{"id":307,"date":"2015-09-18T21:09:38","date_gmt":"2015-09-18T21:09:38","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=307"},"modified":"2015-11-03T23:01:55","modified_gmt":"2015-11-03T23:01:55","slug":"factoring-by-grouping","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/chapter\/factoring-by-grouping\/","title":{"raw":"Factoring by Grouping","rendered":"Factoring by Grouping"},"content":{"raw":"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 x<\/em> term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression. The trinomial [latex]2{x}^{2}+5x+3[\/latex] can be rewritten as [latex]\\left(2x+3\\right)\\left(x+1\\right)[\/latex] using this process. We begin by rewriting the original expression as [latex]2{x}^{2}+2x+3x+3[\/latex] and then factor each portion of the expression to obtain [latex]2x\\left(x+1\\right)+3\\left(x+1\\right)[\/latex]. We then pull out the GCF of [latex]\\left(x+1\\right)[\/latex] to find the factored expression.\r\n
\r\n

A General Note: Factor by Grouping<\/h3>\r\nTo factor a trinomial in the form [latex]a{x}^{2}+bx+c[\/latex] by grouping, we find two numbers with a product of [latex]ac[\/latex] and a sum of [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<\/div>\r\n
\r\n

How To: Given a trinomial in the form [latex]a{x}^{2}+bx+c[\/latex], factor by grouping.<\/h3>\r\n
    \r\n\t
  1. List factors of [latex]ac[\/latex].<\/li>\r\n\t
  2. Find [latex]p[\/latex] and [latex]q[\/latex], a pair of factors of [latex]ac[\/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<\/div>\r\n
    \r\n

    Example 3: Factoring a Trinomial by Grouping<\/h3>\r\nFactor [latex]5{x}^{2}+7x - 6[\/latex] by grouping.\r\n\r\n<\/div>\r\n
    \r\n

    Solution<\/h3>\r\nWe have a trinomial with [latex]a=5,b=7[\/latex], and [latex]c=-6[\/latex]. First, determine [latex]ac=-30[\/latex]. We need to find two numbers with a product of [latex]-30[\/latex] and a sum of [latex]7[\/latex]. In the table, we list factors until we find a pair with the desired sum.\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\n29<\/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\n13<\/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\n7<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSo [latex]p=-3[\/latex] and [latex]q=10[\/latex].\r\n
    [latex]\\begin{array}{cc}5{x}^{2}-3x+10x - 6 \\hfill & \\text{Rewrite the original expression as }a{x}^{2}+px+qx+c.\\hfill \\\\ x\\left(5x - 3\\right)+2\\left(5x - 3\\right)\\hfill & \\text{Factor out the GCF of each part}.\\hfill \\\\ \\left(5x - 3\\right)\\left(x+2\\right)\\hfill & \\text{Factor out the GCF}\\text{ }\\text{ of the expression}.\\hfill \\end{array}[\/latex]<\/div>\r\n<\/div>\r\n
    \r\n

    Analysis of the Solution<\/h3>\r\nWe can check our work by multiplying. Use FOIL to confirm that [latex]\\left(5x - 3\\right)\\left(x+2\\right)=5{x}^{2}+7x - 6[\/latex].\r\n\r\n<\/div>\r\n
    \r\n

    Try It 3<\/h3>\r\nFactor the following.\r\n

    a.[latex]2{x}^{2}+9x+9[\/latex]\r\nb. [latex]6{x}^{2}+x - 1[\/latex]<\/p>\r\nSolution<\/a>\r\n\r\n<\/div>","rendered":"

    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 x<\/em> term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression. The trinomial [latex]2{x}^{2}+5x+3[\/latex] can be rewritten as [latex]\\left(2x+3\\right)\\left(x+1\\right)[\/latex] using this process. We begin by rewriting the original expression as [latex]2{x}^{2}+2x+3x+3[\/latex] and then factor each portion of the expression to obtain [latex]2x\\left(x+1\\right)+3\\left(x+1\\right)[\/latex]. We then pull out the GCF of [latex]\\left(x+1\\right)[\/latex] to find the factored expression.<\/p>\n

    \n

    A General Note: Factor by Grouping<\/h3>\n

    To factor a trinomial in the form [latex]a{x}^{2}+bx+c[\/latex] by grouping, we find two numbers with a product of [latex]ac[\/latex] and a sum of [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<\/div>\n

    \n

    How To: Given a trinomial in the form [latex]a{x}^{2}+bx+c[\/latex], factor by grouping.<\/h3>\n
      \n
    1. List factors of [latex]ac[\/latex].<\/li>\n
    2. Find [latex]p[\/latex] and [latex]q[\/latex], a pair of factors of [latex]ac[\/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<\/div>\n
      \n

      Example 3: Factoring a Trinomial by Grouping<\/h3>\n

      Factor [latex]5{x}^{2}+7x - 6[\/latex] by grouping.<\/p>\n<\/div>\n

      \n

      Solution<\/h3>\n

      We have a trinomial with [latex]a=5,b=7[\/latex], and [latex]c=-6[\/latex]. First, determine [latex]ac=-30[\/latex]. We need to find two numbers with a product of [latex]-30[\/latex] and a sum of [latex]7[\/latex]. 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\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>\n29<\/td>\n<\/tr>\n
      [latex]2,-15[\/latex]<\/td>\n[latex]-13[\/latex]<\/td>\n<\/tr>\n
      [latex]-2,15[\/latex]<\/td>\n13<\/td>\n<\/tr>\n
      [latex]3,-10[\/latex]<\/td>\n[latex]-7[\/latex]<\/td>\n<\/tr>\n
      [latex]-3,10[\/latex]<\/td>\n7<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

      So [latex]p=-3[\/latex] and [latex]q=10[\/latex].<\/p>\n

      [latex]\\begin{array}{cc}5{x}^{2}-3x+10x - 6 \\hfill & \\text{Rewrite the original expression as }a{x}^{2}+px+qx+c.\\hfill \\\\ x\\left(5x - 3\\right)+2\\left(5x - 3\\right)\\hfill & \\text{Factor out the GCF of each part}.\\hfill \\\\ \\left(5x - 3\\right)\\left(x+2\\right)\\hfill & \\text{Factor out the GCF}\\text{ }\\text{ of the expression}.\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n
      \n

      Analysis of the Solution<\/h3>\n

      We can check our work by multiplying. Use FOIL to confirm that [latex]\\left(5x - 3\\right)\\left(x+2\\right)=5{x}^{2}+7x - 6[\/latex].<\/p>\n<\/div>\n

      \n

      Try It 3<\/h3>\n

      Factor the following.<\/p>\n

      a.[latex]2{x}^{2}+9x+9[\/latex]
      \nb. [latex]6{x}^{2}+x - 1[\/latex]<\/p>\n

      Solution<\/a><\/p>\n<\/div>\n\n\t\t\t

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