{"id":7723,"date":"2020-12-13T17:42:16","date_gmt":"2020-12-13T17:42:16","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/slcc-beginalgebra\/?post_type=chapter&p=7723"},"modified":"2023-05-23T02:22:26","modified_gmt":"2023-05-23T02:22:26","slug":"6-6-summary-of-factoring","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-elementaryalgebra\/chapter\/6-6-summary-of-factoring\/","title":{"raw":"6.6: Summary of Factoring","rendered":"6.6: Summary of Factoring"},"content":{"raw":"
\r\n

section 6.6 Learning Objectives<\/h3>\r\n6.6: Summary of Factoring<\/strong>\r\n
    \r\n \t
  • Review the factoring methods presented in this module<\/li>\r\n \t
  • Be able to recognize and apply an appropriate factoring technique to a given problem<\/li>\r\n \t
  • Factor expressions completely<\/li>\r\n<\/ul>\r\n<\/div>\r\n \r\n\r\nIn the previous sections, you have learned several factoring techniques. Now it is time to put it all together.\r\n\r\nIn this section, we focus on two important questions you should ask yourself when encountering any factoring problem:\r\n
      \r\n \t
    1. Which factoring technique should I use for this problem?<\/li>\r\n \t
    2. Can I factor the polynomial more?<\/li>\r\n<\/ol>\r\n

      Choosing the Best Factoring Technique<\/h2>\r\nHere, we present a strategy you can apply to any factoring problem.\r\n
      \r\n

      Factoring strategy<\/h3>\r\n1.<\/strong> If there is a GCF other than [latex]1[\/latex], factor it out first. Don't forget to factor out a [latex]-1[\/latex] if the leading coefficient is negative.\r\n\r\n2<\/strong>. Count the number of terms in the remaining polynomial and select an appropriate technique.\r\n

      I.<\/strong>\u00a0 4 or More Terms:<\/strong>\u00a0 Factor by Grouping (Section 6.2)<\/p>\r\n

      II.<\/strong>\u00a0 3 Terms:<\/strong><\/p>\r\n

      A.<\/strong> If [latex]a=1[\/latex], apply the \"Product and Sum Method\" (Section 6.3)<\/p>\r\n

      B.<\/strong> If [latex]a\\neq 1[\/latex], apply the \"AC-Method\" (Section 6.4)<\/p>\r\n

      C.<\/strong> If it is a perfect square trinomial, use the appropriate formula: [latex]a^2+2ab+b^2=(a+b)^2[\/latex] or [latex]a^2-2ab+b^2=(a-b)^2[\/latex] (Section 6.5)<\/p>\r\n

      III.<\/strong>\u00a0 2 Terms:<\/strong> If the binomial is a difference of squares, use the following formula: [latex]a^2-b^2=(a+b)(a-b)[\/latex] (Section 6.5)<\/p>\r\n3.<\/strong> If none of the above applies, it is possible that the polynomial is not factorable, or \"prime.\"\r\n\r\n<\/div>\r\nLet us try some examples.\r\n

      \r\n

      Example 1<\/h3>\r\nFactor [latex]-3x^2-3x+6[\/latex].\r\n\r\n[reveal-answer q=\"561709\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"561709\"]\r\n\r\nWe first check for a common factor, recognizing that all terms share a factor of [latex]3[\/latex]. Moreover, since the leading coefficient is negative, we will factor out a GCF of [latex]-3[\/latex].\r\n

      [latex]-3(x^2+x-2)[\/latex]<\/p>\r\nNext, we examine the resulting polynomial. It has 3 terms and a leading coefficient of [latex]a=1[\/latex]. So, we can use the Product and Sum Method, searching for two numbers [latex]r[\/latex] and [latex]s[\/latex] that multiply to [latex]c=-2[\/latex] and add to [latex]b=1[\/latex], which are [latex]2[\/latex] and [latex]-1[\/latex].\r\n

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

      Answer<\/span><\/h4>\r\n[latex]-3x^2-3x+6=-3(x+2)(x-1)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
      \r\n

      Example 2<\/h3>\r\nFactor [latex]-3x^2-7x+6[\/latex].\r\n\r\n[reveal-answer q=\"485884\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"485884\"]\r\n\r\nInitially, this looks similar to the previous example. However, this time [latex]3[\/latex] is no longer in common. We will still begin by factoring out [latex]-1[\/latex] though.\r\n

      [latex]-1(3x^2+7x-6)[\/latex]<\/p>\r\nThe resulting polynomial has 3 terms, but a leading coefficient other than 1. So, we use the AC-Method. We need two numbers that multiply to [latex]ac=3(-6)=-18[\/latex] and add to [latex]b=7[\/latex], which are [latex]9[\/latex] and [latex]-2[\/latex].\r\n\r\nRewrite the middle term [latex]7x[\/latex] as [latex]9x-2x[\/latex] and apply the grouping technique.\r\n

      [latex]-(3x^2+9x-2x-6)[\/latex]<\/p>\r\n

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

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

      Answer<\/span><\/h4>\r\n[latex]-3x^2-7x+6=-(x+3)(3x-2)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe next example includes a perfect square trinomial.\r\n
      \r\n

      Example 3<\/h3>\r\nFactor [latex]12x^5+60x^4+75x^3[\/latex]\r\n\r\n[reveal-answer q=\"204805\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"204805\"]\r\n\r\nWe first factor out the GCF of [latex] 3x^3[\/latex].\r\n

      [latex]3x^3(4x^2+20x+25)[\/latex]<\/p>\r\nWe could proceed by applying the AC-Method to the resulting trinomial with leading coefficient other than 1, but let us check whether the result is a Perfect Square Trinomial. Comparing to [latex]a^2+2ab+b^2[\/latex], we have [latex]a=2x[\/latex] and[latex]b=5[\/latex]. Since [latex]2ab=2(2x)(5)=20[\/latex] matches our middle term, we can apply the formula\u00a0[latex]a^2+2ab+b^2=(a+b)^2[\/latex] to factor as\r\n

      [latex]3x^3(2x+5)^2[\/latex]<\/p>\r\n\r\n

      Answer<\/span><\/h4>\r\n[latex]12x^5+60x^4+75x^3=3x^3(2x+5)^2[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n \r\n\r\nIn the next two examples, we review factoring binomials.\r\n
      \r\n

      Example 4<\/h3>\r\nFactor [latex]-2x^2+8[\/latex]\r\n\r\n[reveal-answer q=\"111134\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"111134\"]\r\n\r\nWe first factor out the GCF. Noting that the leading coefficient is negative, we will factor out [latex]-2[\/latex].\r\n

      [latex]-2(x^2-4)[\/latex]<\/p>\r\nThe resulting binomial is a difference of squares, which we factor as\r\n

      [latex]-2(x+2)(x-2)[\/latex].<\/p>\r\n\r\n

      Answer<\/span><\/h4>\r\n[latex]-2x^2+8=-2(x+2)(x-2)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n \r\n
      \r\n

      Example 5<\/h3>\r\nFactor [latex]50x^{2}y^{3}-8y[\/latex]\r\n\r\n[reveal-answer q=\"269108\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"269108\"]\r\n\r\nOnce again, we start by looking for a common factor. We can factor out a GCF of [latex]2y[\/latex].\r\n

      [latex]2y(25x^{2}y^{2}-4)[\/latex]<\/p>\r\nThe resulting polynomial has 2 terms and is in the form of a difference of squares with [latex]a=5xy[\/latex] and [latex]b=2[\/latex].\r\n

      [latex]2y(5xy+2)(5xy-2)[\/latex]<\/p>\r\n\r\n

      Answer<\/span><\/h4>\r\n[latex]50x^{2}y^{3}-8y=2y(5xy+2)(5xy-2)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nDon't forget that we cannot factor every polynomial.\r\n
      \r\n

      Example 6<\/h3>\r\nFactor [latex]5x^2+8x+4[\/latex]\r\n\r\n[reveal-answer q=\"305813\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"305813\"]\r\n\r\nSince there are 3 terms and the leading coefficient is [latex]a=5[\/latex] (which is not a common factor), we attempt to use the AC-Method. We need two numbers that multiply to [latex]ac=5(4)=20[\/latex] and add to [latex]b=8[\/latex]. However, two such integers do not exist. We conclude that the polynomial is not factorable.\r\n

      Answer<\/span><\/h4>\r\nPrime\r\n\r\nNote: This example reminds of of the important role that signs play in this process. You may be tempted to conclude that [latex]10[\/latex] and [latex]2[\/latex] is a potential pair of numbers that would work for this problem. However, to multiply to positive [latex]20[\/latex], the numbers must have the same sign (both positive or both negative), and there is no pair of numbers that will then also add to [latex]8[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n \r\n

      Factoring More<\/h2>\r\nSometimes after we factor, one or more of the resulting polynomials can be factored even further. You will see this more in future classes, but present one example here for you to think about.\r\n
      \r\n

      think about it<\/h3>\r\nFactor completely: [latex]x^4-16[\/latex]\r\n\r\n[reveal-answer q=\"475929\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"475929\"]\r\n\r\nApplying our factoring strategy, we first see that there is no common factor other than 1. Moving onto step 2, the polynomial has two terms and is a difference of squares since it can be written as\r\n

      [latex](x^2)^2-4^2[\/latex]<\/p>\r\nThus, in the difference of square equation, [latex]a=x^2[\/latex] and [latex]b=4[\/latex], which leads to the following factored form:\r\n

      [latex](x^2+4)(x^2-4)[\/latex]<\/p>\r\nWe now have two binomials and should consider whether either of these can be factored using our same factoring strategy again. The first binomial, [latex]x^2+4[\/latex] is a sum of squares, which is prime. But, [latex]x^2-4[\/latex] is another difference of squares, this time with [latex]a=x[\/latex] and [latex]b=2[\/latex]. This leads us to\r\n

      [latex](x^2+4)(x+2)(x-2)[\/latex]<\/p>\r\nWe now have three binomials, but none are the difference of squares and therefore we do not know how to factor them any further.\r\n

      Answer<\/span><\/h4>\r\n[latex]x^4-16=(x^2+4)(x+2)(x-2)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n \r\n\r\nNow that you have mastered several factoring techniques, we finally explore some applications of factoring in the next section.","rendered":"
      \n

      section 6.6 Learning Objectives<\/h3>\n

      6.6: Summary of Factoring<\/strong><\/p>\n

        \n
      • Review the factoring methods presented in this module<\/li>\n
      • Be able to recognize and apply an appropriate factoring technique to a given problem<\/li>\n
      • Factor expressions completely<\/li>\n<\/ul>\n<\/div>\n

         <\/p>\n

        In the previous sections, you have learned several factoring techniques. Now it is time to put it all together.<\/p>\n

        In this section, we focus on two important questions you should ask yourself when encountering any factoring problem:<\/p>\n

          \n
        1. Which factoring technique should I use for this problem?<\/li>\n
        2. Can I factor the polynomial more?<\/li>\n<\/ol>\n

          Choosing the Best Factoring Technique<\/h2>\n

          Here, we present a strategy you can apply to any factoring problem.<\/p>\n

          \n

          Factoring strategy<\/h3>\n

          1.<\/strong> If there is a GCF other than [latex]1[\/latex], factor it out first. Don’t forget to factor out a [latex]-1[\/latex] if the leading coefficient is negative.<\/p>\n

          2<\/strong>. Count the number of terms in the remaining polynomial and select an appropriate technique.<\/p>\n

          I.<\/strong>\u00a0 4 or More Terms:<\/strong>\u00a0 Factor by Grouping (Section 6.2)<\/p>\n

          II.<\/strong>\u00a0 3 Terms:<\/strong><\/p>\n

          A.<\/strong> If [latex]a=1[\/latex], apply the “Product and Sum Method” (Section 6.3)<\/p>\n

          B.<\/strong> If [latex]a\\neq 1[\/latex], apply the “AC-Method” (Section 6.4)<\/p>\n

          C.<\/strong> If it is a perfect square trinomial, use the appropriate formula: [latex]a^2+2ab+b^2=(a+b)^2[\/latex] or [latex]a^2-2ab+b^2=(a-b)^2[\/latex] (Section 6.5)<\/p>\n

          III.<\/strong>\u00a0 2 Terms:<\/strong> If the binomial is a difference of squares, use the following formula: [latex]a^2-b^2=(a+b)(a-b)[\/latex] (Section 6.5)<\/p>\n

          3.<\/strong> If none of the above applies, it is possible that the polynomial is not factorable, or “prime.”<\/p>\n<\/div>\n

          Let us try some examples.<\/p>\n

          \n

          Example 1<\/h3>\n

          Factor [latex]-3x^2-3x+6[\/latex].<\/p>\n

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

          We first check for a common factor, recognizing that all terms share a factor of [latex]3[\/latex]. Moreover, since the leading coefficient is negative, we will factor out a GCF of [latex]-3[\/latex].<\/p>\n

          [latex]-3(x^2+x-2)[\/latex]<\/p>\n

          Next, we examine the resulting polynomial. It has 3 terms and a leading coefficient of [latex]a=1[\/latex]. So, we can use the Product and Sum Method, searching for two numbers [latex]r[\/latex] and [latex]s[\/latex] that multiply to [latex]c=-2[\/latex] and add to [latex]b=1[\/latex], which are [latex]2[\/latex] and [latex]-1[\/latex].<\/p>\n

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

          Answer<\/span><\/h4>\n

          [latex]-3x^2-3x+6=-3(x+2)(x-1)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n

          \n

          Example 2<\/h3>\n

          Factor [latex]-3x^2-7x+6[\/latex].<\/p>\n

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

          Initially, this looks similar to the previous example. However, this time [latex]3[\/latex] is no longer in common. We will still begin by factoring out [latex]-1[\/latex] though.<\/p>\n

          [latex]-1(3x^2+7x-6)[\/latex]<\/p>\n

          The resulting polynomial has 3 terms, but a leading coefficient other than 1. So, we use the AC-Method. We need two numbers that multiply to [latex]ac=3(-6)=-18[\/latex] and add to [latex]b=7[\/latex], which are [latex]9[\/latex] and [latex]-2[\/latex].<\/p>\n

          Rewrite the middle term [latex]7x[\/latex] as [latex]9x-2x[\/latex] and apply the grouping technique.<\/p>\n

          [latex]-(3x^2+9x-2x-6)[\/latex]<\/p>\n

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

          [latex]=-(x+3)(3x-2)[\/latex]<\/p>\n

          Answer<\/span><\/h4>\n

          [latex]-3x^2-7x+6=-(x+3)(3x-2)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n

          The next example includes a perfect square trinomial.<\/p>\n

          \n

          Example 3<\/h3>\n

          Factor [latex]12x^5+60x^4+75x^3[\/latex]<\/p>\n

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

          We first factor out the GCF of [latex]3x^3[\/latex].<\/p>\n

          [latex]3x^3(4x^2+20x+25)[\/latex]<\/p>\n

          We could proceed by applying the AC-Method to the resulting trinomial with leading coefficient other than 1, but let us check whether the result is a Perfect Square Trinomial. Comparing to [latex]a^2+2ab+b^2[\/latex], we have [latex]a=2x[\/latex] and[latex]b=5[\/latex]. Since [latex]2ab=2(2x)(5)=20[\/latex] matches our middle term, we can apply the formula\u00a0[latex]a^2+2ab+b^2=(a+b)^2[\/latex] to factor as<\/p>\n

          [latex]3x^3(2x+5)^2[\/latex]<\/p>\n

          Answer<\/span><\/h4>\n

          [latex]12x^5+60x^4+75x^3=3x^3(2x+5)^2[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n

           <\/p>\n

          In the next two examples, we review factoring binomials.<\/p>\n

          \n

          Example 4<\/h3>\n

          Factor [latex]-2x^2+8[\/latex]<\/p>\n

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

          We first factor out the GCF. Noting that the leading coefficient is negative, we will factor out [latex]-2[\/latex].<\/p>\n

          [latex]-2(x^2-4)[\/latex]<\/p>\n

          The resulting binomial is a difference of squares, which we factor as<\/p>\n

          [latex]-2(x+2)(x-2)[\/latex].<\/p>\n

          Answer<\/span><\/h4>\n

          [latex]-2x^2+8=-2(x+2)(x-2)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n

           <\/p>\n

          \n

          Example 5<\/h3>\n

          Factor [latex]50x^{2}y^{3}-8y[\/latex]<\/p>\n

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

          Once again, we start by looking for a common factor. We can factor out a GCF of [latex]2y[\/latex].<\/p>\n

          [latex]2y(25x^{2}y^{2}-4)[\/latex]<\/p>\n

          The resulting polynomial has 2 terms and is in the form of a difference of squares with [latex]a=5xy[\/latex] and [latex]b=2[\/latex].<\/p>\n

          [latex]2y(5xy+2)(5xy-2)[\/latex]<\/p>\n

          Answer<\/span><\/h4>\n

          [latex]50x^{2}y^{3}-8y=2y(5xy+2)(5xy-2)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n

          Don’t forget that we cannot factor every polynomial.<\/p>\n

          \n

          Example 6<\/h3>\n

          Factor [latex]5x^2+8x+4[\/latex]<\/p>\n

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

          Since there are 3 terms and the leading coefficient is [latex]a=5[\/latex] (which is not a common factor), we attempt to use the AC-Method. We need two numbers that multiply to [latex]ac=5(4)=20[\/latex] and add to [latex]b=8[\/latex]. However, two such integers do not exist. We conclude that the polynomial is not factorable.<\/p>\n

          Answer<\/span><\/h4>\n

          Prime<\/p>\n

          Note: This example reminds of of the important role that signs play in this process. You may be tempted to conclude that [latex]10[\/latex] and [latex]2[\/latex] is a potential pair of numbers that would work for this problem. However, to multiply to positive [latex]20[\/latex], the numbers must have the same sign (both positive or both negative), and there is no pair of numbers that will then also add to [latex]8[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n

           <\/p>\n

          Factoring More<\/h2>\n

          Sometimes after we factor, one or more of the resulting polynomials can be factored even further. You will see this more in future classes, but present one example here for you to think about.<\/p>\n

          \n

          think about it<\/h3>\n

          Factor completely: [latex]x^4-16[\/latex]<\/p>\n

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

          Applying our factoring strategy, we first see that there is no common factor other than 1. Moving onto step 2, the polynomial has two terms and is a difference of squares since it can be written as<\/p>\n

          [latex](x^2)^2-4^2[\/latex]<\/p>\n

          Thus, in the difference of square equation, [latex]a=x^2[\/latex] and [latex]b=4[\/latex], which leads to the following factored form:<\/p>\n

          [latex](x^2+4)(x^2-4)[\/latex]<\/p>\n

          We now have two binomials and should consider whether either of these can be factored using our same factoring strategy again. The first binomial, [latex]x^2+4[\/latex] is a sum of squares, which is prime. But, [latex]x^2-4[\/latex] is another difference of squares, this time with [latex]a=x[\/latex] and [latex]b=2[\/latex]. This leads us to<\/p>\n

          [latex](x^2+4)(x+2)(x-2)[\/latex]<\/p>\n

          We now have three binomials, but none are the difference of squares and therefore we do not know how to factor them any further.<\/p>\n

          Answer<\/span><\/h4>\n

          [latex]x^4-16=(x^2+4)(x+2)(x-2)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n

           <\/p>\n

          Now that you have mastered several factoring techniques, we finally explore some applications of factoring in the next section.<\/p>\n","protected":false},"author":348856,"menu_order":7,"template":"","meta":{"_candela_citation":"[]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-7723","chapter","type-chapter","status-publish","hentry"],"part":949,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-elementaryalgebra\/wp-json\/pressbooks\/v2\/chapters\/7723","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-elementaryalgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-elementaryalgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-elementaryalgebra\/wp-json\/wp\/v2\/users\/348856"}],"version-history":[{"count":29,"href":"https:\/\/courses.lumenlearning.com\/slcc-elementaryalgebra\/wp-json\/pressbooks\/v2\/chapters\/7723\/revisions"}],"predecessor-version":[{"id":9458,"href":"https:\/\/courses.lumenlearning.com\/slcc-elementaryalgebra\/wp-json\/pressbooks\/v2\/chapters\/7723\/revisions\/9458"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-elementaryalgebra\/wp-json\/pressbooks\/v2\/parts\/949"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-elementaryalgebra\/wp-json\/pressbooks\/v2\/chapters\/7723\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-elementaryalgebra\/wp-json\/wp\/v2\/media?parent=7723"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-elementaryalgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=7723"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-elementaryalgebra\/wp-json\/wp\/v2\/contributor?post=7723"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-elementaryalgebra\/wp-json\/wp\/v2\/license?post=7723"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}