{"id":2528,"date":"2024-08-01T22:37:41","date_gmt":"2024-08-01T22:37:41","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/?post_type=chapter&p=2528"},"modified":"2024-11-21T22:36:03","modified_gmt":"2024-11-21T22:36:03","slug":"module-15-background-youll-need-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/chapter\/module-15-background-youll-need-1\/","title":{"raw":"Conic Sections: Background You'll Need 1","rendered":"Conic Sections: Background You’ll Need 1"},"content":{"raw":"
Factor trinomials and perfect square trinomials into binomials.<\/span><\/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].Solution<\/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>
                  \r\n

                  Factoring a Perfect Square Trinomial<\/h2>\r\nA perfect square trinomial is a trinomial that can be written as the square of a binomial. Recall that when a binomial is squared, the result is the square of the first term added to twice the product of the two terms and the square of the last term.\r\n
                  [latex]\\begin{array}{ccc}\\hfill {a}^{2}+2ab+{b}^{2}& =& {\\left(a+b\\right)}^{2}\\hfill \\\\ & \\text{and}& \\\\ \\hfill {a}^{2}-2ab+{b}^{2}& =& {\\left(a-b\\right)}^{2}\\hfill \\end{array}[\/latex]<\/div>\r\n
                  We can use this equation to factor any perfect square trinomial.<\/div>\r\n
                  \r\n

                  perfect square trinomial<\/h3>\r\nA perfect square trinomial can be written as the square of a binomial:\r\n
                  [latex]{a}^{2}+2ab+{b}^{2}={\\left(a+b\\right)}^{2}[\/latex]<\/div>\r\n<\/section>
                  How To: Given a perfect square trinomial, factor it into the square of a binomial\r\n<\/strong>\r\n
                    \r\n \t
                  1. Confirm that the first and last term are perfect squares.<\/li>\r\n \t
                  2. Confirm that the middle term is twice the product of [latex]ab[\/latex].<\/li>\r\n \t
                  3. Write the factored form as [latex]{\\left(a+b\\right)}^{2}[\/latex].<\/li>\r\n<\/ol>\r\n<\/section>
                    Factor [latex]25{x}^{2}+20x+4[\/latex].Solution<\/strong>\r\n\r\nTo factor the quadratic expression [latex]25{x}^{2}+20x+4[\/latex], recognizing it as a perfect square trinomial will streamline the process. This type of expression comes from squaring a binomial and has a special format, [latex]a^2 +2ab+b^2[\/latex], where it can be rewritten as [latex](a+b)^2[\/latex].[latex]\\begin{align*} \\text{Original expression:} & \\quad 25x^2 + 20x + 4 \\\\ \\text{Identify square terms:} & \\quad 25x^2 = (5x)^2 \\quad \\text{and} \\quad 4 = 2^2 \\\\ \\text{Check middle term:} & \\quad 2 \\times 5x \\times 2 = 20x \\\\ \\text{Write as a square of a binomial:} & \\quad (5x + 2)^2 \\end{align*}[\/latex]\r\n\r\n<\/section>
                    Factor [latex]49{x}^{2}-14x+1[\/latex].[reveal-answer q=\"530221\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"530221\"][latex]{\\left(7x - 1\\right)}^{2}[\/latex][\/hidden-answer]<\/section>
                    [ohm2_question hide_question_numbers=1]18884[\/ohm2_question]<\/section>
                    [ohm2_question hide_question_numbers=1]18885[\/ohm2_question]<\/section>\r\n

                    Factoring a Difference of Squares<\/h2>\r\nA difference of squares is a perfect square subtracted from a perfect square. Recall that a difference of squares can be rewritten as factors containing the same terms but opposite signs because the middle terms cancel each other out when the two factors are multiplied.\r\n
                    [latex]{a}^{2}-{b}^{2}=\\left(a+b\\right)\\left(a-b\\right)[\/latex]<\/div>\r\nWe can use this equation to factor any differences of squares.\r\n\r\n
                    \r\n

                    difference of squares<\/h3>\r\nA difference of squares can be rewritten as two factors containing the same terms but opposite signs.\r\n
                    [latex]{a}^{2}-{b}^{2}=\\left(a+b\\right)\\left(a-b\\right)[\/latex]<\/div>\r\n<\/section>
                    How To: Given a difference of squares, factor it into binomials<\/strong>\r\n
                      \r\n \t
                    1. Confirm that the first and last term are perfect squares.<\/li>\r\n \t
                    2. Write the factored form as [latex]\\left(a+b\\right)\\left(a-b\\right)[\/latex].<\/li>\r\n<\/ol>\r\n<\/section>
                      Factor [latex]9{x}^{2}-25[\/latex].Solution<\/strong>\r\n\r\nTo factor the quadratic expression [latex]9{x}^{2}-25[\/latex], we recognize that it is a difference of squares.[latex]\\begin{align*} \\text{Original expression:} & \\quad 9x^2 - 25 \\\\ \\text{Identify square terms:} & \\quad 9x^2 = (3x)^2 \\quad \\text{and} \\quad 25 = 5^2 \\\\ \\text{Apply difference of squares formula:} & \\quad (3x + 5)(3x - 5) \\end{align*}[\/latex]\r\n\r\n<\/section>
                      Factor [latex]81{y}^{2}-100[\/latex].[reveal-answer q=\"979538\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"979538\"][latex]\\left(9y+10\\right)\\left(9y - 10\\right)[\/latex][\/hidden-answer]<\/section>
                      [ohm2_question hide_question_numbers=1]18886[\/ohm2_question]<\/section>
                      [ohm2_question hide_question_numbers=1]18887[\/ohm2_question]<\/section><\/section>","rendered":"
                      Factor trinomials and perfect square trinomials into binomials.<\/span><\/section>\n

                      Factoring by Grouping (Factoring a Trinomial with Leading Coefficient of Not 1)<\/h2>\n

                      Trinomials 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.<\/p>\n

                      \n

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

                      Factoring 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.<\/p>\n

                       <\/p>\n

                      This 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.<\/p>\n<\/section>\n

                      How To: Given a trinomial in the form [latex]a{x}^{2}+bx+c[\/latex], factor by grouping<\/strong><\/p>\n
                        \n
                      1. List factors of [latex]a \\times c[\/latex].<\/li>\n
                      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>\n
                      3. Rewrite the original expression as [latex]a{x}^{2}+px+qx+c[\/latex].<\/li>\n
                      4. Pull out the GCF of [latex]a{x}^{2}+px[\/latex].<\/li>\n
                      5. Pull out the GCF of [latex]qx+c[\/latex].<\/li>\n
                      6. Factor out the GCF of the expression.<\/li>\n<\/ol>\n<\/section>\n
                        Factor [latex]5{x}^{2}+7x - 6[\/latex].Solution<\/strong><\/p>\n
                          \n
                        • List factors of [latex]a \\times c[\/latex].\n
                            \n
                          • Calculate [latex]a \\times c[\/latex]: [latex]a = 5[\/latex] and [latex]c = -6[\/latex], so [latex]a \\times c = -30[\/latex].<\/li>\n
                          • Factors of [latex]-30[\/latex]:
                            \n\n\n\n\n\n\n\n\n\n\n
                            Factors of [latex]-30[\/latex]<\/th>\nSum of Factors<\/th>\n<\/tr>\n<\/thead>\n
                            [latex]1,-30[\/latex]<\/td>\n[latex]-29[\/latex]<\/td>\n<\/tr>\n
                            [latex]-1,30[\/latex]<\/td>\n[latex]29[\/latex]<\/td>\n<\/tr>\n
                            [latex]2,-15[\/latex]<\/td>\n[latex]-13[\/latex]<\/td>\n<\/tr>\n
                            [latex]-2,15[\/latex]<\/td>\n[latex]13[\/latex]<\/td>\n<\/tr>\n
                            [latex]3,-10[\/latex]<\/td>\n[latex]-7[\/latex]<\/td>\n<\/tr>\n
                            [latex]-3,10[\/latex]<\/td>\n[latex]7[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/li>\n<\/ul>\n<\/li>\n
                          • 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].\n
                              \n
                            • Based on the table above, the correct pair is [latex]p = -3[\/latex] and [latex]q = 10[\/latex].<\/li>\n<\/ul>\n<\/li>\n
                            • Rewrite the original expression as [latex]a{x}^{2}+px+qx+c[\/latex].<\/li>\n<\/ul>\n

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

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

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

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

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

                                  Factor the following.<\/p>\n
                                    \n
                                  1. [latex]2{x}^{2}+9x+9[\/latex]<\/li>\n
                                  2. [latex]6{x}^{2}+x - 1[\/latex]<\/li>\n<\/ol>\n