{"id":6787,"date":"2020-10-09T19:00:55","date_gmt":"2020-10-09T19:00:55","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/slcc-beginalgebra\/?post_type=chapter&p=6787"},"modified":"2023-05-22T19:52:42","modified_gmt":"2023-05-22T19:52:42","slug":"6-5-factoring-the-difference-of-squares","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-elementaryalgebra\/chapter\/6-5-factoring-the-difference-of-squares\/","title":{"raw":"6.5: Factoring Perfect Square Trinomials and the Difference of Squares","rendered":"6.5: Factoring Perfect Square Trinomials and the Difference of Squares"},"content":{"raw":"
\r\n

section 6.5 Learning Objectives<\/h3>\r\n6.5: Factoring Perfect Square Trinomials and the Difference of Squares<\/strong>\r\n
    \r\n \t
  • Factor perfect square trinomials<\/li>\r\n \t
  • Factor the difference of two squares<\/li>\r\n \t
  • Factor difference of squares by first factoring out the GCF<\/li>\r\n<\/ul>\r\n<\/div>\r\n \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 {\\left(a+b\\right)}^{2} & = & {a}^{2}+2ab+{b}^{2} \\hfill \\\\ & \\text{and}& \\\\ \\hfill {\\left(a-b\\right)}^{2} & = & {a}^{2}-2ab+{b}^{2}\\hfill \\end{array}[\/latex]<\/div>\r\n
    <\/div>\r\n
    \r\n\r\nWe can use these same equations to factor any perfect square trinomial.\r\n
    \r\n

    Perfect Square Trinomials<\/h3>\r\n

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

    [latex]{a}^{2}+2ab+{b}^{2}={\\left(a+b\\right)}^{2}[\/latex]<\/div>\r\n
    [latex]{a}^{2}-2ab+{b}^{2}={\\left(a-b\\right)}^{2}[\/latex]<\/div>\r\n<\/div>\r\n<\/div>\r\nIn our first example, we solve the problem two different ways to show how to use the above formulas while also demonstrating that this strategy will produce the same result as the techniques we learned earlier for factoring trinomials of this form.\r\n
    \r\n

    Example 1<\/h3>\r\nFactor [latex]x^2-14x+49[\/latex].\r\n\r\n[reveal-answer q=\"104579\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"104579\"]\r\n\r\nWe first note that the Product and Sum Method from Section 6.3 is applicable, so we begin by using that strategy.\r\n\r\nWe need to find two integers that multiply to [latex]49[\/latex] while adding to [latex]-14[\/latex]. The integers are [latex]-7[\/latex] and [latex]-7[\/latex]. Therefore, the polynomial factors as\r\n

    [latex]x^2-14x+49=(x-7)(x-7)[\/latex].<\/p>\r\nNow, let us try the new Perfect Square Trinomial formula. Since our middle term is negative, we focus on the second formula. We begin by comparing our given problem to [latex]a^2-2ab+b^2[\/latex] and identifying [latex]a[\/latex] and [latex]b[\/latex]. Since we must have [latex]a^2=x^2[\/latex] and [latex]b^2=49[\/latex], we conclude that\r\n

    [latex]a=x[\/latex] and [latex]b=7[\/latex].<\/p>\r\nIn order to apply the formula, we must check that [latex]2ab[\/latex] matches our middle term (ignoring the negative sign as that was built into the formula). We have\r\n

    [latex]2ab=2\\cdot x\\cdot 7=14x[\/latex].<\/p>\r\nThis does match the middle term of our problem, and so, we can apply the formula [latex]a^2-2ab+b^2=(a-b)^2[\/latex] to get\r\n

    [latex]x^2-14x+49=\\left(x-7\\right)^2[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the previous example, note that the results do match since [latex](x-7)(x-7)=(x-7)^2[\/latex]. Which way did you prefer? You may have found it easier to simply use the Product and Sum Strategy. However, as the problems become more complicated, the benefits of identifying Perfect Square Trinomials become more apparent.\r\n\r\nThe next example would have required AC-Method, where instead, we will apply our new formula.\r\n

    \r\n

    Example 2<\/h3>\r\nFactor [latex]25{x}^{2}+20x+4[\/latex].\r\n[reveal-answer q=\"119279\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"119279\"]\r\n\r\nFirst, 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].\r\n\r\nThis means that [latex]a=5x\\text{ and }b=2[\/latex]\r\n\r\nNext, check to see if the middle term is equal to [latex]2ab[\/latex], which it is:\r\n

    [latex]2ab = 2\\left(5x\\right)\\left(2\\right)=20x[\/latex]<\/p>\r\n \r\n\r\nTherefore, the trinomial is a perfect square trinomial and can be written as:\r\n

    [latex]{\\left(a+b\\right)}^{2}={\\left(5x+2\\right)}^{2}[\/latex].<\/p>\r\n\r\n

    Answer<\/span><\/h4>\r\n[latex]25x^2+20x+4=(5x+2)^{2}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the next example, we will show that we can use [latex]1 = 1^2[\/latex] to factor a polynomial with a term equal to\u00a0[latex]1[\/latex].\r\n
    \r\n

    Example 3<\/h3>\r\nFactor [latex]49{x}^{2}-14x+1[\/latex].\r\n[reveal-answer q=\"865849\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"865849\"]\r\n\r\nFirst, notice that [latex]49{x}^{2}[\/latex] and [latex]1[\/latex] are perfect squares because [latex]49{x}^{2}={\\left(7x\\right)}^{2}[\/latex] and [latex]1={1}^{2}[\/latex].\r\n\r\nThis means that [latex]a=7x[\/latex] and [latex]b=1[\/latex].\r\n\r\nNext, check to see if the middle term (not including the negative) is equal to [latex]2ab[\/latex], which it is:\r\n

    [latex]2ab = 2\\left(7x\\right)\\left(1\\right)=14x[\/latex]<\/p>\r\nTherefore, the trinomial is a perfect square trinomial and can be written as:\r\n

    [latex]{\\left(a-b\\right)}^{2}={\\left(7x-1\\right)}^{2}[\/latex].<\/p>\r\n\r\n

    Answer<\/span><\/h4>\r\n[latex]49x^2-14x+1=(7x-1)^{2}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the following video, we provide another short description of what a perfect square trinomial is and show how to factor them using a formula.\r\n\r\nhttps:\/\/youtu.be\/UMCVGDTxxTI\r\n\r\nIt is worth noting that these problems could also be factored using the techniques detailed in Section 6.3 (if [latex]a=1[\/latex]) and Section 6.4 (if [latex]a\\neq 1[\/latex]). However, if we recognize that a trinomial is in the form of a perfect square trinomial, this process can serve as a nice shortcut.\r\n\r\nWe can summarize our process in the following way:\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] or [latex]{\\left(a-b\\right)}^{2}[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\nLet us look at one more example. In this one, notice that there are now two variables.\r\n
      \r\n

      Example 4<\/h3>\r\nFactor [latex]x^{2}+4xy+4y^{2}[\/latex].\r\n\r\n[reveal-answer q=\"949165\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"949165\"]\r\n\r\nWe note that the first and last terms are perfect squares, with [latex]4y^2=(2y)^2[\/latex] for the last term. So, [latex]a=x[\/latex] and [latex]b=2y[\/latex].\r\n\r\nFurthermore, [latex]2ab=2x(2y)=4xy[\/latex] matches the middle term.\r\n\r\nAnd so, the trinomial factors as [latex](a+b)^{2}=(x+2y)^{2}[\/latex].\r\n

      Answer<\/span><\/h4>\r\n[latex]x^2+4xy+4y^2=(x+2y)^{2}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

      Difference of Squares<\/h2>\r\nWe would be remiss if we failed to introduce one more type of polynomial that can be factored. This polynomial can be factored into two binomials but has only two terms. \u00a0Let's start from the product of two special binomials to see the pattern.\r\n\r\nConsider the product of the following two binomials, which are identical except for the operation: [latex]\\left(x-2\\right)\\left(x+2\\right)[\/latex]. If we multiply them together, we lose the middle term that we are used to seeing as a result.\r\n

      Multiply:<\/p>\r\n

      [latex]\\begin{array}{l}\\left(x-2\\right)\\left(x+2\\right)\\\\\\text{}\\\\=x^2-2x+2x-2^2\\\\\\text{}\\\\=x^2-2^2\\\\\\text{}\\\\=x^2-4\\end{array}[\/latex]<\/p>\r\nThe polynomial [latex]x^2-4[\/latex] is called a difference of squares because teach term is a perfect square and the operation connecting them is subtraction. \u00a0A difference of squares will always factor in the following way:\r\n

      \r\n

      Factoring the Difference of Squares<\/h3>\r\n

      Given [latex]a^2-b^2[\/latex], its factored form will be [latex]\\left(a+b\\right)\\left(a-b\\right)[\/latex].<\/p>\r\n \r\n

      Try using the following rhyme to help you remember: \"The difference of squares, factors to conjugate pairs!\"<\/em><\/p>\r\n\r\n<\/div>\r\nTo further verify this, it is interesting to note that we could also use the techniques of the previous sections. Let\u2019s factor [latex]x^{2}\u20134[\/latex]\u00a0by writing it as a trinomial, [latex]x^{2}+0x\u20134[\/latex]. \u00a0This\u00a0is similar in format to the\u00a0trinomials we have been factoring so far, so let\u2019s\u00a0use the same method.\r\n

      Find the factors of [latex]a\\cdot{c}[\/latex]\u00a0whose sum is b, <\/i>in this case, 0:<\/i><\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
      Factors of [latex]\u22124[\/latex]<\/th>\r\nSum of the factors<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
      [latex]1\\cdot-4=\u22124[\/latex]<\/td>\r\n[latex]1-4=\u22123[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]2\\cdot\u22122=\u22124[\/latex]<\/td>\r\n[latex]2-2=0[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]-1\\cdot4=\u22124[\/latex]<\/td>\r\n[latex]-1+4=3[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n2 and -2 have a sum of 0. You can use these to factor [latex]x^{2}\u20134[\/latex]. Using the Product and Sum Method in Section 6.3, results in [latex]x^2-4=(x+2)(x-2)[\/latex].\r\n\r\nHowever, applying the formula given above gives us an even quicker shortcut! Let us explore this same example one more time using our difference of squares equation.\r\n
      \r\n

      Example 5<\/h3>\r\nFactor [latex]x^{2}\u20134[\/latex].\r\n[reveal-answer q=\"23133\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"23133\"]\r\n\r\nWe first note that this is a binomial (two terms), and is a difference of squares, since we could write it as\r\n

      [latex]x^2-2^2[\/latex]<\/p>\r\nSo, [latex]a=x[\/latex] and [latex]b=2[\/latex].\r\n\r\nHence, this will factor as\r\n

      [latex](a+b)(a-b)=(x+2)(x-2)[\/latex]<\/p>\r\n\r\n

      Answer<\/h4>\r\n[latex]x^{2}-4=\\left(x+2\\right)\\left(x-2\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nSince order doesn't matter with multiplication, the answer can also be written as [latex]\\left(x-2\\right)\\left(x+2\\right)[\/latex].\r\n\r\nYou can check the answer by multiplying [latex]\\left(x+2\\right)\\left(x-2\\right)=x^{2}+2x\u20132x\u20134=x^{2}\u20134[\/latex].\r\n\r\nThe following video show two more examples of factoring a difference of squares.\r\n\r\nhttps:\/\/youtu.be\/Li9IBp5HrFA\r\n\r\nWe can summarize the process for factoring a difference of squares with the shortcut this way:\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 6<\/h3>\r\nFactor [latex]9{x}^{2}-25[\/latex].\r\n[reveal-answer q=\"960938\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"960938\"]\r\n\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].\r\n\r\nThis means that [latex]a=3x,\\text{ and }b=5[\/latex]\r\n\r\nThe polynomial represents a difference of squares and can be rewritten as [latex]\\left(3x+5\\right)\\left(3x - 5\\right)[\/latex].\r\n\r\nCheck that you are correct by multiplying.\r\n\r\n[latex]\\left(3x+5\\right)\\left(3x - 5\\right)=9x^2-15x+15x-25=9x^2-25[\/latex]\r\n

        Answer<\/span><\/h4>\r\n[latex]9x^2-25=(3x+5)(3x-5)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe most helpful\u00a0thing for recognizing a difference of squares that can be factored with the shortcut is knowing which numbers are perfect squares, as you will see in the next example.\r\n
        \r\n

        Example 7<\/h3>\r\nFactor [latex]49{y}^{2}-144z^{2}[\/latex].\r\n[reveal-answer q=\"193159\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"193159\"]\r\n\r\nNotice that [latex]49{y}^{2}[\/latex] and [latex]144z^{2}[\/latex] are perfect squares because [latex]49{y}^{2}={\\left(7y\\right)}^{2}[\/latex] and [latex]144z^{2}={(12z)}^{2}[\/latex].\r\n\r\nThis means that [latex]a=7y,\\text{ and }b=12z[\/latex]\r\n\r\nThe polynomial represents a difference of squares and can be rewritten as [latex]\\left(7y+12z\\right)\\left(7y - 12z\\right)[\/latex].\r\n\r\nCheck that you are correct by multiplying.\r\n\r\n[latex]\\left(7y+12z\\right)\\left(7y - 12z\\right)=49y^2-84yz+84yz-144z^2=49y^2-144z^2[\/latex]\r\n

        Answer<\/span><\/h4>\r\n[latex]49y^2-144z^2=(7y+12z)(7y-12z)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nAnd as we have seen in the previous sections, we must always be on the lookout for first factoring out a GCF.\r\n
        \r\n

        Example 8<\/h3>\r\nFactor [latex]18x^3-2x[\/latex].\r\n\r\n[reveal-answer q=\"883692\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"883692\"]\r\n\r\nNotice that the terms are not perfect squares. However, they do share a common factor of [latex]2x[\/latex] which can be factored out.\r\n

        [latex]2x(9x^2-1)[\/latex]<\/p>\r\nThe resulting binomial is now a difference of squares. Since [latex]9x^2=(3x)^{2}[\/latex] and [latex]1=1^2[\/latex], we have [latex]a=3x[\/latex] and [latex]b=1[\/latex]. Applying our difference of squares equation, and remembering the factor of [latex]2x[\/latex] in front, this factors as\r\n

        [latex]2x(3x+1)(3x-1)[\/latex]<\/p>\r\n\r\n

        Answer<\/span><\/h4>\r\n

        [latex]18x^3-2x=2x(3x+1)(3x-1)[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nAs we have described, there are two key components to a difference of squares, perfect squares and the difference. We explore the importance of the operation further in the \"Think About It\" below.\r\n

        \r\n

        Think About It<\/h3>\r\nIs there a formula to factor the sum of squares, [latex]a^2+b^2[\/latex], into a product of two binomials?\r\n\r\nWrite down some ideas for how you would answer this in the box below before you look at the answer.\r\n\r\n[practice-area rows=\"1\"][\/practice-area]\r\n[reveal-answer q=\"121734\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"121734\"]\r\n\r\nThere is no way to factor a sum of squares into a product of two binomials with real number coefficients. The middle term needs to \"disappear\" when the two binomials are multiplied and the only way to do that is with opposite signs. However, to get a positive result for the last term, you must multiply two numbers with the same signs.\r\n\r\nThe only time a sum of squares can be factored is if they share any common factors, as in the following case:\r\n\r\n[latex]9x^2+36[\/latex]\r\n\r\nThe only way to factor this expression is by pulling out the GCF which is 9.\r\n\r\n[latex]9x^2+36=9(x^2+4)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

        Summary<\/h2>\r\nIn all factoring problems, we first look for a common factor that can be factored out of all terms. After achieving this (if possible), we have now learned two more factoring techniques in this section.\r\n\r\nIf a trinomial is of the form of a perfect square trinomial, it factors as [latex]a^2+2ab+b^2=(a+b)^2[\/latex] or [latex]a^2-2ab+b^2=(a-b)^2[\/latex].\r\n\r\nIf a binomial is of the form of a difference of squares, [latex]a^2-b^2[\/latex], it will factor as [latex]a^2-b^2=\\left(a+b\\right)\\left(a-b\\right)[\/latex].","rendered":"
        \n

        section 6.5 Learning Objectives<\/h3>\n

        6.5: Factoring Perfect Square Trinomials and the Difference of Squares<\/strong><\/p>\n

          \n
        • Factor perfect square trinomials<\/li>\n
        • Factor the difference of two squares<\/li>\n
        • Factor difference of squares by first factoring out the GCF<\/li>\n<\/ul>\n<\/div>\n

           <\/p>\n

          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 {\\left(a+b\\right)}^{2} & = & {a}^{2}+2ab+{b}^{2} \\hfill \\\\ & \\text{and}& \\\\ \\hfill {\\left(a-b\\right)}^{2} & = & {a}^{2}-2ab+{b}^{2}\\hfill \\end{array}[\/latex]<\/div>\n
          <\/div>\n
          \n

          We can use these same equations to factor any perfect square trinomial.<\/p>\n

          \n

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

          In our first example, we solve the problem two different ways to show how to use the above formulas while also demonstrating that this strategy will produce the same result as the techniques we learned earlier for factoring trinomials of this form.<\/p>\n

          \n

          Example 1<\/h3>\n

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

          Show Solution<\/span><\/p>\n
          \n

          We first note that the Product and Sum Method from Section 6.3 is applicable, so we begin by using that strategy.<\/p>\n

          We need to find two integers that multiply to [latex]49[\/latex] while adding to [latex]-14[\/latex]. The integers are [latex]-7[\/latex] and [latex]-7[\/latex]. Therefore, the polynomial factors as<\/p>\n

          [latex]x^2-14x+49=(x-7)(x-7)[\/latex].<\/p>\n

          Now, let us try the new Perfect Square Trinomial formula. Since our middle term is negative, we focus on the second formula. We begin by comparing our given problem to [latex]a^2-2ab+b^2[\/latex] and identifying [latex]a[\/latex] and [latex]b[\/latex]. Since we must have [latex]a^2=x^2[\/latex] and [latex]b^2=49[\/latex], we conclude that<\/p>\n

          [latex]a=x[\/latex] and [latex]b=7[\/latex].<\/p>\n

          In order to apply the formula, we must check that [latex]2ab[\/latex] matches our middle term (ignoring the negative sign as that was built into the formula). We have<\/p>\n

          [latex]2ab=2\\cdot x\\cdot 7=14x[\/latex].<\/p>\n

          This does match the middle term of our problem, and so, we can apply the formula [latex]a^2-2ab+b^2=(a-b)^2[\/latex] to get<\/p>\n

          [latex]x^2-14x+49=\\left(x-7\\right)^2[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n

          In the previous example, note that the results do match since [latex](x-7)(x-7)=(x-7)^2[\/latex]. Which way did you prefer? You may have found it easier to simply use the Product and Sum Strategy. However, as the problems become more complicated, the benefits of identifying Perfect Square Trinomials become more apparent.<\/p>\n

          The next example would have required AC-Method, where instead, we will apply our new formula.<\/p>\n

          \n

          Example 2<\/h3>\n

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

          Show Solution<\/span><\/p>\n
          \n

          First, 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].<\/p>\n

          This means that [latex]a=5x\\text{ and }b=2[\/latex]<\/p>\n

          Next, check to see if the middle term is equal to [latex]2ab[\/latex], which it is:<\/p>\n

          [latex]2ab = 2\\left(5x\\right)\\left(2\\right)=20x[\/latex]<\/p>\n

           <\/p>\n

          Therefore, the trinomial is a perfect square trinomial and can be written as:<\/p>\n

          [latex]{\\left(a+b\\right)}^{2}={\\left(5x+2\\right)}^{2}[\/latex].<\/p>\n

          Answer<\/span><\/h4>\n

          [latex]25x^2+20x+4=(5x+2)^{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n

          In the next example, we will show that we can use [latex]1 = 1^2[\/latex] to factor a polynomial with a term equal to\u00a0[latex]1[\/latex].<\/p>\n

          \n

          Example 3<\/h3>\n

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

          Show Solution<\/span><\/p>\n
          \n

          First, notice that [latex]49{x}^{2}[\/latex] and [latex]1[\/latex] are perfect squares because [latex]49{x}^{2}={\\left(7x\\right)}^{2}[\/latex] and [latex]1={1}^{2}[\/latex].<\/p>\n

          This means that [latex]a=7x[\/latex] and [latex]b=1[\/latex].<\/p>\n

          Next, check to see if the middle term (not including the negative) is equal to [latex]2ab[\/latex], which it is:<\/p>\n

          [latex]2ab = 2\\left(7x\\right)\\left(1\\right)=14x[\/latex]<\/p>\n

          Therefore, the trinomial is a perfect square trinomial and can be written as:<\/p>\n

          [latex]{\\left(a-b\\right)}^{2}={\\left(7x-1\\right)}^{2}[\/latex].<\/p>\n

          Answer<\/span><\/h4>\n

          [latex]49x^2-14x+1=(7x-1)^{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n

          In the following video, we provide another short description of what a perfect square trinomial is and show how to factor them using a formula.<\/p>\n