{"id":965,"date":"2016-02-15T22:16:24","date_gmt":"2016-02-15T22:16:24","guid":{"rendered":"https:\/\/courses.candelalearning.com\/nrocarithmetic\/?post_type=chapter&p=965"},"modified":"2023-03-15T20:21:41","modified_gmt":"2023-03-15T20:21:41","slug":"5-2-1-factor-trinomials","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-elementaryalgebra\/chapter\/5-2-1-factor-trinomials\/","title":{"raw":"6.2: Factor by Grouping","rendered":"6.2: Factor by Grouping"},"content":{"raw":"
\r\n

section 6.2 Learning Objectives<\/h3>\r\n6.2: Factoring by Grouping<\/strong>\r\n
    \r\n \t
  • Factor a four-term polynomial by grouping<\/li>\r\n<\/ul>\r\n<\/div>\r\n \r\n\r\nWhen we learned to multiply two binomials, we found that the result, before combining like terms, was a four term polynomial, as in this 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 \u00a0to return a four term polynomial to the product of two binomials. Why would we even want to do this?\r\n\r\n[caption id=\"attachment_4825\" align=\"aligncenter\" width=\"391\"]\"Thought Why Should I Care?[\/caption]\r\n\r\nBecause it is an important step in learning techniques for factoring trinomials, such as the one you get when you simplify the product of the two binomials from above:\r\n

    [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, factoring by grouping is a technique that allows us to factor a polynomial whose terms don't all share a GCF. In the following example, we will introduce you to the technique. Remember, one of the main reasons to factor is because it will help solve polynomial equations.\r\n

    \r\n

    Example 1<\/h3>\r\nFactor [latex]a^2+3a+5a+15[\/latex]\r\n[reveal-answer q=\"437455\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"437455\"]\r\n\r\nNotice that there is no GCF other than 1 for all four terms. Resist the temptation to combine the like terms, as instead we are going to group the problem into two pairs of terms and then search for the GCF of each.\r\n

    [latex]\\left(a^2+3a\\right)+\\left(5a+15\\right)[\/latex]<\/p>\r\nFind the GCF of the first pair of terms.\r\n

    [latex]\\begin{array}{l}\\,\\,\\,\\,a^2=a\\cdot{a}\\\\\\,\\,\\,\\,3a=3\\cdot{a}\\\\\\text{GCF}=a\\end{array}[\/latex]<\/p>\r\nFactor the GCF, a<\/i>, out of the first group.\r\n

    [latex]\\begin{array}{r}a\\left(a+3\\right)+\\left(5a+15\\right)\\end{array}[\/latex]<\/p>\r\nFind the GCF of the second pair of terms.\r\n

    [latex]\\begin{array}{r}5a=5\\cdot{a}\\\\15=5\\cdot3\\\\\\text{GCF}=5\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\r\nFactor 5 out of the second group.\r\n

    [latex]\\begin{array}{l}a\\left(a+3\\right)+5\\left(a+3\\right)\\end{array}[\/latex]<\/p>\r\nNotice that the two terms have a common factor [latex]\\left(a+3\\right)[\/latex].\r\n

    [latex]a\\left(a+3\\right)+5\\left(a+3\\right)[\/latex]<\/p>\r\nFactor out the common factor [latex]\\left(a+3\\right)[\/latex] from the two terms.\r\n

    [latex]\\left(a+3\\right)\\left(a+5\\right)[\/latex]<\/p>\r\nNote how the [latex]a[\/latex] and [latex]5[\/latex] become a binomial sum, along with the common factor of [latex](a+3)[\/latex] in front.\r\n

    Answer<\/h4>\r\n[latex]a^2+3a+5a+15=\\left(a+3\\right)\\left(a+5\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nNotice that with this process, we have been able to factor a four-term polynomial into the product of two binomials.\r\n\r\nThis process is called the grouping technique<\/i> . Broken down into individual steps, here's how we apply this technique to a four-term polynomial (you can also follow this process in the example below).\r\n
      \r\n \t
    • Group the terms of the polynomial into pairs.<\/li>\r\n \t
    • Find the greatest common factor of each pair and factor it out.<\/li>\r\n \t
    • Look for the common binomial\u00a0between the factored terms<\/li>\r\n \t
    • Factor the common binomial\u00a0out of the groups, the other factors will make the other binomial<\/li>\r\n<\/ul>\r\nIt is worth noting that grouping can be used for polynomials with more than four terms, but the steps above and our examples will focus on four-term polynomials.\u00a0 Let\u2019s try factoring a few more. Note how there is a now a constant in front of the [latex]x^2[\/latex] term. We will just consider this another factor when we are finding the GCF.\r\n
      \r\n

      Example 2<\/h3>\r\nFactor [latex]2x^{2}+4x+5x+10[\/latex].\r\n[reveal-answer q=\"313122\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"313122\"]Group terms of the polynomial into pairs.\r\n

      [latex]\\left(2x^{2}+4x\\right)+\\left(5x+10\\right)[\/latex]<\/p>\r\nFactor out the like factor, 2x<\/i>, from the first group.\r\n

      [latex]2x\\left(x+2\\right)+\\left(5x+10\\right)[\/latex]<\/p>\r\nFactor out the like factor, 5, from the second group.\r\n

      [latex]2x\\left(x+2\\right)+5\\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

      [latex]\\left(x+2\\right)\\left(2x+5\\right)[\/latex]<\/p>\r\nThe polynomial is now factored.\r\n

      Answer<\/h4>\r\n[latex]2x^{2}+4x+5x+10=\\left(x+2\\right)\\left(2x+5\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\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
      \r\n

      Example 3<\/h3>\r\nFactor [latex]2x^{2}\u20133x+8x\u201312[\/latex].\r\n[reveal-answer q=\"715080\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"715080\"]Group terms into pairs.\r\n

      [latex](2x^{2}\u20133x)+(8x\u201312)[\/latex]<\/p>\r\nFactor the common factor, x<\/i>, out of the first group and the common factor, 4, out of the second group.\r\n

      [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

      [latex]\\left(x+4\\right)\\left(2x\u20133\\right)[\/latex]<\/p>\r\n\r\n

      Answer<\/h4>\r\n[latex]2x^{2}\u20133x+8x\u201312=\\left(x+4\\right)\\left(2x-3\\right)[\/latex]\r\n\r\n[\/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 factor out a negative with the GCF. So what is the difference here compared to the last example. Recall that we looked at an example of this in Section 6.1, where we decided it can be useful to factor out the negative when it is part of the leading coefficient. As we group in the next example, the negative will be \"in front\" of the second pair. Factoring out the negative plays an important role in grouping as the resulting binomials will not match if we fail to factor out the negative.\r\n
      \r\n

      Example 4<\/h3>\r\nFactor [latex]3x^{2}+3x\u20132x\u20132[\/latex].\r\n[reveal-answer q=\"744005\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"744005\"]Group terms into pairs.\r\n

      [latex]\\left(3x^{2}+3x\\right)+\\left(-2x-2\\right)[\/latex]<\/p>\r\nFactor the common factor 3x<\/i> out of first group.\r\n

      [latex]3x\\left(x+1\\right)+\\left(-2x-2\\right)[\/latex]<\/p>\r\nFactor the common factor [latex]\u22122[\/latex] out of the second group. Notice what happens to the signs within the parentheses once [latex]\u22122[\/latex] is factored out.\r\n

      [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], from both terms.\r\n

      [latex]\\left(x+1\\right)\\left(3x-2\\right)[\/latex]<\/p>\r\n\r\n

      Answer<\/h4>\r\n[latex]3x^{2}+3x\u20132x\u20132=\\left(x+1\\right)\\left(3x-2\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video we present another example of factoring by grouping when the GCF of one of the pairs of terms is negative.\r\n\r\nhttps:\/\/youtu.be\/0dvGmDGVC5U\r\n\r\nSo far, every polynomial we have factored has had degree 2. The next example shows that we can still try applying the grouping technique to a four-term polynomial regardless of degree.\r\n
      \r\n

      Example 5<\/h3>\r\nFactor [latex]4x^{3}+12x^{2}+x+3[\/latex].\r\n\r\n[reveal-answer q=\"44633\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44633\"]\r\n\r\nGroup the expression into pairs.\r\n

      [latex]\\left(4x^{3}+12x^{2}\\right)+\\left(x+3\\right)[\/latex]<\/p>\r\nFactor the common factor of [latex]4x^{2}[\/latex] out of the first group.\r\n

      [latex]4x^{2}\\left(x+3\\right)+\\left(x+3\\right)[\/latex]<\/p>\r\nThis is the first time we have encountered a group where the only factor in common is 1. It is recommended that we write the 1, as this will make the proceeding step more obvious.\r\n

      [latex]4x^{2}\\left(x+3\\right)+1\\left(x+3\\right)[\/latex]<\/p>\r\nIt is now more apparent that there is a common factor of [latex](x+3)[\/latex] and that the remaining binomial will be the sum of [latex]4x^{2}[\/latex] and 1.\r\n

      [latex]\\left(x+3\\right)\\left(4x^{2}+1\\right)[\/latex]<\/p>\r\n\r\n

      Answer<\/h4>\r\n[latex]4x^{3}+12x^{2}+x+3=\\left(x+3\\right)\\left(4x^{2}+1\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nAnother feature we may encounter is a polynomial with more than one variable, as shown in the next two examples.\r\n
      \r\n

      Example 6<\/h3>\r\nFactor [latex]6pq+9qr-2p-3r[\/latex]\r\n\r\n[reveal-answer q=\"393796\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"393796\"]\r\n\r\nGroup the expression into pairs.\r\n

      [latex](6pq+9qr)+(-2p-3r)[\/latex]<\/p>\r\nFactor out the GCF of each pair of terms. For the first pair, the GCF is [latex]3q[\/latex]. For the second pair, there appears to be nothing in common, but we will factor out [latex]-1[\/latex] since the negative is \"in front.\"\r\n

      [latex]3q(2p+3r)-1(2p+3r)[\/latex]<\/p>\r\nFactor out the common factor of [latex](2p+3r)[\/latex].\r\n

      [latex](2p+3r)(3q-1)[\/latex]<\/p>\r\n\r\n

      Answer<\/h4>\r\n[latex]6pq+9qr-2p-3r=(2p+3r)(3q-1)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
      \r\n

      Example 7<\/h3>\r\nFactor [latex]x^2y-xy^2-2x+2y[\/latex]\r\n\r\n[reveal-answer q=\"917147\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"917147\"]\r\n\r\nGroup the expression into pairs.\r\n

      [latex](x^2y-xy^2)+(-2x+2y)[\/latex]<\/p>\r\nFactor out the GCF of each pair of terms. For the first pair, the GCF is [latex]xy[\/latex], while for the second pair, we factor out a GCF of [latex]-2[\/latex] (remembering that we want to include the negative when it is part of the first term).\r\n

      [latex]xy(x-y)-2(x-y)[\/latex]<\/p>\r\nFactor out the common factor of [latex](x-y)[\/latex].\r\n

      [latex](x-y)(xy-2)[\/latex]<\/p>\r\n\r\n

      Answer<\/h4>\r\n[latex]x^2y-xy^2-2x+2y=(x-y)(xy-2)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n \r\n\r\nSometimes, you will encounter polynomials that, despite your best efforts, cannot be factored into the product of two binomials.\r\n
      \r\n

      Example 8<\/h3>\r\nFactor [latex]7x^{2}\u201321x+5x\u20135[\/latex].\r\n[reveal-answer q=\"262926\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"262926\"]Group terms into pairs.\r\n

      [latex]\\left(7x^{2}\u201321x\\right)+\\left(5x\u20135\\right)[\/latex]<\/p>\r\nFactor the common factor 7x<\/i> out of the first group.\r\n

      [latex]7x\\left(x-3\\right)+\\left(5x-5\\right)[\/latex]<\/p>\r\nFactor the common factor 5 out of the second group.\r\n

      [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, so this polynomial cannot be factored.\r\n

      [latex]7x\\left(x\u20133\\right)+5\\left(x\u20131\\right)[\/latex]<\/p>\r\n\r\n

      Answer<\/h4>\r\nCannot be factored (prime)\r\n\r\n[\/hidden-answer]\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! Despite factoring each pair, that does not result in a factoring of the original polynomial. Don't forget that \"to factor\" means to \"express as a product.\" Where we got stuck, the two separate pieces were still attached by addition. That is why the final answer is \"prime.\"\r\n\r\nNow that you are an expert on grouping, we will see in the next two sections how this technique can help us to factor trinomials.\r\n

      <\/h2>","rendered":"
      \n

      section 6.2 Learning Objectives<\/h3>\n

      6.2: Factoring by Grouping<\/strong><\/p>\n

        \n
      • Factor a four-term polynomial by grouping<\/li>\n<\/ul>\n<\/div>\n

         <\/p>\n

        When we learned to multiply two binomials, we found that the result, before combining like terms, was a four term polynomial, as in this example:\u00a0[latex]\\left(x+4\\right)\\left(x+2\\right)=x^{2}+2x+4x+8[\/latex].<\/p>\n

        We can apply what we have learned about factoring out a common monomial \u00a0to return a four term polynomial to the product of two binomials. Why would we even want to do this?<\/p>\n

        \"Thought<\/p>\n

        Why Should I Care?<\/p>\n<\/div>\n

        Because it is an important step in learning techniques for factoring trinomials, such as the one you get when you simplify the product of the two binomials from above:<\/p>\n

        [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

        Additionally, factoring by grouping is a technique that allows us to factor a polynomial whose terms don’t all share a GCF. In the following example, we will introduce you to the technique. Remember, one of the main reasons to factor is because it will help solve polynomial equations.<\/p>\n

        \n

        Example 1<\/h3>\n

        Factor [latex]a^2+3a+5a+15[\/latex]<\/p>\n

        Show Solution<\/span><\/p>\n
        \n

        Notice that there is no GCF other than 1 for all four terms. Resist the temptation to combine the like terms, as instead we are going to group the problem into two pairs of terms and then search for the GCF of each.<\/p>\n

        [latex]\\left(a^2+3a\\right)+\\left(5a+15\\right)[\/latex]<\/p>\n

        Find the GCF of the first pair of terms.<\/p>\n

        [latex]\\begin{array}{l}\\,\\,\\,\\,a^2=a\\cdot{a}\\\\\\,\\,\\,\\,3a=3\\cdot{a}\\\\\\text{GCF}=a\\end{array}[\/latex]<\/p>\n

        Factor the GCF, a<\/i>, out of the first group.<\/p>\n

        [latex]\\begin{array}{r}a\\left(a+3\\right)+\\left(5a+15\\right)\\end{array}[\/latex]<\/p>\n

        Find the GCF of the second pair of terms.<\/p>\n

        [latex]\\begin{array}{r}5a=5\\cdot{a}\\\\15=5\\cdot3\\\\\\text{GCF}=5\\,\\,\\,\\,\\,\\,\\,\\end{array}[\/latex]<\/p>\n

        Factor 5 out of the second group.<\/p>\n

        [latex]\\begin{array}{l}a\\left(a+3\\right)+5\\left(a+3\\right)\\end{array}[\/latex]<\/p>\n

        Notice that the two terms have a common factor [latex]\\left(a+3\\right)[\/latex].<\/p>\n

        [latex]a\\left(a+3\\right)+5\\left(a+3\\right)[\/latex]<\/p>\n

        Factor out the common factor [latex]\\left(a+3\\right)[\/latex] from the two terms.<\/p>\n

        [latex]\\left(a+3\\right)\\left(a+5\\right)[\/latex]<\/p>\n

        Note how the [latex]a[\/latex] and [latex]5[\/latex] become a binomial sum, along with the common factor of [latex](a+3)[\/latex] in front.<\/p>\n

        Answer<\/h4>\n

        [latex]a^2+3a+5a+15=\\left(a+3\\right)\\left(a+5\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n

        Notice that with this process, we have been able to factor a four-term polynomial into the product of two binomials.<\/p>\n

        This process is called the grouping technique<\/i> . Broken down into individual steps, here’s how we apply this technique to a four-term polynomial (you can also follow this process in the example below).<\/p>\n

          \n
        • Group the terms of the polynomial into pairs.<\/li>\n
        • Find the greatest common factor of each pair and factor it out.<\/li>\n
        • Look for the common binomial\u00a0between the factored terms<\/li>\n
        • Factor the common binomial\u00a0out of the groups, the other factors will make the other binomial<\/li>\n<\/ul>\n

          It is worth noting that grouping can be used for polynomials with more than four terms, but the steps above and our examples will focus on four-term polynomials.\u00a0 Let\u2019s try factoring a few more. Note how there is a now a constant in front of the [latex]x^2[\/latex] term. We will just consider this another factor when we are finding the GCF.<\/p>\n

          \n

          Example 2<\/h3>\n

          Factor [latex]2x^{2}+4x+5x+10[\/latex].<\/p>\n

          Show Solution<\/span><\/p>\n
          Group terms of the polynomial into pairs.<\/p>\n

          [latex]\\left(2x^{2}+4x\\right)+\\left(5x+10\\right)[\/latex]<\/p>\n

          Factor out the like factor, 2x<\/i>, from the first group.<\/p>\n

          [latex]2x\\left(x+2\\right)+\\left(5x+10\\right)[\/latex]<\/p>\n

          Factor out the like factor, 5, from the second group.<\/p>\n

          [latex]2x\\left(x+2\\right)+5\\left(x+2\\right)[\/latex]<\/p>\n

          Look for common factors between the factored forms of the paired terms. Here, the common factor is [latex](x+2)[\/latex].<\/p>\n

          Factor out the common factor, [latex]\\left(x+2\\right)[\/latex], from both terms.<\/p>\n

          [latex]\\left(x+2\\right)\\left(2x+5\\right)[\/latex]<\/p>\n

          The polynomial is now factored.<\/p>\n

          Answer<\/h4>\n

          [latex]2x^{2}+4x+5x+10=\\left(x+2\\right)\\left(2x+5\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n

          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

          \n

          Example 3<\/h3>\n

          Factor [latex]2x^{2}\u20133x+8x\u201312[\/latex].<\/p>\n

          Show Solution<\/span><\/p>\n
          Group terms into pairs.<\/p>\n

          [latex](2x^{2}\u20133x)+(8x\u201312)[\/latex]<\/p>\n

          Factor the common factor, x<\/i>, out of the first group and the common factor, 4, out of the second group.<\/p>\n

          [latex]x\\left(2x\u20133\\right)+4\\left(2x\u20133\\right)[\/latex]<\/p>\n

          Factor out the common factor, [latex]\\left(2x\u20133\\right)[\/latex], from both terms.<\/p>\n

          [latex]\\left(x+4\\right)\\left(2x\u20133\\right)[\/latex]<\/p>\n

          Answer<\/h4>\n

          [latex]2x^{2}\u20133x+8x\u201312=\\left(x+4\\right)\\left(2x-3\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n

          The video that follows provides another example of factoring by grouping.<\/p>\n