{"id":876,"date":"2024-04-30T18:21:59","date_gmt":"2024-04-30T18:21:59","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/?post_type=chapter&p=876"},"modified":"2024-11-20T02:43:41","modified_gmt":"2024-11-20T02:43:41","slug":"factoring-polynomials-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/chapter\/factoring-polynomials-learn-it-4\/","title":{"raw":"Factoring Polynomials: Learn It 4","rendered":"Factoring Polynomials: Learn It 4"},"content":{"raw":"

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\r\n
\r\n

Sum and Difference of Cubes<\/h3>\r\nThe sum and difference of cubes are algebraic formulas used to factor and solve polynomial equations involving cubic terms.\r\n
    \r\n \t
  • Sum of Cubes:<\/strong> [latex]{a}^{3}+{b}^{3}=\\left(a+b\\right)\\left({a}^{2}-ab+{b}^{2}\\right)[\/latex]<\/li>\r\n \t
  • Difference of Cubes:<\/strong> [latex]{a}^{3}-{b}^{3}=\\left(a-b\\right)\\left({a}^{2}+ab+{b}^{2}\\right)[\/latex]<\/li>\r\n<\/ul>\r\n<\/section>
    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.<\/section>
    How To: Given a sum of cubes or difference of cubes, factor it<\/strong>\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<\/section>
      Factor [latex]{x}^{3}+512[\/latex].Solution<\/strong>\r\n\r\nTo factor the cubic expression [latex]{x}^{3}+512[\/latex], we recognize that [latex]512[\/latex] is a perfect cube, specifically [latex]8^3[\/latex]. This allows us to apply the sum of cubes formula.\r\n\r\n[latex]\\begin{align*} \\text{Original expression:} & \\quad x^3 + 512 \\\\ \\text{Identify cubes:} & \\quad x^3 \\text{ and } 512 = 8^3 \\\\ \\text{Apply sum of cubes formula:} & \\quad x^3 + 8^3 = (x + 8)(x^2 - 8x + 64) \\\\ \\text{Factored form:} & \\quad (x + 8)(x^2 - 8x + 64) \\end{align*}[\/latex]\r\n\r\n<\/section>
      Factor [latex]8{x}^{3}-125[\/latex].\r\n<\/strong>[reveal-answer q=\"769186\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"769186\"][latex]\\begin{align*} \\text{Original expression:} & \\quad 8x^3 - 125 \\\\ \\text{Identify cubes:} & \\quad 8x^3 = (2x)^3, \\quad 125 = 5^3 \\\\ \\text{Apply difference of cubes formula:} & \\quad (2x)^3 - 5^3 = (2x - 5)(4x^2 + 10x + 25) \\\\ \\text{Factored form:} & \\quad (2x - 5)(4x^2 + 10x + 25) \\end{align*}[\/latex][\/hidden-answer]<\/section>
      [ohm2_question hide_question_numbers=1]18888[\/ohm2_question]<\/section>\r\n

      Factoring Expressions with Fractional or Negative Exponents<\/h2>\r\nExpressions with fractional or negative exponents can be factored by pulling out a GCF. Look for the variable or exponent that is common to each term of the expression and pull out that variable or exponent raised to the lowest power. These expressions follow the same factoring rules as those with integer exponents. For instance, [latex]2{x}^{\\frac{1}{4}}+5{x}^{\\frac{3}{4}}[\/latex] can be factored by pulling out [latex]{x}^{\\frac{1}{4}}[\/latex] and being rewritten as [latex]{x}^{\\frac{1}{4}}\\left(2+5{x}^{\\frac{1}{2}}\\right)[\/latex].\r\n\r\n
      Factor [latex]3x{\\left(x+2\\right)}^{\\frac{-1}{3}}+4{\\left(x+2\\right)}^{\\frac{2}{3}}[\/latex].[reveal-answer q=\"164967\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"164967\"]Factor out the term with the lowest value of the exponent. In this case, that would be [latex]{\\left(x+2\\right)}^{-\\frac{1}{3}}[\/latex].\r\n
      [latex]\\begin{array}{cc}{\\left(x+2\\right)}^{-\\frac{1}{3}}\\left(3x+4\\left(x+2\\right)\\right)\\hfill & \\text{Factor out the GCF}.\\hfill \\\\ {\\left(x+2\\right)}^{-\\frac{1}{3}}\\left(3x+4x+8\\right)\\hfill & \\text{Simplify}.\\hfill \\\\ {\\left(x+2\\right)}^{-\\frac{1}{3}}\\left(7x+8\\right)\\hfill & \\end{array}[\/latex]<\/div>\r\n
      [\/hidden-answer]<\/div>\r\n<\/section>
      [ohm2_question hide_question_numbers=1]18889[\/ohm2_question]<\/section>
      [ohm2_question hide_question_numbers=1]18890[\/ohm2_question]<\/section>","rendered":"

      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

      \n

      Sum and Difference of Cubes<\/h3>\n

      The sum and difference of cubes are algebraic formulas used to factor and solve polynomial equations involving cubic terms.<\/p>\n

        \n
      • Sum of Cubes:<\/strong> [latex]{a}^{3}+{b}^{3}=\\left(a+b\\right)\\left({a}^{2}-ab+{b}^{2}\\right)[\/latex]<\/li>\n
      • Difference of Cubes:<\/strong> [latex]{a}^{3}-{b}^{3}=\\left(a-b\\right)\\left({a}^{2}+ab+{b}^{2}\\right)[\/latex]<\/li>\n<\/ul>\n<\/section>\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.<\/section>\n
        How To: Given a sum of cubes or difference of cubes, factor it<\/strong><\/p>\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<\/section>\n
          Factor [latex]{x}^{3}+512[\/latex].Solution<\/strong><\/p>\n

          To factor the cubic expression [latex]{x}^{3}+512[\/latex], we recognize that [latex]512[\/latex] is a perfect cube, specifically [latex]8^3[\/latex]. This allows us to apply the sum of cubes formula.<\/p>\n

          [latex]\\begin{align*} \\text{Original expression:} & \\quad x^3 + 512 \\\\ \\text{Identify cubes:} & \\quad x^3 \\text{ and } 512 = 8^3 \\\\ \\text{Apply sum of cubes formula:} & \\quad x^3 + 8^3 = (x + 8)(x^2 - 8x + 64) \\\\ \\text{Factored form:} & \\quad (x + 8)(x^2 - 8x + 64) \\end{align*}[\/latex]<\/p>\n<\/section>\n

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