{"id":648,"date":"2019-04-29T16:21:05","date_gmt":"2019-04-29T20:21:05","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/?post_type=chapter&p=648"},"modified":"2019-12-05T11:28:51","modified_gmt":"2019-12-05T16:28:51","slug":"8-3-least-common-denominators","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/chapter\/8-3-least-common-denominators\/","title":{"raw":"8.3 Least Common Denominators","rendered":"8.3 Least Common Denominators"},"content":{"raw":"[latexpage]\r\n\r\nFinding the least common denominator, or LCD, is very important to working with rational expressions. The process used depends on finding what is common to each rational expression and identifying what is not common. These common and not common factors are then combined to form the LCD.\r\n
\r\n

Example 8.3.1<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nFind the LCD of the numbers 12, 8, and 6.\r\n\r\nFirst, break these three numbers into primes:\r\n\r\n\\[\\begin{array}{rrl}\r\n12&=&2\\cdot 2\\cdot 3\\text{ or }2^2\\cdot 3 \\\\\r\n8&=&2\\cdot 2\\cdot 2\\text{ or }2^3 \\\\\r\n6&=&2\\cdot 3\r\n\\end{array}\\]\r\n\r\nThen write out the primes for the first number, 12, and set the LCD to \\(2^2\\cdot 3\\).\r\n\r\nNotice the factorization of 8 includes \\(2^3\\), yet the LCD currently only has \\(2^2\\), so you add one 2.\r\n\r\nNow the LCD = \\(2^3\\cdot 3\\).\r\n\r\nChecking \\(6 = 2\\cdot 3\\), there already is a \\(2\\cdot 3\\) in the LCD, so we need not add any more primes.\r\n\r\nThe LCD = \\(2^3\\cdot 3\\) or 24.\r\n\r\n<\/div>\r\n<\/div>\r\n

This process can be duplicated with variables. <\/span><\/p>\r\n\r\n

\r\n

Example 8.3.2<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nFind the LCD of \\(4x^2y^5\\) and \\(6x^4y^3z^6\\).\r\n\r\nFirst, break both terms into primes:\r\n\r\n\\[\\begin{array}{rrl}\r\n4x^2y^5&=&2^2\\cdot x^2\\cdot y^5 \\\\\r\n6x^4y^3z^6&=&2\\cdot 3\\cdot x^4\\cdot y^3\\cdot z^6\r\n\\end{array}\\]\r\n\r\nThen write out the primes for the first term, \\(4x^2y^5\\), and set the LCD to \\(2^2\\cdot x^2\\cdot y^5\\).\r\n\r\nThe LCD for \\(6x^4y^3z^6=2\\cdot 3\\cdot x^4\\cdot y^3\\cdot z^6\\) has an extra 3, \\(x^2\\), and \\(z^6\\), which we add to the LCD that we are constructing.\r\n\r\nThis yields LCD = \\(2^2\\cdot 3\\cdot x^{2+2}\\cdot y^5\\cdot z^6\\), or LCD = \\(12x^4y^5z^6\\).\r\n\r\n<\/div>\r\n<\/div>\r\n

This process can also be duplicated with polynomials.<\/span><\/p>\r\n\r\n

\r\n

Example 8.3.3<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nFind the LCD of \\(x^2 + 2x - 3\\) and \\(x^2 - x - 12\\).\r\n\r\nFirst, we factor both of these polynomials, much like finding the primes of the above terms:\r\n\r\n\\[\\begin{array}{rrl}\r\nx^2 + 2x - 3&=&(x-1)(x+3) \\\\\r\nx^2 - x - 12&=&(x-4)(x+3)\r\n\\end{array}\\]\r\n\r\nThe LCD is constructed as we did before, except this time, we write out the factored terms from the first polynomial, so the LCD = \\((x - 1)(x + 3)\\).\r\n\r\nNotice that \\(x^2 - x - 12 = (x - 4)(x + 3)\\), where the \\((x + 3)\\) is already in the LCD, which means that we only need to add \\((x - 4)\\).\r\n\r\nThe LCD = \\((x - 1)(x + 3)(x - 4)\\).\r\n\r\n<\/div>\r\n<\/div>\r\n

Questions<\/h1>\r\n

<\/span>For Questions 1 to 10, find each Least Common Denominator.<\/p>\r\n\r\n

    \r\n \t
  1. \\(2a^3, 6a^4b^2, 4a^3b^5\\)<\/li>\r\n \t
  2. \\(5x^2y, 25x^3y^5z\\)<\/li>\r\n \t
  3. \\(x^2-3x, x-3, x\\)<\/li>\r\n \t
  4. \\(4x-8, x-2, 4\\)<\/li>\r\n \t
  5. \\(x+2, x-4\\)<\/li>\r\n \t
  6. \\(x, x-7, x+1\\)<\/li>\r\n \t
  7. \\(x^2-25, x+5\\)<\/li>\r\n \t
  8. \\(x^2-9, x^2-6x+9\\)<\/li>\r\n \t
  9. \\(x^2+3x+2, x^2+5x+6\\)<\/li>\r\n \t
  10. \\(x^2-7x+10, x^2-2x-15, x^2+x-6\\)<\/li>\r\n<\/ol>\r\nFor Questions 11 to 20, find the LCD of each fraction and place each expression over the same common denominator.\r\n
      \r\n \t
    1. \\(\\dfrac{3a}{5b^2}, \\dfrac{2}{10a^3b}\\)<\/li>\r\n \t
    2. \\(\\dfrac{3x}{x-4}, \\dfrac{2}{x+2}\\)<\/li>\r\n \t
    3. \\(\\dfrac{x+2}{x-3}, \\dfrac{x-3}{x+2}\\)<\/li>\r\n \t
    4. \\(\\dfrac{5}{x^2-6x}, \\dfrac{2}{x}, \\dfrac{-3}{x-6}\\)<\/li>\r\n \t
    5. \\(\\dfrac{x}{x^2-16}, \\dfrac{3x}{x^2-8x+16}\\)<\/li>\r\n \t
    6. \\(\\dfrac{5x+1}{x^2-3x-10}, \\dfrac{4}{x-5}\\)<\/li>\r\n \t
    7. \\(\\dfrac{x+1}{x^2-36}, \\dfrac{2x+3}{x^2+12x+36}\\)<\/li>\r\n \t
    8. \\(\\dfrac{3x+1}{x^2-x-12}, \\dfrac{2x}{x^2+4x+3}\\)<\/li>\r\n \t
    9. \\(\\dfrac{4x}{x^2-x-6}, \\dfrac{x+2}{x-3}\\)<\/li>\r\n \t
    10. \\(\\dfrac{3x}{x^2-6x+8}, \\dfrac{x-2}{x^2+x-20}, \\dfrac{5}{x^2+3x-10}\\)<\/li>\r\n<\/ol>\r\nAnswer Key 8.3<\/a>","rendered":"

      Finding the least common denominator, or LCD, is very important to working with rational expressions. The process used depends on finding what is common to each rational expression and identifying what is not common. These common and not common factors are then combined to form the LCD.<\/p>\n

      \n
      \n

      Example 8.3.1<\/p>\n<\/header>\n

      \n

      Find the LCD of the numbers 12, 8, and 6.<\/p>\n

      First, break these three numbers into primes:<\/p>\n

        <\/span>   <\/span>\"\[\begin{array}{rrl}<\/p>\n

      Then write out the primes for the first number, 12, and set the LCD to \"2^2\cdot 3\".<\/p>\n

      Notice the factorization of 8 includes \"2^3\", yet the LCD currently only has \"2^2\", so you add one 2.<\/p>\n

      Now the LCD = \"2^3\cdot 3\".<\/p>\n

      Checking \"6 = 2\cdot 3\", there already is a \"2\cdot 3\" in the LCD, so we need not add any more primes.<\/p>\n

      The LCD = \"2^3\cdot 3\" or 24.<\/p>\n<\/div>\n<\/div>\n

      This process can be duplicated with variables. <\/span><\/p>\n

      \n
      \n

      Example 8.3.2<\/p>\n<\/header>\n

      \n

      Find the LCD of \"4x^2y^5\" and \"6x^4y^3z^6\".<\/p>\n

      First, break both terms into primes:<\/p>\n

        <\/span>   <\/span>\"\[\begin{array}{rrl}<\/p>\n

      Then write out the primes for the first term, \"4x^2y^5\", and set the LCD to \"2^2\cdot x^2\cdot y^5\".<\/p>\n

      The LCD for \"6x^4y^3z^6=2\cdot 3\cdot x^4\cdot y^3\cdot z^6\" has an extra 3, \"x^2\", and \"z^6\", which we add to the LCD that we are constructing.<\/p>\n

      This yields LCD = \"2^2\cdot 3\cdot x^{2+2}\cdot y^5\cdot z^6\", or LCD = \"12x^4y^5z^6\".<\/p>\n<\/div>\n<\/div>\n

      This process can also be duplicated with polynomials.<\/span><\/p>\n

      \n
      \n

      Example 8.3.3<\/p>\n<\/header>\n

      \n

      Find the LCD of \"x^2 + 2x - 3\" and \"x^2 - x - 12\".<\/p>\n

      First, we factor both of these polynomials, much like finding the primes of the above terms:<\/p>\n

        <\/span>   <\/span>\"\[\begin{array}{rrl}<\/p>\n

      The LCD is constructed as we did before, except this time, we write out the factored terms from the first polynomial, so the LCD = \"(x - 1)(x + 3)\".<\/p>\n

      Notice that \"x^2 - x - 12 = (x - 4)(x + 3)\", where the \"(x + 3)\" is already in the LCD, which means that we only need to add \"(x - 4)\".<\/p>\n

      The LCD = \"(x - 1)(x + 3)(x - 4)\".<\/p>\n<\/div>\n<\/div>\n

      Questions<\/h1>\n

      <\/span>For Questions 1 to 10, find each Least Common Denominator.<\/p>\n

        \n
      1. \"2a^3, 6a^4b^2, 4a^3b^5\"<\/li>\n
      2. \"5x^2y, 25x^3y^5z\"<\/li>\n
      3. \"x^2-3x, x-3, x\"<\/li>\n
      4. \"4x-8, x-2, 4\"<\/li>\n
      5. \"x+2, x-4\"<\/li>\n
      6. \"x, x-7, x+1\"<\/li>\n
      7. \"x^2-25, x+5\"<\/li>\n
      8. \"x^2-9, x^2-6x+9\"<\/li>\n
      9. \"x^2+3x+2, x^2+5x+6\"<\/li>\n
      10. \"x^2-7x+10, x^2-2x-15, x^2+x-6\"<\/li>\n<\/ol>\n

        For Questions 11 to 20, find the LCD of each fraction and place each expression over the same common denominator.<\/p>\n

          \n
        1. \"\dfrac{3a}{5b^2}, \dfrac{2}{10a^3b}\"<\/li>\n
        2. \"\dfrac{3x}{x-4}, \dfrac{2}{x+2}\"<\/li>\n
        3. \"\dfrac{x+2}{x-3}, \dfrac{x-3}{x+2}\"<\/li>\n
        4. \"\dfrac{5}{x^2-6x}, \dfrac{2}{x}, \dfrac{-3}{x-6}\"<\/li>\n
        5. \"\dfrac{x}{x^2-16}, \dfrac{3x}{x^2-8x+16}\"<\/li>\n
        6. \"\dfrac{5x+1}{x^2-3x-10}, \dfrac{4}{x-5}\"<\/li>\n
        7. \"\dfrac{x+1}{x^2-36}, \dfrac{2x+3}{x^2+12x+36}\"<\/li>\n
        8. \"\dfrac{3x+1}{x^2-x-12}, \dfrac{2x}{x^2+4x+3}\"<\/li>\n
        9. \"\dfrac{4x}{x^2-x-6}, \dfrac{x+2}{x-3}\"<\/li>\n
        10. \"\dfrac{3x}{x^2-6x+8}, \dfrac{x-2}{x^2+x-20}, \dfrac{5}{x^2+3x-10}\"<\/li>\n<\/ol>\n

          Answer Key 8.3<\/a><\/p>\n","protected":false},"author":540,"menu_order":19,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-648","chapter","type-chapter","status-publish","hentry"],"part":386,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/648","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/users\/540"}],"version-history":[{"count":8,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/648\/revisions"}],"predecessor-version":[{"id":3628,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/648\/revisions\/3628"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/parts\/386"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/648\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/media?parent=648"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapter-type?post=648"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/contributor?post=648"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/license?post=648"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}