td>[latex]4+6=10[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThere is only one combination where the product is 24 and the sum is 11, and that is when [latex]r=3[\/latex], and [latex]s=8[\/latex]. Let\u2019s use these values to factor the original trinomial.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFactor [latex]6z^{2}+11z+4[\/latex].\r\n[reveal-answer q=\"796129\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"796129\"]Rewrite the middle term, [latex]11z[\/latex], as [latex]3z + 8z[\/latex] (from the chart above.)\r\n<p style=\"text-align: center\">[latex]6z^{2}+3z+8z+4[\/latex]<\/p>\r\nGroup pairs. Use grouping to consider the terms in pairs.\r\n<p style=\"text-align: center\">[latex]\\left(6z^{2}+3z\\right)+\\left(8z+4\\right)[\/latex]<\/p>\r\nFactor 3<i>z<\/i> out of the first group and 4 out of the second group.\r\n<p style=\"text-align: center\">[latex]3z\\left(2z+1\\right)+4\\left(2z+1\\right)[\/latex]<\/p>\r\nFactor out [latex]\\left(2z+1\\right)[\/latex].\r\n<p style=\"text-align: center\">[latex]\\left(2z+1\\right)\\left(3z+4\\right)[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]\\left(2z+1\\right)\\left(3z+4\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, we present another example of factoring a trinomial using grouping. \u00a0In this example, the middle term, b, is negative. Note how having a negative middle term and a positive c term influence the options for r and s when factoring.\r\n\r\nhttps:\/\/youtu.be\/agDaQ_cZnNc\r\n\r\nBefore going any further, it is worth mentioning that not all trinomials can be factored using integer pairs. Take the trinomial [latex]2z^{2}+35z+7[\/latex], for instance. Can you think of two integers whose sum is [latex]b=35[\/latex] and whose product is [latex]a\\cdot{c}=2\\cdot7=14[\/latex]? There are none! This type of trinomial, which cannot be factored using integers, is called a prime trinomial.\r\n\r\nIn some situations, <i>a<\/i> is negative, as in [latex]\u22124h^{2}+11h+3[\/latex]. It often makes sense to factor out [latex]\u22121[\/latex] as the first step in factoring, as doing so will change the sign of [latex]ax^{2}[\/latex]\u00a0from negative to positive, making the remaining trinomial easier to factor.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFactor\u00a0[latex]\u22124h^{2}+11h+3[\/latex]\r\n[reveal-answer q=\"471034\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"471034\"]Factor [latex]\u22121[\/latex] out of the trinomial. Notice that the signs of all three terms have changed.\r\n<p style=\"text-align: center\">[latex]\u22121\\left(4h^{2}\u201311h\u20133\\right)[\/latex]<\/p>\r\nTo factor the trinomial, you need to figure out how to rewrite [latex]\u221211h[\/latex].\u00a0The product of [latex]rs=4\\cdot\u22123=\u221212[\/latex], and the sum of [latex]rs=\u221211[\/latex].\r\n<table style=\"width: 20%\">\r\n<tbody>\r\n<tr>\r\n<td>[latex]r\\cdot{s}=\u221212[\/latex]<\/td>\r\n<td>[latex]r+s=\u221211[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\u221212\\cdot1=\u221212[\/latex]<\/td>\r\n<td>[latex]\u221212+1=\u221211[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\u22126\\cdot2=\u221212[\/latex]<\/td>\r\n<td>[latex]\u22126+2=\u22124[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>[latex]\u22124\\cdot3=\u221212[\/latex]<\/td>\r\n<td>[latex]\u22124+3=\u22121[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nRewrite the middle term\u00a0[latex]\u221211h[\/latex] as [latex]\u221212h+1h[\/latex].\r\n<p style=\"text-align: center\">[latex]\u22121\\left(4h^{2}\u201312h+1h\u20133\\right)[\/latex]<\/p>\r\nGroup terms.\r\n<p style=\"text-align: center\">[latex]\u22121\\left[\\left(4h^{2}\u201312h\\right)+\\left(1h\u20133\\right)\\right][\/latex]<\/p>\r\nFactor out 4<i>h<\/i> from the first pair. The second group cannot be factored further, but you can write it as [latex]+1\\left(h\u20133\\right)[\/latex]\u00a0since [latex]+1\\left(h\u20133\\right)=\\left(h\u20133\\right)[\/latex]. This helps with factoring in the next step.\r\n<p style=\"text-align: center\">[latex]\u22121\\left[4h\\left(h\u20133\\right)+1\\left(h\u20133\\right)\\right][\/latex]<\/p>\r\nFactor out a common factor of [latex]\\left(h\u20133\\right)[\/latex]. Notice you are left with [latex]\\left(h\u20133\\right)\\left(4h+1\\right)[\/latex]; the [latex]+1[\/latex] comes from the term [latex]+1\\left(h\u20133\\right)[\/latex]\u00a0in the previous step.\r\n<p style=\"text-align: center\">[latex]\u22121\\left[\\left(h\u20133\\right)\\left(4h+1\\right)\\right][\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]\u22121\\left(h\u20133\\right)\\left(4h+1\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNote that the answer above can also be written as [latex]\\left(\u2212h+3\\right)\\left(4h+1\\right)[\/latex] or [latex]\\left(h\u20133\\right)\\left(\u22124h\u20131\\right)[\/latex] if you multiply [latex]\u22121[\/latex]\u00a0times one of the other factors.\r\n\r\nIn the following video we present another example of factoring a trinomial in the form [latex]-ax^2+bx+c[\/latex] using the grouping technique.\r\n\r\nhttps:\/\/youtu.be\/zDAMjdBfkDs\r\n<h2>Difference of Squares<\/h2>\r\nWe would be remiss if we failed to introduce one more type of polynomial that can be factored. This polynomial can be factored into two binomials but has only two terms. \u00a0Let's 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 teach term can be written as something squared. \u00a0A difference of squares will always factor in the following way:\r\n<div class=\"textbox shaded\">\r\n<h3>Factor a Difference of Squares<\/h3>\r\nGiven [latex]a^2-b^2[\/latex], it's factored form will be [latex]\\left(a+b\\right)\\left(a-b\\right)[\/latex]\r\n\r\n<\/div>\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\n2, and -2 have a sum of 0. 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 2 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(3x\u20132\\right)\\left(3x+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\r\nThe following video show two more examples of factoring a difference of squares.\r\n\r\nhttps:\/\/youtu.be\/Li9IBp5HrFA\r\n<h2>Summary<\/h2>\r\nWhen a trinomial is in the form of [latex]ax^{2}+bx+c[\/latex], where <i>a<\/i> is a coefficient other than 1, look first for common factors for all three terms. Factor out the common factor first, then factor the remaining simpler trinomial. If the remaining trinomial is still of the form [latex]ax^{2}+bx+c[\/latex], find two integers, <i>r<\/i> and <i>s<\/i>, whose sum is <i>b<\/i> and whose product is <i>ac.<\/i> Then rewrite the trinomial as [latex]ax^{2}+rx+sx+c[\/latex]\u00a0and use grouping and the distributive property to factor the polynomial.\r\n\r\nWhen [latex]ax^{2}[\/latex] is negative, you can factor [latex]\u22121[\/latex] out of the whole trinomial before continuing.\r\n\r\nA difference of squares [latex]a^2-b^2[\/latex] will factor in this way: [latex]\\left(a+b\\right)\\left(a-b\\right)[\/latex].","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Factor trinomials of the form [latex]ax^2+bx+c[\/latex]<\/li>\n<li>Recognize where to place negative signs when factoring a trinomial<\/li>\n<li>Recognize when a polynomial is a difference of squares, and how it would factor as the product of two binomials<\/li>\n<\/ul>\n<\/div>\n<p>Because we couldn&#8217;t get enough factoring, we thought we would give an encore.<\/p>\n<div id=\"attachment_4874\" style=\"width: 446px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-4874\" class=\"wp-image-4874\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2011\/2017\/06\/07185702\/Screen-Shot-2016-06-13-at-5.22.39-PM-300x205.png\" alt=\"The word &quot;Encore&quot; in lights\" width=\"436\" height=\"298\" \/><\/p>\n<p id=\"caption-attachment-4874\" class=\"wp-caption-text\">Encore<\/p>\n<\/div>\n<p>In this section we will continue to factor trinomials so you can become proficient with using the techniques presented in the last section. \u00a0The goal is for you to be comfortable with recognizing where to place negative signs, and whether a trinomial can even be factored. \u00a0Additionally, we will introduce one special case to look out for at the end of this page.<\/p>\n<p>Not all trinomials look like [latex]x^{2}+5x+6[\/latex], where the coefficient in front of the [latex]x^{2}[\/latex]\u00a0term is 1. In these cases, your first step should be to look for common factors for the three terms.<\/p>\n<table>\n<thead>\n<tr>\n<th>Trinomial<\/th>\n<th>Factor out Common Factor<\/th>\n<th>Factored<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]2x^{2}+10x+12[\/latex]<\/td>\n<td>[latex]2(x^{2}+5x+6)[\/latex]<\/td>\n<td>[latex]2\\left(x+2\\right)\\left(x+3\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\u22125a^{2}\u221215a\u221210[\/latex]<\/td>\n<td>[latex]\u22125(a^{2}+3a+2)[\/latex]<\/td>\n<td>[latex]\u22125\\left(a+2\\right)\\left(a+1\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]c^{3}\u20138c^{2}+15c[\/latex]<\/td>\n<td>[latex]c\\left(c^{2}\u20138c+15\\right)[\/latex]<\/td>\n<td>[latex]c\\left(c\u20135\\right)\\left(c\u20133\\right)[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]y^{4}\u20139y^{3}\u201310y^{2}[\/latex]<\/td>\n<td>[latex]y^{2}\\left(y^{2}\u20139y\u201310\\right)[\/latex]<\/td>\n<td>[latex]y^{2}\\left(y\u201310\\right)\\left(y+1\\right)[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Notice that once you have identified and pulled out the common factor, you can factor the remaining trinomial as usual. This process is shown below.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Factor\u00a0[latex]3x^{3}\u20133x^{2}\u201390x[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q298928\">Show Solution<\/span><\/p>\n<div id=\"q298928\" class=\"hidden-answer\" style=\"display: none\">Since 3 is a common factor for the three terms, factor out the 3.<\/p>\n<p style=\"text-align: center\">[latex]3\\left(x^{3}\u2013x^{2}\u201330x\\right)[\/latex]<\/p>\n<p><i>x<\/i> is also a common factor, so factor out <i>x<\/i>.<\/p>\n<p style=\"text-align: center\">[latex]3x\\left(x^{2}\u2013x\u201330\\right)[\/latex]<\/p>\n<p>Now you can factor the trinomial\u00a0[latex]x^{2}\u2013x\u201330[\/latex]. To find <i>r<\/i> and <i>s<\/i>, identify two numbers whose product is [latex]\u221230[\/latex] and whose sum is [latex]\u22121[\/latex].<\/p>\n<p>The pair of factors is [latex]\u22126[\/latex] and [latex]5[\/latex]. So replace [latex]\u2013x[\/latex] with [latex]\u22126x+5x[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]3x\\left(x^{2}\u20136x+5x\u201330\\right)[\/latex]<\/p>\n<p>Use grouping to consider the terms in pairs.<\/p>\n<p style=\"text-align: center\">[latex]3x\\left[\\left(x^{2}\u20136x\\right)+\\left(5x\u201330\\right)\\right][\/latex]<\/p>\n<p>Factor <i>x<\/i> out of the first group and factor 5 out of the second group.<\/p>\n<p style=\"text-align: center\">[latex]3x\\left[\\left(x\\left(x\u20136\\right)\\right)+5\\left(x\u20136\\right)\\right][\/latex]<\/p>\n<p>Then factor out [latex]x\u20136[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]3x\\left(x\u20136\\right)\\left(x+5\\right)[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]3x\\left(x\u20136\\right)\\left(x+5\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The following video contains two more examples of factoring a quadratic trinomial where the first step is to factor out a GCF. We use the shortcut method instead of factoring by grouping.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Factor a Trinomial With A Common Factor Using the Shortcut Method - Form x^2+bx+c\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/pgH77rAtsbs?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>The general form of trinomials with a leading coefficient of <i>a<\/i> is [latex]ax^{2}+bx+c[\/latex]. Sometimes the factor of <i>a<\/i> can be factored as you saw above; this happens when <i>a<\/i> can be factored out of all three terms. The remaining trinomial that still needs factoring will then be simpler, with the leading term only being an [latex]x^{2}[\/latex]\u00a0term, instead of an [latex]ax^{2}[\/latex]\u00a0term.<\/p>\n<p>However, if the coefficients of all three terms of a trinomial don\u2019t have a common factor, then you will need to factor the trinomial with a coefficient of something other than 1.<\/p>\n<div class=\"textbox shaded\">\n<h3>Factoring Trinomials in the form [latex]ax^{2}+bx+c[\/latex]<\/h3>\n<p>To factor a trinomial in the form [latex]ax^{2}+bx+c[\/latex], find two integers, <i>r<\/i> and <i>s<\/i>, whose sum is <i>b<\/i> and whose product is <i>ac.<\/i><\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}r\\cdot{s}=a\\cdot{c}\\\\r+s=b\\end{array}[\/latex]<\/p>\n<p>Rewrite the trinomial as [latex]ax^{2}+rx+sx+c[\/latex] and then use grouping and the distributive property to factor the polynomial.<\/p>\n<\/div>\n<p>This is almost the same as factoring trinomials in the form [latex]x^{2}+bx+c[\/latex],<i> <\/i>as in this form\u00a0[latex]a=1[\/latex].<i> <\/i>Now you are looking for two factors whose product is [latex]a\\cdot{c}[\/latex],\u00a0and whose sum is <i>b<\/i>.<\/p>\n<p>Let\u2019s see how this strategy works by factoring [latex]6z^{2}+11z+4[\/latex].<\/p>\n<p>In this trinomial, [latex]a=6[\/latex], [latex]b=11[\/latex], and [latex]c=4[\/latex]. According to the strategy, you need to find two factors, <i>r<\/i> and <i>s<\/i>, whose sum is [latex]b=11[\/latex]\u00a0and whose product is [latex]a\\cdot{c}=6\\cdot4=24[\/latex]. You can make a chart to organize the possible factor combinations. (Notice that this chart only has positive numbers. Since <i>ac<\/i> is positive and <i>b<\/i> is positive, you can be certain that the two factors you&#8217;re looking for are also positive numbers.)<\/p>\n<table style=\"width: 20%\">\n<thead>\n<tr>\n<th>Factors whose product is 24<\/th>\n<th>Sum of the factors<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>[latex]1\\cdot24=24[\/latex]<\/td>\n<td>[latex]1+24=25[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]2\\cdot12=24[\/latex]<\/td>\n<td>[latex]2+12=14[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]3\\cdot8=24[\/latex]<\/td>\n<td>[latex]3+8=11[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]4\\cdot6=24[\/latex]<\/td>\n<td>[latex]4+6=10[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>There is only one combination where the product is 24 and the sum is 11, and that is when [latex]r=3[\/latex], and [latex]s=8[\/latex]. Let\u2019s use these values to factor the original trinomial.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Factor [latex]6z^{2}+11z+4[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q796129\">Show Solution<\/span><\/p>\n<div id=\"q796129\" class=\"hidden-answer\" style=\"display: none\">Rewrite the middle term, [latex]11z[\/latex], as [latex]3z + 8z[\/latex] (from the chart above.)<\/p>\n<p style=\"text-align: center\">[latex]6z^{2}+3z+8z+4[\/latex]<\/p>\n<p>Group pairs. Use grouping to consider the terms in pairs.<\/p>\n<p style=\"text-align: center\">[latex]\\left(6z^{2}+3z\\right)+\\left(8z+4\\right)[\/latex]<\/p>\n<p>Factor 3<i>z<\/i> out of the first group and 4 out of the second group.<\/p>\n<p style=\"text-align: center\">[latex]3z\\left(2z+1\\right)+4\\left(2z+1\\right)[\/latex]<\/p>\n<p>Factor out [latex]\\left(2z+1\\right)[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\left(2z+1\\right)\\left(3z+4\\right)[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\left(2z+1\\right)\\left(3z+4\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the following video, we present another example of factoring a trinomial using grouping. \u00a0In this example, the middle term, b, is negative. Note how having a negative middle term and a positive c term influence the options for r and s when factoring.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Factor a Trinomial in the Form ax^2+bx+c Using the Grouping Technique\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/agDaQ_cZnNc?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>Before going any further, it is worth mentioning that not all trinomials can be factored using integer pairs. Take the trinomial [latex]2z^{2}+35z+7[\/latex], for instance. Can you think of two integers whose sum is [latex]b=35[\/latex] and whose product is [latex]a\\cdot{c}=2\\cdot7=14[\/latex]? There are none! This type of trinomial, which cannot be factored using integers, is called a prime trinomial.<\/p>\n<p>In some situations, <i>a<\/i> is negative, as in [latex]\u22124h^{2}+11h+3[\/latex]. It often makes sense to factor out [latex]\u22121[\/latex] as the first step in factoring, as doing so will change the sign of [latex]ax^{2}[\/latex]\u00a0from negative to positive, making the remaining trinomial easier to factor.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Factor\u00a0[latex]\u22124h^{2}+11h+3[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q471034\">Show Solution<\/span><\/p>\n<div id=\"q471034\" class=\"hidden-answer\" style=\"display: none\">Factor [latex]\u22121[\/latex] out of the trinomial. Notice that the signs of all three terms have changed.<\/p>\n<p style=\"text-align: center\">[latex]\u22121\\left(4h^{2}\u201311h\u20133\\right)[\/latex]<\/p>\n<p>To factor the trinomial, you need to figure out how to rewrite [latex]\u221211h[\/latex].\u00a0The product of [latex]rs=4\\cdot\u22123=\u221212[\/latex], and the sum of [latex]rs=\u221211[\/latex].<\/p>\n<table style=\"width: 20%\">\n<tbody>\n<tr>\n<td>[latex]r\\cdot{s}=\u221212[\/latex]<\/td>\n<td>[latex]r+s=\u221211[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\u221212\\cdot1=\u221212[\/latex]<\/td>\n<td>[latex]\u221212+1=\u221211[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\u22126\\cdot2=\u221212[\/latex]<\/td>\n<td>[latex]\u22126+2=\u22124[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>[latex]\u22124\\cdot3=\u221212[\/latex]<\/td>\n<td>[latex]\u22124+3=\u22121[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Rewrite the middle term\u00a0[latex]\u221211h[\/latex] as [latex]\u221212h+1h[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\u22121\\left(4h^{2}\u201312h+1h\u20133\\right)[\/latex]<\/p>\n<p>Group terms.<\/p>\n<p style=\"text-align: center\">[latex]\u22121\\left[\\left(4h^{2}\u201312h\\right)+\\left(1h\u20133\\right)\\right][\/latex]<\/p>\n<p>Factor out 4<i>h<\/i> from the first pair. The second group cannot be factored further, but you can write it as [latex]+1\\left(h\u20133\\right)[\/latex]\u00a0since [latex]+1\\left(h\u20133\\right)=\\left(h\u20133\\right)[\/latex]. This helps with factoring in the next step.<\/p>\n<p style=\"text-align: center\">[latex]\u22121\\left[4h\\left(h\u20133\\right)+1\\left(h\u20133\\right)\\right][\/latex]<\/p>\n<p>Factor out a common factor of [latex]\\left(h\u20133\\right)[\/latex]. Notice you are left with [latex]\\left(h\u20133\\right)\\left(4h+1\\right)[\/latex]; the [latex]+1[\/latex] comes from the term [latex]+1\\left(h\u20133\\right)[\/latex]\u00a0in the previous step.<\/p>\n<p style=\"text-align: center\">[latex]\u22121\\left[\\left(h\u20133\\right)\\left(4h+1\\right)\\right][\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\u22121\\left(h\u20133\\right)\\left(4h+1\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Note that the answer above can also be written as [latex]\\left(\u2212h+3\\right)\\left(4h+1\\right)[\/latex] or [latex]\\left(h\u20133\\right)\\left(\u22124h\u20131\\right)[\/latex] if you multiply [latex]\u22121[\/latex]\u00a0times one of the other factors.<\/p>\n<p>In the following video we present another example of factoring a trinomial in the form [latex]-ax^2+bx+c[\/latex] using the grouping technique.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Factor a Trinomial in the Form -ax^2+bx+c Using the Grouping Technique\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/zDAMjdBfkDs?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Difference of Squares<\/h2>\n<p>We would be remiss if we failed to introduce one more type of polynomial that can be factored. This polynomial can be factored into two binomials but has only two terms. \u00a0Let&#8217;s 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 teach term can be written as something squared. \u00a0A difference of squares will always factor in the following way:<\/p>\n<div class=\"textbox shaded\">\n<h3>Factor a Difference of Squares<\/h3>\n<p>Given [latex]a^2-b^2[\/latex], it&#8217;s factored form will be [latex]\\left(a+b\\right)\\left(a-b\\right)[\/latex]<\/p>\n<\/div>\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>2, and -2 have a sum of 0. 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 2 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(3x\u20132\\right)\\left(3x+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<p>The following video show two more examples of factoring a difference of squares.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" 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<h2>Summary<\/h2>\n<p>When a trinomial is in the form of [latex]ax^{2}+bx+c[\/latex], where <i>a<\/i> is a coefficient other than 1, look first for common factors for all three terms. Factor out the common factor first, then factor the remaining simpler trinomial. If the remaining trinomial is still of the form [latex]ax^{2}+bx+c[\/latex], find two integers, <i>r<\/i> and <i>s<\/i>, whose sum is <i>b<\/i> and whose product is <i>ac.<\/i> Then rewrite the trinomial as [latex]ax^{2}+rx+sx+c[\/latex]\u00a0and use grouping and the distributive property to factor the polynomial.<\/p>\n<p>When [latex]ax^{2}[\/latex] is negative, you can factor [latex]\u22121[\/latex] out of the whole trinomial before continuing.<\/p>\n<p>A difference of squares [latex]a^2-b^2[\/latex] will factor in this way: [latex]\\left(a+b\\right)\\left(a-b\\right)[\/latex].<\/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-4779\">\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>Factor a Trinomial With A Common Factor Using the Shortcut Method - Form x^2+bx+c. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/pgH77rAtsbs\">https:\/\/youtu.be\/pgH77rAtsbs<\/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 a Trinomial in the Form ax^2+bx+c Using the Grouping Technique. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/agDaQ_cZnNc\">https:\/\/youtu.be\/agDaQ_cZnNc<\/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 a Trinomial in the Form -ax^2+bx+c Using the Grouping Technique. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/zDAMjdBfkDs\">https:\/\/youtu.be\/zDAMjdBfkDs<\/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>Screenshot: Encore. <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>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":17533,"menu_order":21,"template":"","meta":{"_candela_citation":"[{\"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\":\"Factor a Trinomial With A Common Factor Using the Shortcut Method - 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