{"id":363,"date":"2021-06-04T00:05:51","date_gmt":"2021-06-04T00:05:51","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/chapter\/5-2-1-factor-trinomials\/"},"modified":"2022-01-04T23:55:23","modified_gmt":"2022-01-04T23:55:23","slug":"5-2-1-factor-trinomials","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/chapter\/5-2-1-factor-trinomials\/","title":{"raw":"8.6.2: Factoring by Grouping","rendered":"8.6.2: Factoring by Grouping"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Factor a four term polynomial by grouping terms<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Key words<\/h3>\r\n<ul>\r\n \t<li><strong>Prime polynomial<\/strong>: a polynomial that does not factor; it is divisible by only 1 and itself.<\/li>\r\n \t<li><strong>Factoring by grouping<\/strong>:\u00a0a technique to factor polynomials with four or more terms<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Factoring by Grouping<\/h2>\r\nWhen we multiply two binomials, the result, before combining like terms, is a four term polynomial. For example:\u00a0[latex]\\left(x+4\\right)\\left(x+2\\right)=x^{2}+2x+4x+8[\/latex].\r\n\r\nWe can apply what we have learned about factoring out a common monomial to help us factor a four-term polynomial into the product of two binomials.\r\n\r\n<img class=\"alignright wp-image-4825\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/06\/12195953\/Screen-Shot-2016-06-12-at-12.59.11-PM-300x138.png\" alt=\"Thought bubble with the words .....and i should care why?\" width=\"391\" height=\"180\" \/>\r\n\r\n<span style=\"font-size: 1em;\">Why would we even want to do this?\u00a0<\/span>Because it is an important step in learning techniques for factoring trinomials, such as the one we get when we simplify the product of the two binomials from above:\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\left(x+4\\right)\\left(x+2\\right)\\\\=x^{2}+2x+4x+8\\\\=x^2+6x+8\\end{array}[\/latex]<\/p>\r\nAdditionally, <em><strong>factoring by grouping<\/strong><\/em> is a technique that allows us to factor a polynomial whose terms don't all share a GCF.\u00a0 We will introduce the technique in the following example. Remember, one of the main reasons to factor is because it will help us solve polynomial equations.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFactor [latex]a^2+3a+5a+15[\/latex]\r\n<h4>Solution<\/h4>\r\nThe only common factor between all four terms is [latex]1[\/latex], so we will <strong>group the terms into pairs<\/strong>\u00a0and find a GCF for each pair.\r\n<p style=\"text-align: center;\">[latex]\\left(a^2+3a\\right)+\\left(5a+15\\right)[\/latex]<\/p>\r\nThe GCF for the first pair is [latex]a[\/latex] and the GCF for the second pair is [latex]5[\/latex].\r\n\r\nFactor each pair using the GCF:\r\n<p style=\"text-align: center;\">[latex](a^2+3a)+(5a+15)\\\\=a(a+3)+5(a+3)[\/latex]<\/p>\r\nNotice that the two terms now have a common factor of [latex]\\left(a+3\\right)[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\color{red}{a}\\color{blue}{\\left(a+3\\right)}\\color{green}{+5}\\color{blue}{\\left(a+3\\right)}[\/latex]<\/p>\r\nFactor out the common factor [latex]\\left(a+3\\right)[\/latex] from the two terms.\r\n<p style=\"text-align: center;\">[latex]\\color{blue}{\\left(a+3\\right)}\\left(\\color{red}{a}\\color{green}{+5}\\right)[\/latex]<\/p>\r\nNote how the [latex]a[\/latex] and [latex]5[\/latex] become a binomial sum, and the other factor.\r\n<h4>Answer<\/h4>\r\n[latex]a^2+3a+5a+15=\\left(a+3\\right)\\left(a+5\\right)[\/latex]\r\n\r\n<\/div>\r\nThis process is called <em><strong>factoring by\u00a0<\/strong><\/em><i><strong>grouping<\/strong>.<\/i>\u00a0Broken down into individual steps, here's how to do it (you can also follow this process in the example below).\r\n<div class=\"textbox shaded\">\r\n<h3>FACTOR BY GROUPING<\/h3>\r\n<ul>\r\n \t<li>Group the terms of the polynomial into pairs that share a GCF.<\/li>\r\n \t<li>Find the greatest common factor of each pair and use the distributive property to factor each pair<\/li>\r\n \t<li>Look for a common binomial\u00a0between the factored terms<\/li>\r\n \t<li>Factor the common binomial\u00a0out of the groups, the other factors will make the other binomial<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFactor [latex]2x^{2}+4x+5x+10[\/latex].\r\n<h4>Solution<\/h4>\r\nGroup terms of the polynomial into pairs:\r\n<p style=\"text-align: center;\">[latex]\\left(2x^{2}+4x\\right)+\\left(5x+10\\right)[\/latex]<\/p>\r\nFactor out the like factor, [latex]2x[\/latex], from the first group, and<span style=\"font-size: 1rem; text-align: initial;\">\u00a0[latex]5[\/latex] from the second group.<\/span>\r\n<p style=\"text-align: center;\">[latex]2x\\color{blue}{\\left(x+2\\right)}+5\\color{blue}{\\left(x+2\\right)}[\/latex]<\/p>\r\nLook for common factors between the factored forms of the paired terms. Here, the common factor is [latex](x+2)[\/latex].\r\n\r\nFactor out the common factor, [latex]\\left(x+2\\right)[\/latex], from both terms.\r\n<p style=\"text-align: center;\">[latex]\\color{blue}{\\left(x+2\\right)}\\left(2x+5\\right)[\/latex]<\/p>\r\nThe polynomial is now factored.\r\n<h4>Answer<\/h4>\r\n[latex]\\left(x+2\\right)\\left(2x+5\\right)[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\nFactor [latex]3x^{2}+6x+7x+14[\/latex].\r\n\r\n[reveal-answer q=\"hjm647\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm647\"][latex](x+2)(3x+7)[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nAnother example follows that contains subtraction. Note how we choose a positive GCF\u00a0from each group of terms, and the negative signs stay.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFactor [latex]2x^{2}\u20133x+8x\u201312[\/latex].\r\n<h4>Solution<\/h4>\r\nGroup terms into pairs:\r\n<p style=\"text-align: center;\">[latex](2x^{2}\u20133x)+(8x\u201312)[\/latex]<\/p>\r\nFactor the common factor, [latex]x[\/latex], out of the first group and the common factor, [latex]4[\/latex], out of the second group.\r\n<p style=\"text-align: center;\">[latex]x\\left(2x\u20133\\right)+4\\left(2x\u20133\\right)[\/latex]<\/p>\r\nFactor out the common factor, [latex]\\left(2x\u20133\\right)[\/latex], from both terms.\r\n<p style=\"text-align: center;\">[latex]\\left(x+4\\right)\\left(2x\u20133\\right)[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]\\left(x+4\\right)\\left(2x-3\\right)[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\nFactor [latex]2x^{2}\u20133x+10x\u201315[\/latex].\r\n\r\n[reveal-answer q=\"hjm910\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm910\"][latex](2x-3)(x+5)[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\nThe video that follows provides another example of factoring by grouping.\r\n\r\nhttps:\/\/youtu.be\/RR5nj7RFSiU\r\n\r\nIn the next example, we will have a GCF that is negative. \u00a0It is important to pay attention to what happens to the resulting binomial when the GCF is negative.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFactor [latex]3x^{2}+3x\u20132x\u20132[\/latex].\r\n<h4>Solution<\/h4>\r\nGroup terms into pairs:\r\n<p style=\"text-align: center;\">[latex]\\left(3x^{2}+3x\\right)+\\left(-2x-2\\right)[\/latex]<\/p>\r\nFactor the common factor \u00a0[latex]3x[\/latex] out of first group, and<span style=\"font-size: 1rem; text-align: initial;\">\u00a0[latex]\u22122[\/latex] out of the second group. Notice what happens to the signs within the parentheses once [latex]\u22122[\/latex] is factored out.<\/span>\r\n<p style=\"text-align: center;\">[latex]3x\\left(x+1\\right)-2\\left(x+1\\right)[\/latex]<\/p>\r\nFactor out the common factor, [latex]\\left(x+1\\right)[\/latex]:\r\n<p style=\"text-align: center;\">[latex]\\left(x+1\\right)\\left(3x-2\\right)[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\n[latex]\\left(x+1\\right)\\left(3x-2\\right)[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\nFactor [latex]5x^{2}+5x\u20137x\u20137[\/latex].\r\n\r\n[reveal-answer q=\"hjm755\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm755\"][latex](x+1)(5x-7)[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\nThe following video presents another example of factoring by grouping when one of the GCF is negative.\r\n\r\nhttps:\/\/youtu.be\/0dvGmDGVC5U\r\n\r\nSometimes, we will encounter polynomials that, despite our best efforts, cannot be factored into the product of two binomials.\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Example<\/h3>\r\nFactor [latex]7x^{2}\u201321x+5x\u20135[\/latex].\r\n<h4>Solution<\/h4>\r\nGroup terms into pairs:\r\n<p style=\"text-align: center;\">[latex]\\left(7x^{2}\u201321x\\right)+\\left(5x\u20135\\right)[\/latex]<\/p>\r\nFactor the common factor [latex]7x[\/latex] out of the first group, and\u00a0<span style=\"font-size: 1rem; text-align: initial;\">the common factor [latex]5[\/latex] out of the second group:<\/span>\r\n<p style=\"text-align: center;\">[latex]7x\\left(x-3\\right)+5\\left(x-1\\right)[\/latex]<\/p>\r\nThe two groups [latex]7x\\left(x\u20133\\right)[\/latex] and [latex]5\\left(x\u20131\\right)[\/latex] do not have any common factors, other than 1, so this polynomial cannot be factored any further.\r\n<p style=\"text-align: center;\">[latex]7x\\left(x\u20133\\right)+5\\left(x\u20131\\right)[\/latex]<\/p>\r\n\r\n<h4>Answer<\/h4>\r\nCannot be factored\r\n\r\n<\/div>\r\nIn the example above, each pair can be factored, but then there is no common factor between the pairs! This means that the polynomial cannot be factored. In this case, the polynomial is said to be <em><strong>prime<\/strong><\/em>.\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\n1. Factor [latex]-2x^{2}\u20132x+5x+5[\/latex].\r\n\r\n[reveal-answer q=\"hjm009\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm009\"][latex](-2x+5)(x+1)[\/latex][\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n2. Factor [latex]6x^2+3x-8x-4[\/latex]\r\n\r\n[reveal-answer q=\"hjm154\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm154\"][latex](3x-4)(2x+1)[\/latex][\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n3. Factor [latex]x^2-3x+x-3[\/latex]\r\n\r\n[reveal-answer q=\"hjm991\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm991\"][latex](x-3)(x+1)[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\nIn the next section, we will see how factoring by grouping can be used to factor a trinomial.","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Factor a four term polynomial by grouping terms<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Key words<\/h3>\n<ul>\n<li><strong>Prime polynomial<\/strong>: a polynomial that does not factor; it is divisible by only 1 and itself.<\/li>\n<li><strong>Factoring by grouping<\/strong>:\u00a0a technique to factor polynomials with four or more terms<\/li>\n<\/ul>\n<\/div>\n<h2>Factoring by Grouping<\/h2>\n<p>When we multiply two binomials, the result, before combining like terms, is a four term polynomial. For example:\u00a0[latex]\\left(x+4\\right)\\left(x+2\\right)=x^{2}+2x+4x+8[\/latex].<\/p>\n<p>We can apply what we have learned about factoring out a common monomial to help us factor a four-term polynomial into the product of two binomials.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignright wp-image-4825\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/117\/2016\/06\/12195953\/Screen-Shot-2016-06-12-at-12.59.11-PM-300x138.png\" alt=\"Thought bubble with the words .....and i should care why?\" width=\"391\" height=\"180\" \/><\/p>\n<p><span style=\"font-size: 1em;\">Why would we even want to do this?\u00a0<\/span>Because it is an important step in learning techniques for factoring trinomials, such as the one we get when we simplify the product of the two binomials from above:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}\\left(x+4\\right)\\left(x+2\\right)\\\\=x^{2}+2x+4x+8\\\\=x^2+6x+8\\end{array}[\/latex]<\/p>\n<p>Additionally, <em><strong>factoring by grouping<\/strong><\/em> is a technique that allows us to factor a polynomial whose terms don&#8217;t all share a GCF.\u00a0 We will introduce the technique in the following example. Remember, one of the main reasons to factor is because it will help us solve polynomial equations.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Factor [latex]a^2+3a+5a+15[\/latex]<\/p>\n<h4>Solution<\/h4>\n<p>The only common factor between all four terms is [latex]1[\/latex], so we will <strong>group the terms into pairs<\/strong>\u00a0and find a GCF for each pair.<\/p>\n<p style=\"text-align: center;\">[latex]\\left(a^2+3a\\right)+\\left(5a+15\\right)[\/latex]<\/p>\n<p>The GCF for the first pair is [latex]a[\/latex] and the GCF for the second pair is [latex]5[\/latex].<\/p>\n<p>Factor each pair using the GCF:<\/p>\n<p style=\"text-align: center;\">[latex](a^2+3a)+(5a+15)\\\\=a(a+3)+5(a+3)[\/latex]<\/p>\n<p>Notice that the two terms now have a common factor of [latex]\\left(a+3\\right)[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\color{red}{a}\\color{blue}{\\left(a+3\\right)}\\color{green}{+5}\\color{blue}{\\left(a+3\\right)}[\/latex]<\/p>\n<p>Factor out the common factor [latex]\\left(a+3\\right)[\/latex] from the two terms.<\/p>\n<p style=\"text-align: center;\">[latex]\\color{blue}{\\left(a+3\\right)}\\left(\\color{red}{a}\\color{green}{+5}\\right)[\/latex]<\/p>\n<p>Note how the [latex]a[\/latex] and [latex]5[\/latex] become a binomial sum, and the other factor.<\/p>\n<h4>Answer<\/h4>\n<p>[latex]a^2+3a+5a+15=\\left(a+3\\right)\\left(a+5\\right)[\/latex]<\/p>\n<\/div>\n<p>This process is called <em><strong>factoring by\u00a0<\/strong><\/em><i><strong>grouping<\/strong>.<\/i>\u00a0Broken down into individual steps, here&#8217;s how to do it (you can also follow this process in the example below).<\/p>\n<div class=\"textbox shaded\">\n<h3>FACTOR BY GROUPING<\/h3>\n<ul>\n<li>Group the terms of the polynomial into pairs that share a GCF.<\/li>\n<li>Find the greatest common factor of each pair and use the distributive property to factor each pair<\/li>\n<li>Look for a common binomial\u00a0between the factored terms<\/li>\n<li>Factor the common binomial\u00a0out of the groups, the other factors will make the other binomial<\/li>\n<\/ul>\n<\/div>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Factor [latex]2x^{2}+4x+5x+10[\/latex].<\/p>\n<h4>Solution<\/h4>\n<p>Group terms of the polynomial into pairs:<\/p>\n<p style=\"text-align: center;\">[latex]\\left(2x^{2}+4x\\right)+\\left(5x+10\\right)[\/latex]<\/p>\n<p>Factor out the like factor, [latex]2x[\/latex], from the first group, and<span style=\"font-size: 1rem; text-align: initial;\">\u00a0[latex]5[\/latex] from the second group.<\/span><\/p>\n<p style=\"text-align: center;\">[latex]2x\\color{blue}{\\left(x+2\\right)}+5\\color{blue}{\\left(x+2\\right)}[\/latex]<\/p>\n<p>Look for common factors between the factored forms of the paired terms. Here, the common factor is [latex](x+2)[\/latex].<\/p>\n<p>Factor out the common factor, [latex]\\left(x+2\\right)[\/latex], from both terms.<\/p>\n<p style=\"text-align: center;\">[latex]\\color{blue}{\\left(x+2\\right)}\\left(2x+5\\right)[\/latex]<\/p>\n<p>The polynomial is now factored.<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\left(x+2\\right)\\left(2x+5\\right)[\/latex]<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p>Factor [latex]3x^{2}+6x+7x+14[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm647\">Show Answer<\/span><\/p>\n<div id=\"qhjm647\" class=\"hidden-answer\" style=\"display: none\">[latex](x+2)(3x+7)[\/latex]<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>Another example follows that contains subtraction. Note how we choose a positive GCF\u00a0from each group of terms, and the negative signs stay.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Factor [latex]2x^{2}\u20133x+8x\u201312[\/latex].<\/p>\n<h4>Solution<\/h4>\n<p>Group terms into pairs:<\/p>\n<p style=\"text-align: center;\">[latex](2x^{2}\u20133x)+(8x\u201312)[\/latex]<\/p>\n<p>Factor the common factor, [latex]x[\/latex], out of the first group and the common factor, [latex]4[\/latex], out of the second group.<\/p>\n<p style=\"text-align: center;\">[latex]x\\left(2x\u20133\\right)+4\\left(2x\u20133\\right)[\/latex]<\/p>\n<p>Factor out the common factor, [latex]\\left(2x\u20133\\right)[\/latex], from both terms.<\/p>\n<p style=\"text-align: center;\">[latex]\\left(x+4\\right)\\left(2x\u20133\\right)[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\left(x+4\\right)\\left(2x-3\\right)[\/latex]<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p>Factor [latex]2x^{2}\u20133x+10x\u201315[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm910\">Show Answer<\/span><\/p>\n<div id=\"qhjm910\" class=\"hidden-answer\" style=\"display: none\">[latex](2x-3)(x+5)[\/latex]<\/div>\n<\/div>\n<\/div>\n<p>The video that follows provides another example of factoring by grouping.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex 2: Intro to Factor By Grouping Technique\" width=\"500\" height=\"375\" src=\"https:\/\/www.youtube.com\/embed\/RR5nj7RFSiU?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>In the next example, we will have a GCF that is negative. \u00a0It is important to pay attention to what happens to the resulting binomial when the GCF is negative.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Factor [latex]3x^{2}+3x\u20132x\u20132[\/latex].<\/p>\n<h4>Solution<\/h4>\n<p>Group terms into pairs:<\/p>\n<p style=\"text-align: center;\">[latex]\\left(3x^{2}+3x\\right)+\\left(-2x-2\\right)[\/latex]<\/p>\n<p>Factor the common factor \u00a0[latex]3x[\/latex] out of first group, and<span style=\"font-size: 1rem; text-align: initial;\">\u00a0[latex]\u22122[\/latex] out of the second group. Notice what happens to the signs within the parentheses once [latex]\u22122[\/latex] is factored out.<\/span><\/p>\n<p style=\"text-align: center;\">[latex]3x\\left(x+1\\right)-2\\left(x+1\\right)[\/latex]<\/p>\n<p>Factor out the common factor, [latex]\\left(x+1\\right)[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\left(x+1\\right)\\left(3x-2\\right)[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>[latex]\\left(x+1\\right)\\left(3x-2\\right)[\/latex]<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p>Factor [latex]5x^{2}+5x\u20137x\u20137[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm755\">Show Answer<\/span><\/p>\n<div id=\"qhjm755\" class=\"hidden-answer\" style=\"display: none\">[latex](x+1)(5x-7)[\/latex]<\/div>\n<\/div>\n<\/div>\n<p>The following video presents another example of factoring by grouping when one of the GCF is negative.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex 1: Intro to Factor By Grouping Technique\" width=\"500\" height=\"375\" src=\"https:\/\/www.youtube.com\/embed\/0dvGmDGVC5U?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>Sometimes, we will encounter polynomials that, despite our best efforts, cannot be factored into the product of two binomials.<\/p>\n<div class=\"bcc-box bcc-info\">\n<h3>Example<\/h3>\n<p>Factor [latex]7x^{2}\u201321x+5x\u20135[\/latex].<\/p>\n<h4>Solution<\/h4>\n<p>Group terms into pairs:<\/p>\n<p style=\"text-align: center;\">[latex]\\left(7x^{2}\u201321x\\right)+\\left(5x\u20135\\right)[\/latex]<\/p>\n<p>Factor the common factor [latex]7x[\/latex] out of the first group, and\u00a0<span style=\"font-size: 1rem; text-align: initial;\">the common factor [latex]5[\/latex] out of the second group:<\/span><\/p>\n<p style=\"text-align: center;\">[latex]7x\\left(x-3\\right)+5\\left(x-1\\right)[\/latex]<\/p>\n<p>The two groups [latex]7x\\left(x\u20133\\right)[\/latex] and [latex]5\\left(x\u20131\\right)[\/latex] do not have any common factors, other than 1, so this polynomial cannot be factored any further.<\/p>\n<p style=\"text-align: center;\">[latex]7x\\left(x\u20133\\right)+5\\left(x\u20131\\right)[\/latex]<\/p>\n<h4>Answer<\/h4>\n<p>Cannot be factored<\/p>\n<\/div>\n<p>In the example above, each pair can be factored, but then there is no common factor between the pairs! This means that the polynomial cannot be factored. In this case, the polynomial is said to be <em><strong>prime<\/strong><\/em>.<\/p>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p>1. Factor [latex]-2x^{2}\u20132x+5x+5[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm009\">Show Answer<\/span><\/p>\n<div id=\"qhjm009\" class=\"hidden-answer\" style=\"display: none\">[latex](-2x+5)(x+1)[\/latex]<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>2. Factor [latex]6x^2+3x-8x-4[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm154\">Show Answer<\/span><\/p>\n<div id=\"qhjm154\" class=\"hidden-answer\" style=\"display: none\">[latex](3x-4)(2x+1)[\/latex]<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>3. Factor [latex]x^2-3x+x-3[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm991\">Show Answer<\/span><\/p>\n<div id=\"qhjm991\" class=\"hidden-answer\" style=\"display: none\">[latex](x-3)(x+1)[\/latex]<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>In the next section, we will see how factoring by grouping can be used to factor a trinomial.<\/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-363\">\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: Why Should I Care?. <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>Try It hjm991; hjm154; hjm009; hjm755; hjm910; hjm647. <strong>Authored by<\/strong>: Hazel McKenna. <strong>Provided by<\/strong>: Utah Valley University. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Ex 2: Intro to Factor By Grouping Technique. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/RR5nj7RFSiU\">https:\/\/youtu.be\/RR5nj7RFSiU<\/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 1: Intro to Factor By Grouping Technique Mathispower4u . <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/0dvGmDGVC5U\">https:\/\/youtu.be\/0dvGmDGVC5U<\/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>Revised and adapted: 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\/resources\/downloads\/nroc-math-open-textbook-units-1-12-pdf-and-word-formats\/\">http:\/\/nrocnetwork.org\/resources\/downloads\/nroc-math-open-textbook-units-1-12-pdf-and-word-formats\/<\/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":23485,"menu_order":11,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Screenshot: Why Should I Care?\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex 2: Intro to Factor By Grouping Technique\",\"author\":\"James Sousa (Mathispower4u.com) for Lumen Learning\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/RR5nj7RFSiU\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex 1: Intro to Factor By Grouping Technique Mathispower4u \",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/0dvGmDGVC5U\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Revised and adapted: Unit 12: Factoring, from Developmental Math: An Open Program\",\"author\":\"\",\"organization\":\"Monterey Institute of Technology and Education\",\"url\":\" http:\/\/nrocnetwork.org\/resources\/downloads\/nroc-math-open-textbook-units-1-12-pdf-and-word-formats\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Try It hjm991; hjm154; hjm009; hjm755; hjm910; hjm647\",\"author\":\"Hazel McKenna\",\"organization\":\"Utah Valley University\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"8076008d880d480494979e137e1b2ada, d6cdd768d9aa4089a7f8c4ae2cc03db4","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-363","chapter","type-chapter","status-publish","hentry"],"part":663,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/chapters\/363","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/wp\/v2\/users\/23485"}],"version-history":[{"count":6,"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/chapters\/363\/revisions"}],"predecessor-version":[{"id":2059,"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/chapters\/363\/revisions\/2059"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/parts\/663"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/chapters\/363\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/wp\/v2\/media?parent=363"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=363"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/wp\/v2\/contributor?post=363"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/wp-json\/wp\/v2\/license?post=363"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}