{"id":309,"date":"2015-09-18T21:11:09","date_gmt":"2015-09-18T21:11:09","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=309"},"modified":"2015-11-03T23:26:24","modified_gmt":"2015-11-03T23:26:24","slug":"factoring-trinomials-with-squares","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/chapter\/factoring-trinomials-with-squares\/","title":{"raw":"Special Products","rendered":"Special Products"},"content":{"raw":"

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\nWe can use this equation to factor any perfect square trinomial.\r\n
\r\n

A General Note: Perfect Square Trinomials<\/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<\/div>\r\n
\r\n

How To: Given a perfect square trinomial, factor it into the square of a binomial.\r\n<\/strong><\/h3>\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<\/div>\r\n
    \r\n

    Example 4: Factoring a Perfect Square Trinomial<\/h3>\r\nFactor [latex]25{x}^{2}+20x+4[\/latex].\r\n\r\n<\/div>\r\n
    \r\n

    Solution<\/h3>\r\nNotice that [latex]25{x}^{2}[\/latex] and [latex]4[\/latex] are perfect squares because [latex]25{x}^{2}={\\left(5x\\right)}^{2}[\/latex] and [latex]4={2}^{2}[\/latex]. Then check to see if the middle term is twice the product of [latex]5x[\/latex] and [latex]2[\/latex]. The middle term is, indeed, twice the product: [latex]2\\left(5x\\right)\\left(2\\right)=20x[\/latex]. Therefore, the trinomial is a perfect square trinomial and can be written as [latex]{\\left(5x+2\\right)}^{2}[\/latex].\r\n\r\n<\/div>\r\n
    \r\n

    Try It 4<\/h3>\r\nFactor [latex]49{x}^{2}-14x+1[\/latex].\r\n\r\nSolution<\/a>\r\n\r\n<\/div>\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

    A General Note: Differences 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<\/div>\r\n
    \r\n

    How To: Given a difference of squares, factor it into binomials.<\/h3>\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<\/div>\r\n
      \r\n

      Example 5: Factoring a Difference of Squares<\/h3>\r\nFactor [latex]9{x}^{2}-25[\/latex].\r\n\r\n<\/div>\r\n
      \r\n

      Solution<\/h3>\r\nNotice that [latex]9{x}^{2}[\/latex] and [latex]25[\/latex] are perfect squares because [latex]9{x}^{2}={\\left(3x\\right)}^{2}[\/latex] and [latex]25={5}^{2}[\/latex]. The polynomial represents a difference of squares and can be rewritten as [latex]\\left(3x+5\\right)\\left(3x - 5\\right)[\/latex].\r\n\r\n<\/div>\r\n
      \r\n

      Try It 5<\/h3>\r\nFactor [latex]81{y}^{2}-100[\/latex].\r\n\r\nSolution<\/a>\r\n\r\n<\/div>\r\n
      \r\n

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

      Is there a formula to factor the sum of squares?<\/h3>\r\nNo. A sum of squares cannot be factored.<\/em>\r\n\r\n<\/div>\r\n

      Factoring the Sum and Difference of Cubes<\/h2>\r\nNow, we will look at two new special products: the sum and difference of cubes. Although the sum of squares cannot be factored, the sum of cubes can be factored into a binomial and a trinomial.\r\n
      [latex]{a}^{3}+{b}^{3}=\\left(a+b\\right)\\left({a}^{2}-ab+{b}^{2}\\right)[\/latex]<\/div>\r\nSimilarly, the sum of cubes can be factored into a binomial and a trinomial, but with different signs.\r\n
      [latex]{a}^{3}-{b}^{3}=\\left(a-b\\right)\\left({a}^{2}+ab+{b}^{2}\\right)[\/latex]<\/div>\r\nWe can use the acronym SOAP to remember the signs when factoring the sum or difference of cubes. The first letter of each word relates to the signs: S<\/strong>ame O<\/strong>pposite A<\/strong>lways P<\/strong>ositive. For example, consider the following example.\r\n
      [latex]{x}^{3}-{2}^{3}=\\left(x - 2\\right)\\left({x}^{2}+2x+4\\right)[\/latex]<\/div>\r\nThe sign of the first 2 is the same<\/em> as the sign between [latex]{x}^{3}-{2}^{3}[\/latex]. The sign of the [latex]2x[\/latex] term is opposite<\/em> the sign between [latex]{x}^{3}-{2}^{3}[\/latex]. And the sign of the last term, 4, is always positive<\/em>.\r\n
      \r\n

      A General Note: Sum and Difference of Cubes<\/h3>\r\nWe can factor the sum of two cubes as\r\n
      [latex]{a}^{3}+{b}^{3}=\\left(a+b\\right)\\left({a}^{2}-ab+{b}^{2}\\right)[\/latex]<\/div>\r\nWe can factor the difference of two cubes as\r\n
      [latex]{a}^{3}-{b}^{3}=\\left(a-b\\right)\\left({a}^{2}+ab+{b}^{2}\\right)[\/latex]<\/div>\r\n<\/div>\r\n
      \r\n

      How To: Given a sum of cubes or difference of cubes, factor it.<\/h3>\r\n
        \r\n\t
      1. Confirm that the first and last term are cubes, [latex]{a}^{3}+{b}^{3}[\/latex] or [latex]{a}^{3}-{b}^{3}[\/latex].<\/li>\r\n\t
      2. For a sum of cubes, write the factored form as [latex]\\left(a+b\\right)\\left({a}^{2}-ab+{b}^{2}\\right)[\/latex]. For a difference of cubes, write the factored form as [latex]\\left(a-b\\right)\\left({a}^{2}+ab+{b}^{2}\\right)[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n
        \r\n

        Example 6: Factoring a Sum of Cubes<\/h3>\r\nFactor [latex]{x}^{3}+512[\/latex].\r\n\r\n<\/div>\r\n
        \r\n

        Solution<\/h3>\r\nNotice that [latex]{x}^{3}[\/latex] and [latex]512[\/latex] are cubes because [latex]{8}^{3}=512[\/latex]. Rewrite the sum of cubes as [latex]\\left(x+8\\right)\\left({x}^{2}-8x+64\\right)[\/latex].\r\n\r\n<\/div>\r\n
        \r\n

        Analysis of the Solution<\/h3>\r\nAfter writing the sum of cubes this way, we might think we should check to see if the trinomial portion can be factored further. However, the trinomial portion cannot be factored, so we do not need to check.\r\n\r\n<\/div>\r\n
        \r\n

        Try It 6<\/h3>\r\nFactor the sum of cubes: [latex]216{a}^{3}+{b}^{3}[\/latex].\r\n\r\nSolution<\/a>\r\n\r\n<\/div>\r\n
        \r\n

        Example 7: Factoring a Difference of Cubes<\/h3>\r\nFactor [latex]8{x}^{3}-125[\/latex].\r\n\r\n<\/div>\r\n
        \r\n

        Solution<\/h3>\r\nNotice that [latex]8{x}^{3}[\/latex] and [latex]125[\/latex] are cubes because [latex]8{x}^{3}={\\left(2x\\right)}^{3}[\/latex] and [latex]125={5}^{3}[\/latex]. Write the difference of cubes as [latex]\\left(2x - 5\\right)\\left(4{x}^{2}+10x+25\\right)[\/latex].\r\n\r\n<\/div>\r\n
        \r\n

        Analysis of the Solution<\/h3>\r\nJust as with the sum of cubes, we will not be able to further factor the trinomial portion.\r\n\r\n<\/div>\r\n
        \r\n

        Try It 7<\/h3>\r\nFactor the difference of cubes: [latex]1,000{x}^{3}-1[\/latex].\r\n\r\nSolution<\/a>\r\n\r\n<\/div>","rendered":"

        Factoring a Perfect Square Trinomial<\/h2>\n

        A 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.<\/p>\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>\n

        We can use this equation to factor any perfect square trinomial.<\/p>\n

        \n

        A General Note: Perfect Square Trinomials<\/h3>\n

        A perfect square trinomial can be written as the square of a binomial:<\/p>\n

        [latex]{a}^{2}+2ab+{b}^{2}={\\left(a+b\\right)}^{2}[\/latex]<\/div>\n<\/div>\n
        \n

        How To: Given a perfect square trinomial, factor it into the square of a binomial.
        \n<\/strong><\/h3>\n
          \n
        1. Confirm that the first and last term are perfect squares.<\/li>\n
        2. Confirm that the middle term is twice the product of [latex]ab[\/latex].<\/li>\n
        3. Write the factored form as [latex]{\\left(a+b\\right)}^{2}[\/latex].<\/li>\n<\/ol>\n<\/div>\n
          \n

          Example 4: Factoring a Perfect Square Trinomial<\/h3>\n

          Factor [latex]25{x}^{2}+20x+4[\/latex].<\/p>\n<\/div>\n

          \n

          Solution<\/h3>\n

          Notice that [latex]25{x}^{2}[\/latex] and [latex]4[\/latex] are perfect squares because [latex]25{x}^{2}={\\left(5x\\right)}^{2}[\/latex] and [latex]4={2}^{2}[\/latex]. Then check to see if the middle term is twice the product of [latex]5x[\/latex] and [latex]2[\/latex]. The middle term is, indeed, twice the product: [latex]2\\left(5x\\right)\\left(2\\right)=20x[\/latex]. Therefore, the trinomial is a perfect square trinomial and can be written as [latex]{\\left(5x+2\\right)}^{2}[\/latex].<\/p>\n<\/div>\n

          \n

          Try It 4<\/h3>\n

          Factor [latex]49{x}^{2}-14x+1[\/latex].<\/p>\n

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

          Factoring a Difference of Squares<\/h2>\n

          A 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.<\/p>\n

          [latex]{a}^{2}-{b}^{2}=\\left(a+b\\right)\\left(a-b\\right)[\/latex]<\/div>\n

          We can use this equation to factor any differences of squares.<\/p>\n

          \n

          A General Note: Differences of Squares<\/h3>\n

          A difference of squares can be rewritten as two factors containing the same terms but opposite signs.<\/p>\n

          [latex]{a}^{2}-{b}^{2}=\\left(a+b\\right)\\left(a-b\\right)[\/latex]<\/div>\n<\/div>\n
          \n

          How To: Given a difference of squares, factor it into binomials.<\/h3>\n
            \n
          1. Confirm that the first and last term are perfect squares.<\/li>\n
          2. Write the factored form as [latex]\\left(a+b\\right)\\left(a-b\\right)[\/latex].<\/li>\n<\/ol>\n<\/div>\n
            \n

            Example 5: Factoring a Difference of Squares<\/h3>\n

            Factor [latex]9{x}^{2}-25[\/latex].<\/p>\n<\/div>\n

            \n

            Solution<\/h3>\n

            Notice that [latex]9{x}^{2}[\/latex] and [latex]25[\/latex] are perfect squares because [latex]9{x}^{2}={\\left(3x\\right)}^{2}[\/latex] and [latex]25={5}^{2}[\/latex]. The polynomial represents a difference of squares and can be rewritten as [latex]\\left(3x+5\\right)\\left(3x - 5\\right)[\/latex].<\/p>\n<\/div>\n

            \n

            Try It 5<\/h3>\n

            Factor [latex]81{y}^{2}-100[\/latex].<\/p>\n

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

            \n

            Q & A<\/h3>\n

            Is there a formula to factor the sum of squares?<\/h3>\n

            No. A sum of squares cannot be factored.<\/em><\/p>\n<\/div>\n

            Factoring the Sum and Difference of Cubes<\/h2>\n

            Now, we will look at two new special products: the sum and difference of cubes. Although the sum of squares cannot be factored, the sum of cubes can be factored into a binomial and a trinomial.<\/p>\n

            [latex]{a}^{3}+{b}^{3}=\\left(a+b\\right)\\left({a}^{2}-ab+{b}^{2}\\right)[\/latex]<\/div>\n

            Similarly, the sum of cubes can be factored into a binomial and a trinomial, but with different signs.<\/p>\n

            [latex]{a}^{3}-{b}^{3}=\\left(a-b\\right)\\left({a}^{2}+ab+{b}^{2}\\right)[\/latex]<\/div>\n

            We can use the acronym SOAP to remember the signs when factoring the sum or difference of cubes. The first letter of each word relates to the signs: S<\/strong>ame O<\/strong>pposite A<\/strong>lways P<\/strong>ositive. For example, consider the following example.<\/p>\n

            [latex]{x}^{3}-{2}^{3}=\\left(x - 2\\right)\\left({x}^{2}+2x+4\\right)[\/latex]<\/div>\n

            The sign of the first 2 is the same<\/em> as the sign between [latex]{x}^{3}-{2}^{3}[\/latex]. The sign of the [latex]2x[\/latex] term is opposite<\/em> the sign between [latex]{x}^{3}-{2}^{3}[\/latex]. And the sign of the last term, 4, is always positive<\/em>.<\/p>\n

            \n

            A General Note: Sum and Difference of Cubes<\/h3>\n

            We can factor the sum of two cubes as<\/p>\n

            [latex]{a}^{3}+{b}^{3}=\\left(a+b\\right)\\left({a}^{2}-ab+{b}^{2}\\right)[\/latex]<\/div>\n

            We can factor the difference of two cubes as<\/p>\n

            [latex]{a}^{3}-{b}^{3}=\\left(a-b\\right)\\left({a}^{2}+ab+{b}^{2}\\right)[\/latex]<\/div>\n<\/div>\n
            \n

            How To: Given a sum of cubes or difference of cubes, factor it.<\/h3>\n
              \n
            1. Confirm that the first and last term are cubes, [latex]{a}^{3}+{b}^{3}[\/latex] or [latex]{a}^{3}-{b}^{3}[\/latex].<\/li>\n
            2. For a sum of cubes, write the factored form as [latex]\\left(a+b\\right)\\left({a}^{2}-ab+{b}^{2}\\right)[\/latex]. For a difference of cubes, write the factored form as [latex]\\left(a-b\\right)\\left({a}^{2}+ab+{b}^{2}\\right)[\/latex].<\/li>\n<\/ol>\n<\/div>\n
              \n

              Example 6: Factoring a Sum of Cubes<\/h3>\n

              Factor [latex]{x}^{3}+512[\/latex].<\/p>\n<\/div>\n

              \n

              Solution<\/h3>\n

              Notice that [latex]{x}^{3}[\/latex] and [latex]512[\/latex] are cubes because [latex]{8}^{3}=512[\/latex]. Rewrite the sum of cubes as [latex]\\left(x+8\\right)\\left({x}^{2}-8x+64\\right)[\/latex].<\/p>\n<\/div>\n

              \n

              Analysis of the Solution<\/h3>\n

              After writing the sum of cubes this way, we might think we should check to see if the trinomial portion can be factored further. However, the trinomial portion cannot be factored, so we do not need to check.<\/p>\n<\/div>\n

              \n

              Try It 6<\/h3>\n

              Factor the sum of cubes: [latex]216{a}^{3}+{b}^{3}[\/latex].<\/p>\n

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

              \n

              Example 7: Factoring a Difference of Cubes<\/h3>\n

              Factor [latex]8{x}^{3}-125[\/latex].<\/p>\n<\/div>\n

              \n

              Solution<\/h3>\n

              Notice that [latex]8{x}^{3}[\/latex] and [latex]125[\/latex] are cubes because [latex]8{x}^{3}={\\left(2x\\right)}^{3}[\/latex] and [latex]125={5}^{3}[\/latex]. Write the difference of cubes as [latex]\\left(2x - 5\\right)\\left(4{x}^{2}+10x+25\\right)[\/latex].<\/p>\n<\/div>\n

              \n

              Analysis of the Solution<\/h3>\n

              Just as with the sum of cubes, we will not be able to further factor the trinomial portion.<\/p>\n<\/div>\n

              \n

              Try It 7<\/h3>\n

              Factor the difference of cubes: [latex]1,000{x}^{3}-1[\/latex].<\/p>\n

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

              \n\t\t\t

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