{"id":305,"date":"2015-09-18T21:09:02","date_gmt":"2015-09-18T21:09:02","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=305"},"modified":"2015-11-03T22:52:12","modified_gmt":"2015-11-03T22:52:12","slug":"factoring-a-trinomial-with-leading-coefficient-1","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/chapter\/factoring-a-trinomial-with-leading-coefficient-1\/","title":{"raw":"Factoring a Trinomial with Leading Coefficient 1","rendered":"Factoring a Trinomial with Leading Coefficient 1"},"content":{"raw":"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].\r\n\r\nTrinomials of the form [latex]{x}^{2}+bx+c[\/latex] can be factored by finding two numbers with a product of [latex]c[\/latex] and a sum of [latex]b[\/latex]. The trinomial [latex]{x}^{2}+10x+16[\/latex], for example, can be factored using the numbers [latex]2[\/latex] and [latex]8[\/latex] because the product of those numbers is [latex]16[\/latex] and their sum is [latex]10[\/latex]. The trinomial can be rewritten as the product of [latex]\\left(x+2\\right)[\/latex] and [latex]\\left(x+8\\right)[\/latex].\r\n
\r\n

A General Note: Factoring a Trinomial with Leading Coefficient 1<\/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<\/div>\r\n
\r\n

Q & A<\/h3>\r\n

Can every trinomial be factored as a product of binomials?<\/h3>\r\nNo. Some polynomials cannot be factored. These polynomials are said to be prime.<\/em>\r\n\r\n<\/div>\r\n
\r\n

How To: Given a trinomial in the form [latex]{x}^{2}+bx+c[\/latex], factor it.<\/h3>\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<\/div>\r\n
    \r\n

    Example 2: Factoring a Trinomial with Leading Coefficient 1<\/h3>\r\nFactor [latex]{x}^{2}+2x - 15[\/latex].\r\n\r\n<\/div>\r\n
    \r\n

    Solution<\/h3>\r\nWe have a trinomial with leading coefficient [latex]1,b=2[\/latex], and [latex]c=-15[\/latex]. We need to find two numbers with a product of [latex]-15[\/latex] and a sum of [latex]2[\/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
    \u00a0<\/caption>\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\n14<\/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]<\/td>\r\n2<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNow 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].\r\n\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(x - 3\\right)\\left(x+5\\right)={x}^{2}+2x - 15[\/latex].\r\n\r\n<\/div>\r\n
    \r\n

    Q & A<\/h3>\r\n

    Does the order of the factors matter?<\/h3>\r\nNo. Multiplication is commutative, so the order of the factors does not matter.<\/em>\r\n\r\n<\/div>\r\n
    \r\n

    Try It 2<\/h3>\r\nFactor [latex]{x}^{2}-7x+6[\/latex].\r\n\r\nSolution<\/a>\r\n\r\n<\/div>","rendered":"

    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

    Trinomials of the form [latex]{x}^{2}+bx+c[\/latex] can be factored by finding two numbers with a product of [latex]c[\/latex] and a sum of [latex]b[\/latex]. The trinomial [latex]{x}^{2}+10x+16[\/latex], for example, can be factored using the numbers [latex]2[\/latex] and [latex]8[\/latex] because the product of those numbers is [latex]16[\/latex] and their sum is [latex]10[\/latex]. The trinomial can be rewritten as the product of [latex]\\left(x+2\\right)[\/latex] and [latex]\\left(x+8\\right)[\/latex].<\/p>\n

    \n

    A General Note: Factoring a Trinomial with Leading Coefficient 1<\/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<\/div>\n

    \n

    Q & A<\/h3>\n

    Can every trinomial be factored as a product of binomials?<\/h3>\n

    No. Some polynomials cannot be factored. These polynomials are said to be prime.<\/em><\/p>\n<\/div>\n

    \n

    How To: Given a trinomial in the form [latex]{x}^{2}+bx+c[\/latex], factor it.<\/h3>\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<\/div>\n
      \n

      Example 2: Factoring a Trinomial with Leading Coefficient 1<\/h3>\n

      Factor [latex]{x}^{2}+2x - 15[\/latex].<\/p>\n<\/div>\n

      \n

      Solution<\/h3>\n

      We have a trinomial with leading coefficient [latex]1,b=2[\/latex], and [latex]c=-15[\/latex]. We need to find two numbers with a product of [latex]-15[\/latex] and a sum of [latex]2[\/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
      \u00a0<\/caption>\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>\n14<\/td>\n<\/tr>\n
      [latex]3,-5[\/latex]<\/td>\n[latex]-2[\/latex]<\/td>\n<\/tr>\n
      [latex]-3,5[\/latex]<\/td>\n2<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\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].<\/p>\n<\/div>\n

      \n

      Analysis of the Solution<\/h3>\n

      We can check our work by multiplying. Use FOIL to confirm that [latex]\\left(x - 3\\right)\\left(x+5\\right)={x}^{2}+2x - 15[\/latex].<\/p>\n<\/div>\n

      \n

      Q & A<\/h3>\n

      Does the order of the factors matter?<\/h3>\n

      No. Multiplication is commutative, so the order of the factors does not matter.<\/em><\/p>\n<\/div>\n

      \n

      Try It 2<\/h3>\n

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

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

      \n\t\t\t

      Candela Citations<\/h3>\n\t\t\t\t\t
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