{"id":854,"date":"2024-04-29T22:37:00","date_gmt":"2024-04-29T22:37:00","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/?post_type=chapter&p=854"},"modified":"2025-02-18T15:56:56","modified_gmt":"2025-02-18T15:56:56","slug":"factoring-polynomials-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/chapter\/factoring-polynomials-learn-it-2\/","title":{"raw":"Factoring Polynomials: Learn It 2","rendered":"Factoring Polynomials: Learn It 2"},"content":{"raw":"

Factoring a Trinomial with Leading Coefficient of 1<\/h2>\r\nAlthough we should always begin by looking for a GCF, pulling out the GCF is not the only way that polynomial expressions can be factored. The polynomial [latex]{x}^{2}+5x+6[\/latex] has a GCF of 1, but it can be written as the product of the factors [latex]\\left(x+2\\right)[\/latex] and [latex]\\left(x+3\\right)[\/latex].\r\n\r\n
\r\n

Factoring a Trinomial (Leading Coefficient\u00a0 [latex]= 1[\/latex])<\/h3>\r\nA trinomial of the form [latex]{x}^{2}+bx+c[\/latex] can be written in factored form as [latex]\\left(x+p\\right)\\left(x+q\\right)[\/latex] where [latex]pq=c[\/latex] and [latex]p+q=b[\/latex].\r\n\r\n<\/section>
Some polynomials cannot be factored. These polynomials are said to be prime.<\/section>
How To: Given a trinomial in the form [latex]{x}^{2}+bx+c[\/latex], factor it<\/strong>\r\n
    \r\n \t
  1. List factors of [latex]c[\/latex].<\/li>\r\n \t
  2. Find [latex]p[\/latex] and [latex]q[\/latex], a pair of factors of [latex]c[\/latex] with a sum of [latex]b[\/latex].<\/li>\r\n \t
  3. Write the factored expression [latex]\\left(x+p\\right)\\left(x+q\\right)[\/latex].<\/li>\r\n<\/ol>\r\n<\/section>
    Factor [latex]{x}^{2}+2x - 15[\/latex].\r\n\r\n
    \r\n\r\nSolution<\/strong>\r\n
      \r\n \t
    • We have a trinomial with leading coefficient [latex]1,b=2[\/latex], and [latex]c=-15[\/latex].<\/li>\r\n \t
    • We need to find two numbers with a product of [latex]-15[\/latex] and a sum of [latex]2[\/latex].<\/li>\r\n<\/ul>\r\n

      In the table, we list factors until we find a pair with the desired sum.<\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
      Factors of [latex]-15[\/latex]<\/th>\r\nSum of Factors<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
      [latex]1,-15[\/latex]<\/td>\r\n[latex]-14[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]-1,15[\/latex]<\/td>\r\n[latex]14[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]3,-5[\/latex]<\/td>\r\n[latex]-2[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]-3,5[\/latex]<\/strong><\/td>\r\n[latex]2[\/latex]<\/strong><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n
        \r\n \t
      • Now that we have identified [latex]p[\/latex] and [latex]q[\/latex] as [latex]-3[\/latex] and [latex]5[\/latex], write the factored form as [latex]\\left(x - 3\\right)\\left(x+5\\right)[\/latex].<\/li>\r\n<\/ul>\r\nThus, [latex]{x}^{2}+2x - 15 = \\left(x - 3\\right)\\left(x+5\\right)[\/latex]\r\n\r\n[reveal-answer q=\"523566\"]Does the order of the factors matter?[\/reveal-answer]\r\n[hidden-answer a=\"523566\"]No. Multiplication is commutative, so the order of the factors does not matter. [latex]{x}^{2}+2x - 15 = \\left(x - 3\\right)\\left(x+5\\right) = \\left(x+5\\right)\\left(x - 3\\right)[\/latex][\/hidden-answer]\r\n\r\n<\/section>
        [ohm2_question hide_question_numbers=1]18881[\/ohm2_question]<\/section>\r\n

        Factoring by Grouping (Factoring a Trinomial with Leading Coefficient of Not 1)<\/h2>\r\nTrinomials with leading coefficients other than 1 are slightly more complicated to factor. For these trinomials, we can factor by grouping<\/strong> by dividing the [latex]x[\/latex] term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression.\r\n\r\n
        \r\n

        Factoring Trinomial (Leading Coefficient [latex]\\ne 1[\/latex])<\/h3>\r\nFactoring by grouping<\/strong> is a method used to decompose a trinomial of the form [latex]ax^2+bx+c[\/latex] into a product of two binomials.\r\n\r\n \r\n\r\nThis method involves finding two numbers that combine to give the product of the leading coefficient and the constant term ([latex]a \\times c[\/latex]) and the sum of the middle coefficient ([latex]b[\/latex]). We use these numbers to divide the [latex]x[\/latex] term into the sum of two terms and factor each portion of the expression separately then factor out the GCF of the entire expression.\r\n\r\n<\/section>
        How To: Given a trinomial in the form [latex]a{x}^{2}+bx+c[\/latex], factor by grouping<\/strong>\r\n
          \r\n \t
        1. List factors of [latex]a \\times c[\/latex].<\/li>\r\n \t
        2. Find [latex]p[\/latex] and [latex]q[\/latex], a pair of factors of [latex]a \\times c[\/latex] with a sum of [latex]b[\/latex].<\/li>\r\n \t
        3. Rewrite the original expression as [latex]a{x}^{2}+px+qx+c[\/latex].<\/li>\r\n \t
        4. Pull out the GCF of [latex]a{x}^{2}+px[\/latex].<\/li>\r\n \t
        5. Pull out the GCF of [latex]qx+c[\/latex].<\/li>\r\n \t
        6. Factor out the GCF of the expression.<\/li>\r\n<\/ol>\r\n<\/section>
          Factor [latex]5{x}^{2}+7x - 6[\/latex].\r\n\r\n
          \r\n\r\nSolution<\/strong>\r\n
            \r\n \t
          • List factors of [latex]a \\times c[\/latex].\r\n
              \r\n \t
            • Calculate [latex]a \\times c[\/latex]: [latex]a = 5[\/latex] and [latex]c = -6[\/latex], so [latex]a \\times c = -30[\/latex].<\/li>\r\n \t
            • Factors of [latex]-30[\/latex]:\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
              Factors of [latex]-30[\/latex]<\/th>\r\nSum of Factors<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
              [latex]1,-30[\/latex]<\/td>\r\n[latex]-29[\/latex]<\/td>\r\n<\/tr>\r\n
              [latex]-1,30[\/latex]<\/td>\r\n[latex]29[\/latex]<\/td>\r\n<\/tr>\r\n
              [latex]2,-15[\/latex]<\/td>\r\n[latex]-13[\/latex]<\/td>\r\n<\/tr>\r\n
              [latex]-2,15[\/latex]<\/td>\r\n[latex]13[\/latex]<\/td>\r\n<\/tr>\r\n
              [latex]3,-10[\/latex]<\/td>\r\n[latex]-7[\/latex]<\/td>\r\n<\/tr>\r\n
              [latex]-3,10[\/latex]<\/td>\r\n[latex]7[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t
            • Find [latex]p[\/latex] and [latex]q[\/latex], a pair of factors of [latex]a \\times c[\/latex] with a sum of [latex]b[\/latex].\r\n
                \r\n \t
              • Based on the table above, the correct pair is [latex]p = -3[\/latex] and [latex]q = 10[\/latex].<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t
              • Rewrite the original expression as [latex]a{x}^{2}+px+qx+c[\/latex].<\/li>\r\n<\/ul>\r\n

                [latex]5{x}^{2}+7x - 6 = 5{x}^{2} -3x+10x-6[\/latex]<\/p>\r\n\r\n

                  \r\n \t
                • Group the first 2 terms and the last 2 terms. Then, pull out the GCF of each group.<\/li>\r\n<\/ul>\r\n

                  [latex]5{x}^{2}+7x - 6 = (5{x}^{2} -3x)+(10x-6) = x(5x-3)+2(5x-3)[\/latex]<\/p>\r\n\r\n

                    \r\n \t
                  • Factor out the GCF of the expression.<\/li>\r\n<\/ul>\r\n

                    [latex]5{x}^{2}+7x - 6 = (x+2)(5x-3)[\/latex]<\/p>\r\n\r\n<\/section>

                    Factor the following.\r\n
                      \r\n \t
                    1. [latex]2{x}^{2}+9x+9[\/latex]<\/li>\r\n \t
                    2. [latex]6{x}^{2}+x - 1[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"343485\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"343485\"]\r\n
                        \r\n \t
                      1. [latex]\\left(2x+3\\right)\\left(x+3\\right)[\/latex]<\/li>\r\n \t
                      2. [latex]\\left(3x - 1\\right)\\left(2x+1\\right)[\/latex][\/hidden-answer]<\/li>\r\n<\/ol>\r\n<\/section>
                        [ohm2_question hide_question_numbers=1]18883[\/ohm2_question]<\/section>
                        [ohm2_question hide_question_numbers=1]18882[\/ohm2_question]<\/section>","rendered":"

                        Factoring a Trinomial with Leading Coefficient of 1<\/h2>\n

                        Although we should always begin by looking for a GCF, pulling out the GCF is not the only way that polynomial expressions can be factored. The polynomial [latex]{x}^{2}+5x+6[\/latex] has a GCF of 1, but it can be written as the product of the factors [latex]\\left(x+2\\right)[\/latex] and [latex]\\left(x+3\\right)[\/latex].<\/p>\n

                        \n

                        Factoring a Trinomial (Leading Coefficient\u00a0 [latex]= 1[\/latex])<\/h3>\n

                        A trinomial of the form [latex]{x}^{2}+bx+c[\/latex] can be written in factored form as [latex]\\left(x+p\\right)\\left(x+q\\right)[\/latex] where [latex]pq=c[\/latex] and [latex]p+q=b[\/latex].<\/p>\n<\/section>\n

                        Some polynomials cannot be factored. These polynomials are said to be prime.<\/section>\n
                        How To: Given a trinomial in the form [latex]{x}^{2}+bx+c[\/latex], factor it<\/strong><\/p>\n
                          \n
                        1. List factors of [latex]c[\/latex].<\/li>\n
                        2. Find [latex]p[\/latex] and [latex]q[\/latex], a pair of factors of [latex]c[\/latex] with a sum of [latex]b[\/latex].<\/li>\n
                        3. Write the factored expression [latex]\\left(x+p\\right)\\left(x+q\\right)[\/latex].<\/li>\n<\/ol>\n<\/section>\n
                          Factor [latex]{x}^{2}+2x - 15[\/latex].<\/p>\n
                          \n

                          Solution<\/strong><\/p>\n

                            \n
                          • We have a trinomial with leading coefficient [latex]1,b=2[\/latex], and [latex]c=-15[\/latex].<\/li>\n
                          • We need to find two numbers with a product of [latex]-15[\/latex] and a sum of [latex]2[\/latex].<\/li>\n<\/ul>\n

                            In the table, we list factors until we find a pair with the desired sum.<\/p>\n\n\n\n\n\n\n\n\n
                            Factors of [latex]-15[\/latex]<\/th>\nSum of Factors<\/th>\n<\/tr>\n<\/thead>\n
                            [latex]1,-15[\/latex]<\/td>\n[latex]-14[\/latex]<\/td>\n<\/tr>\n
                            [latex]-1,15[\/latex]<\/td>\n[latex]14[\/latex]<\/td>\n<\/tr>\n
                            [latex]3,-5[\/latex]<\/td>\n[latex]-2[\/latex]<\/td>\n<\/tr>\n
                            [latex]-3,5[\/latex]<\/strong><\/td>\n[latex]2[\/latex]<\/strong><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n
                              \n
                            • Now that we have identified [latex]p[\/latex] and [latex]q[\/latex] as [latex]-3[\/latex] and [latex]5[\/latex], write the factored form as [latex]\\left(x - 3\\right)\\left(x+5\\right)[\/latex].<\/li>\n<\/ul>\n

                              Thus, [latex]{x}^{2}+2x - 15 = \\left(x - 3\\right)\\left(x+5\\right)[\/latex]<\/p>\n