{"id":685,"date":"2019-04-29T16:37:52","date_gmt":"2019-04-29T20:37:52","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/?post_type=chapter&p=685"},"modified":"2020-02-03T12:55:10","modified_gmt":"2020-02-03T17:55:10","slug":"9-4-multiplication-and-division-of-radicals","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/chapter\/9-4-multiplication-and-division-of-radicals\/","title":{"raw":"9.4 Multiplication and Division of Radicals","rendered":"9.4 Multiplication and Division of Radicals"},"content":{"raw":"[latexpage]\r\n\r\nMultiplying radicals is very simple if the index on all the radicals match. The product rule of radicals, which is already been used, can be generalized as follows:\r\n

\\(\\text{Product Rule of Radicals: }a \\sqrt[m]{b}\\cdot c\\sqrt[m]{d} = ac \\sqrt[m]{bd}\\)<\/p>\r\nThis means that, if the index on the radicals match, then simply multiply the factors outside the radical and also multiply the factors inside the radicals. An example showing this is as follows.\r\n

\r\n

Example 9.4.1<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nMultiply \\(-5\\sqrt{14}\\cdot 4\\sqrt{6}\\).\r\n

\\(\\begin{array}{ll}\r\n\\text{This results in} & -5\\cdot 4\\sqrt{14\\cdot 6} \\\\\r\n\\text{Which simplifies to}&-20\\sqrt{84} \\\\\r\n\\text{Reducing inside the radical leaves}& -20\\sqrt{4\\cdot 21} \\\\\r\n\\text{Yielding}& -20\\cdot 2\\sqrt{21} \\\\\r\n\\text{Or}& -40\\sqrt{21}\r\n\\end{array}\\)<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\nThis same process works with any higher root radicals having matching indices.\r\n

\r\n

Example 9.4.2<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nMultiply \\(2\\sqrt[3]{18}\\cdot 6\\sqrt[3]{15}\\).\r\n

\\(\\begin{array}{ll}\r\n\\text{This results in} & 2\\cdot 6\\sqrt[3]{18\\cdot 15} \\\\\r\n\\text{Which simplifies to}& 12\\sqrt[3]{270} \\\\\r\n\\text{Reducing inside the radical leaves} & 12\\sqrt[3]{27\\cdot 10} \\\\\r\n\\text{Yielding} & 12\\cdot 3\\sqrt[3]{10} \\\\\r\n\\text{Or} & 36\\sqrt[3]{10}\r\n\\end{array}\\)<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\nThis process of multiplying radicals is the same when multiplying monomial radicals by binomial radicals, binomial radicals by binomial radicals, trinomial radicals (although these are not shown here), and so on.\r\n

\r\n

Example 9.4.3<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nMultiply \\(7\\sqrt{6}(3\\sqrt{10} - 5\\sqrt{15})\\).\r\n

\\(\\begin{array}{ll}\r\n\\text{Foiling the radicals will leave you with} & 21\\sqrt{60}-35\\sqrt{90} \\\\\r\n\\text{Reducing inside the radical leaves} & 21\\sqrt{4\\cdot 15}-35\\sqrt{9\\cdot 10}\\\\\r\n\\text{Yielding} & 21\\cdot 2\\sqrt{15} - 35\\cdot 3\\sqrt{10} \\\\\r\n\\text{Or} & 42\\sqrt{15} - 105\\sqrt{10}\r\n\\end{array}\\)<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

\r\n

Example 9.4.4<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nMultiply \\((\\sqrt{5} - 2\\sqrt{3})(4\\sqrt{10} + 6\\sqrt{6})\\).\r\n\r\nMultiplying the factors inside and outside the radicals yields:\r\n

\\(4\\sqrt{50} + 6\\sqrt{30} - 8\\sqrt{30} - 12\\sqrt{18}\\)<\/p>\r\n

\\(\\begin{array}{ll}\r\n\\text{Reducing inside these radicals leaves} & 4\\sqrt{25\\cdot 2}+6\\sqrt{30}-8\\sqrt{30}-12\\sqrt{9\\cdot 2}\\\\\r\n\\text{Yielding} & 4\\cdot 5\\sqrt{2}+6\\sqrt{30}-8\\sqrt{30}-12\\cdot 3\\sqrt{2}\\\\\r\n\\text{Or} & 20\\sqrt{2}+6\\sqrt{30}-8\\sqrt{30}-36\\sqrt{2} \\\\\r\n\\text{Which simplifies to} & -16\\sqrt{2}-2\\sqrt{30}\r\n\\end{array}\\)<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\nDivision with radicals is very similar to multiplication. If you think about division as reducing fractions, you can reduce the coefficients outside the radicals and reduce the values inside the radicals to get our final solution. There is one catch to dividing with radicals: it is considered bad practice to have a radical in the denominator of a final answer, so if there is a radical in the denominator, it should be rationalized by cancelling or multiplying the radicals.\r\n

\\(\\text{Quotient Rule of Radicals: }\\dfrac{a\\sqrt[m]{b}}{c\\sqrt[m]{d}} = \\left(\\dfrac{a}{c}\\right)\\sqrt[m]{\\dfrac{b}{d}}\\)<\/p>\r\nThe quotient rule means that factors outside the radical are divided by each other and the factors inside the radical are also divided by each other. To see this illustrated, consider the following:\r\n

\r\n

Example 9.4.5<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nReduce \\(\\dfrac{15 \\sqrt[3]{108}}{20\\sqrt[3]{2}}\\).\r\n\r\nUsing the quotient rule of radicals, this problem is separated into factors inside and outside the radicals. This results in the following:\r\n

\\(\\begin{array}{ll}\r\n& \\left(\\dfrac{15}{20}\\right) \\sqrt[3]{\\dfrac{108}{2}} \\\\ \\\\\r\n\\text{Simplifying the two resulting divisions leaves us with} & \\left(\\dfrac{3}{4}\\right) \\sqrt[3]{54} \\\\ \\\\\r\n\\text{Which we can further reduce to}& \\left(\\dfrac{3}{4}\\right) \\sqrt[3]{27\\cdot 2} \\\\ \\\\\r\n\\text{Taking the cube root of 27 leaves us with} & \\left(\\dfrac{3}{4}\\right) 3\\sqrt[3]{2} \\\\ \\\\\r\n\\text{Or} & \\left(\\dfrac{9}{4}\\right) \\sqrt[3]{2} \\\\ \\\\\r\n\\text{Which can also be written as} & \\dfrac{9\\sqrt[3]{2}}{4}\r\n\\end{array}\\)<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\nRemoving radicals from the denominator that cannot be divided out by using the numerator is often simply done by multiplying the numerator and denominator by a common radical. This is easily done and is shown by the following examples.\r\n

\r\n

Example 9.4.6<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nRationalize the denominator of \\(\\dfrac{\\sqrt{6}}{\\sqrt{5}}\\).\r\n\r\nFor this pair of radicals, the denominator \\(\\sqrt{5}\\) cannot be cancelled by the \\(\\sqrt{6}\\), so the solution requires that \\(\\sqrt{5}\\) be rationalized through multiplication. This is done as follows:\r\n

\\(\\dfrac{\\sqrt{6\\cdot 5}}{\\sqrt{5\\cdot 5}}\\)<\/p>\r\n

This now simplifies to:<\/p>\r\n

\\(\\dfrac{\\sqrt{30}}{5}\\)<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\nThis process is similar for radicals in which the index is greater than 2.\r\n

\r\n

Example 9.4.7<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nRationalize the denominator of \\(\\dfrac{4 \\sqrt[3]{6}}{5 \\sqrt[3]{25}}\\).\r\n\r\nTo rationalize the denominator, we need to get a cube root of 125, which will leave us with a denominator of 5 \u00d7 5. This requires that both the numerator and the denominator to be multiplied by the cube root of 5. This looks like:\r\n

\\(\\dfrac{4 \\sqrt[3]{6\\cdot 5}}{5 \\sqrt[3]{25\\cdot 5}}=\\dfrac{4 \\sqrt[3]{30}}{5 \\sqrt[3]{125}}\\)<\/p>\r\nThis simplifies to:\r\n

\\(\\dfrac{4\\sqrt[3]{30}}{5\\cdot 5}\\)<\/p>\r\nOr:\r\n

\\(\\dfrac{4\\sqrt[3]{30}}{25}\\)<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\nThe last example to be considered involves rationalizing denominators that have variables. Remeber to always reduce any fractions (inside and outside of the radical) before rationalizing.\r\n

\r\n

Example 9.4.8<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nRationalize the denominator of \\(\\dfrac{18 \\sqrt[4]{6x^3y^4z}}{8 \\sqrt[4]{10xy^6z^3}}\\).\r\n\r\nThe first thing to do is cancel all common factors both inside and outside the radicals. This leaves:\r\n

\\(\\dfrac{9 \\sqrt[4]{3x^2}}{4 \\sqrt[4]{5y^2z^2}}\\)<\/p>\r\nThe next step is to multiply both the numerator and denominator to rationalize the denominator:\r\n

\\(\\dfrac{9 \\sqrt[4]{3x^2}}{4 \\sqrt[4]{5y^2z^2}}\\cdot \\dfrac{\\sqrt[4]{125y^2z^2}}{\\sqrt[4]{125y^2z^2}}\\)<\/p>\r\nMultiplying these yields:\r\n

\\(\\dfrac{9 \\sqrt[4]{375x^2y^2z^2}}{4 \\sqrt[4]{625x^4y^4z^4}}\\)<\/p>\r\nTaking the fourth root of the denominator leaves:\r\n

\\(\\dfrac{9 \\sqrt[4]{375x^2y^2z^2}}{4\\cdot 5xyz}\\)<\/p>\r\nOr:\r\n

\\(\\dfrac{9 \\sqrt[4]{375x^2y^2z^2}}{20xyz}\\)<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

Questions<\/h1>\r\nSimplify.\r\n
    \r\n \t
  1. \\(3\\sqrt{5}\\cdot 4\\sqrt{16}\\)<\/li>\r\n \t
  2. \\(-5\\sqrt{10}\\cdot \\sqrt{15}\\)<\/li>\r\n \t
  3. \\(\\sqrt{12m}\\cdot \\sqrt{15m}\\)<\/li>\r\n \t
  4. \\(\\sqrt{5r^3}-5\\sqrt{10r^2}\\)<\/li>\r\n \t
  5. \\(\\sqrt[3]{4x^3}\\cdot \\sqrt[3]{2x^4}\\)<\/li>\r\n \t
  6. \\(3\\sqrt[3]{4a^4}\\cdot \\sqrt[3]{10a^3}\\)<\/li>\r\n \t
  7. \\(\\sqrt{6}(\\sqrt{2}+2)\\)<\/li>\r\n \t
  8. \\(\\sqrt{10}(\\sqrt{5}+\\sqrt{2})\\)<\/li>\r\n \t
  9. \\(-5\\sqrt{15}(3\\sqrt{3}+2)\\)<\/li>\r\n \t
  10. \\(5\\sqrt{15}(3\\sqrt{3}+2)\\)<\/li>\r\n \t
  11. \\(5\\sqrt{10}(5n+\\sqrt{2})\\)<\/li>\r\n \t
  12. \\(\\sqrt{15}(\\sqrt{5}-3\\sqrt{3v})\\)<\/li>\r\n \t
  13. \\((2+2\\sqrt{2})(-3+\\sqrt{2})\\)<\/li>\r\n \t
  14. \\((-2+\\sqrt{3})(-5+2\\sqrt{3})\\)<\/li>\r\n \t
  15. \\((\\sqrt{5}-5)(2\\sqrt{5}-1)\\)<\/li>\r\n \t
  16. \\((2\\sqrt{3}+\\sqrt{5})(5\\sqrt{3}+2\\sqrt{4})\\)<\/li>\r\n \t
  17. \\((\\sqrt{2a}+2\\sqrt{3a})(3\\sqrt{2a}+\\sqrt{5a})\\)<\/li>\r\n \t
  18. \\((-2\\sqrt{2p}+5\\sqrt{5})(\\sqrt{5p}+\\sqrt{5p})\\)<\/li>\r\n \t
  19. \\((-5-4\\sqrt{3})(-3-4\\sqrt{3})\\)<\/li>\r\n \t
  20. \\((5\\sqrt{2}-1)(-\\sqrt{2m}+5)\\)<\/li>\r\n \t
  21. \\(\\dfrac{\\sqrt{12}}{5\\sqrt{100}}\\)<\/li>\r\n \t
  22. \\(\\dfrac{\\sqrt{15}}{2\\sqrt{4}}\\)<\/li>\r\n \t
  23. \\(\\dfrac{\\sqrt{5}}{4\\sqrt{125}}\\)<\/li>\r\n \t
  24. \\(\\dfrac{\\sqrt{12}}{\\sqrt{3}}\\)<\/li>\r\n \t
  25. \\(\\dfrac{\\sqrt{10}}{\\sqrt{6}}\\)<\/li>\r\n \t
  26. \\(\\dfrac{\\sqrt{2}}{3\\sqrt{5}}\\)<\/li>\r\n \t
  27. \\(\\dfrac{5x^2}{4\\sqrt{3x^3y^3}}\\)<\/li>\r\n \t
  28. \\(\\dfrac{4}{5\\sqrt{3xy^4}}\\)<\/li>\r\n \t
  29. \\(\\dfrac{\\sqrt{2p^2}}{\\sqrt{3p}}\\)<\/li>\r\n \t
  30. \\(\\dfrac{\\sqrt{8n^2}}{\\sqrt{10n}}\\)<\/li>\r\n<\/ol>\r\nAnswer Key 9.4<\/a>","rendered":"

    Multiplying radicals is very simple if the index on all the radicals match. The product rule of radicals, which is already been used, can be generalized as follows:<\/p>\n

    \"\text{Product Rule of Radicals: }a \sqrt[m]{b}\cdot c\sqrt[m]{d} = ac \sqrt[m]{bd}\"<\/p>\n

    This means that, if the index on the radicals match, then simply multiply the factors outside the radical and also multiply the factors inside the radicals. An example showing this is as follows.<\/p>\n

    \n
    \n

    Example 9.4.1<\/p>\n<\/header>\n

    \n

    Multiply \"-5\sqrt{14}\cdot 4\sqrt{6}\".<\/p>\n

    \"\begin{array}{ll}<\/p>\n<\/div>\n<\/div>\n

    This same process works with any higher root radicals having matching indices.<\/p>\n

    \n
    \n

    Example 9.4.2<\/p>\n<\/header>\n

    \n

    Multiply \"2\sqrt[3]{18}\cdot 6\sqrt[3]{15}\".<\/p>\n

    \"\begin{array}{ll}<\/p>\n<\/div>\n<\/div>\n

    This process of multiplying radicals is the same when multiplying monomial radicals by binomial radicals, binomial radicals by binomial radicals, trinomial radicals (although these are not shown here), and so on.<\/p>\n

    \n
    \n

    Example 9.4.3<\/p>\n<\/header>\n

    \n

    Multiply \"7\sqrt{6}(3\sqrt{10} - 5\sqrt{15})\".<\/p>\n

    \"\begin{array}{ll}<\/p>\n<\/div>\n<\/div>\n

    \n
    \n

    Example 9.4.4<\/p>\n<\/header>\n

    \n

    Multiply \"(\sqrt{5} - 2\sqrt{3})(4\sqrt{10} + 6\sqrt{6})\".<\/p>\n

    Multiplying the factors inside and outside the radicals yields:<\/p>\n

    \"4\sqrt{50} + 6\sqrt{30} - 8\sqrt{30} - 12\sqrt{18}\"<\/p>\n

    \"\begin{array}{ll}<\/p>\n<\/div>\n<\/div>\n

    Division with radicals is very similar to multiplication. If you think about division as reducing fractions, you can reduce the coefficients outside the radicals and reduce the values inside the radicals to get our final solution. There is one catch to dividing with radicals: it is considered bad practice to have a radical in the denominator of a final answer, so if there is a radical in the denominator, it should be rationalized by cancelling or multiplying the radicals.<\/p>\n

    \"\text{Quotient Rule of Radicals: }\dfrac{a\sqrt[m]{b}}{c\sqrt[m]{d}} = \left(\dfrac{a}{c}\right)\sqrt[m]{\dfrac{b}{d}}\"<\/p>\n

    The quotient rule means that factors outside the radical are divided by each other and the factors inside the radical are also divided by each other. To see this illustrated, consider the following:<\/p>\n

    \n
    \n

    Example 9.4.5<\/p>\n<\/header>\n

    \n

    Reduce \"\dfrac{15 \sqrt[3]{108}}{20\sqrt[3]{2}}\".<\/p>\n

    Using the quotient rule of radicals, this problem is separated into factors inside and outside the radicals. This results in the following:<\/p>\n

    \"\begin{array}{ll}<\/p>\n<\/div>\n<\/div>\n

    Removing radicals from the denominator that cannot be divided out by using the numerator is often simply done by multiplying the numerator and denominator by a common radical. This is easily done and is shown by the following examples.<\/p>\n

    \n
    \n

    Example 9.4.6<\/p>\n<\/header>\n

    \n

    Rationalize the denominator of \"\dfrac{\sqrt{6}}{\sqrt{5}}\".<\/p>\n

    For this pair of radicals, the denominator \"\sqrt{5}\" cannot be cancelled by the \"\sqrt{6}\", so the solution requires that \"\sqrt{5}\" be rationalized through multiplication. This is done as follows:<\/p>\n

    \"\dfrac{\sqrt{6\cdot 5}}{\sqrt{5\cdot 5}}\"<\/p>\n

    This now simplifies to:<\/p>\n

    \"\dfrac{\sqrt{30}}{5}\"<\/p>\n<\/div>\n<\/div>\n

    This process is similar for radicals in which the index is greater than 2.<\/p>\n

    \n
    \n

    Example 9.4.7<\/p>\n<\/header>\n

    \n

    Rationalize the denominator of \"\dfrac{4 \sqrt[3]{6}}{5 \sqrt[3]{25}}\".<\/p>\n

    To rationalize the denominator, we need to get a cube root of 125, which will leave us with a denominator of 5 \u00d7 5. This requires that both the numerator and the denominator to be multiplied by the cube root of 5. This looks like:<\/p>\n

    \"\dfrac{4 \sqrt[3]{6\cdot 5}}{5 \sqrt[3]{25\cdot 5}}=\dfrac{4 \sqrt[3]{30}}{5 \sqrt[3]{125}}\"<\/p>\n

    This simplifies to:<\/p>\n

    \"\dfrac{4\sqrt[3]{30}}{5\cdot 5}\"<\/p>\n

    Or:<\/p>\n

    \"\dfrac{4\sqrt[3]{30}}{25}\"<\/p>\n<\/div>\n<\/div>\n

    The last example to be considered involves rationalizing denominators that have variables. Remeber to always reduce any fractions (inside and outside of the radical) before rationalizing.<\/p>\n

    \n
    \n

    Example 9.4.8<\/p>\n<\/header>\n

    \n

    Rationalize the denominator of \"\dfrac{18 \sqrt[4]{6x^3y^4z}}{8 \sqrt[4]{10xy^6z^3}}\".<\/p>\n

    The first thing to do is cancel all common factors both inside and outside the radicals. This leaves:<\/p>\n

    \"\dfrac{9 \sqrt[4]{3x^2}}{4 \sqrt[4]{5y^2z^2}}\"<\/p>\n

    The next step is to multiply both the numerator and denominator to rationalize the denominator:<\/p>\n

    \"\dfrac{9 \sqrt[4]{3x^2}}{4 \sqrt[4]{5y^2z^2}}\cdot \dfrac{\sqrt[4]{125y^2z^2}}{\sqrt[4]{125y^2z^2}}\"<\/p>\n

    Multiplying these yields:<\/p>\n

    \"\dfrac{9 \sqrt[4]{375x^2y^2z^2}}{4 \sqrt[4]{625x^4y^4z^4}}\"<\/p>\n

    Taking the fourth root of the denominator leaves:<\/p>\n

    \"\dfrac{9 \sqrt[4]{375x^2y^2z^2}}{4\cdot 5xyz}\"<\/p>\n

    Or:<\/p>\n

    \"\dfrac{9 \sqrt[4]{375x^2y^2z^2}}{20xyz}\"<\/p>\n<\/div>\n<\/div>\n

    Questions<\/h1>\n

    Simplify.<\/p>\n

      \n
    1. \"3\sqrt{5}\cdot 4\sqrt{16}\"<\/li>\n
    2. \"-5\sqrt{10}\cdot \sqrt{15}\"<\/li>\n
    3. \"\sqrt{12m}\cdot \sqrt{15m}\"<\/li>\n
    4. \"\sqrt{5r^3}-5\sqrt{10r^2}\"<\/li>\n
    5. \"\sqrt[3]{4x^3}\cdot \sqrt[3]{2x^4}\"<\/li>\n
    6. \"3\sqrt[3]{4a^4}\cdot \sqrt[3]{10a^3}\"<\/li>\n
    7. \"\sqrt{6}(\sqrt{2}+2)\"<\/li>\n
    8. \"\sqrt{10}(\sqrt{5}+\sqrt{2})\"<\/li>\n
    9. \"-5\sqrt{15}(3\sqrt{3}+2)\"<\/li>\n
    10. \"5\sqrt{15}(3\sqrt{3}+2)\"<\/li>\n
    11. \"5\sqrt{10}(5n+\sqrt{2})\"<\/li>\n
    12. \"\sqrt{15}(\sqrt{5}-3\sqrt{3v})\"<\/li>\n
    13. \"(2+2\sqrt{2})(-3+\sqrt{2})\"<\/li>\n
    14. \"(-2+\sqrt{3})(-5+2\sqrt{3})\"<\/li>\n
    15. \"(\sqrt{5}-5)(2\sqrt{5}-1)\"<\/li>\n
    16. \"(2\sqrt{3}+\sqrt{5})(5\sqrt{3}+2\sqrt{4})\"<\/li>\n
    17. \"(\sqrt{2a}+2\sqrt{3a})(3\sqrt{2a}+\sqrt{5a})\"<\/li>\n
    18. \"(-2\sqrt{2p}+5\sqrt{5})(\sqrt{5p}+\sqrt{5p})\"<\/li>\n
    19. \"(-5-4\sqrt{3})(-3-4\sqrt{3})\"<\/li>\n
    20. \"(5\sqrt{2}-1)(-\sqrt{2m}+5)\"<\/li>\n
    21. \"\dfrac{\sqrt{12}}{5\sqrt{100}}\"<\/li>\n
    22. \"\dfrac{\sqrt{15}}{2\sqrt{4}}\"<\/li>\n
    23. \"\dfrac{\sqrt{5}}{4\sqrt{125}}\"<\/li>\n
    24. \"\dfrac{\sqrt{12}}{\sqrt{3}}\"<\/li>\n
    25. \"\dfrac{\sqrt{10}}{\sqrt{6}}\"<\/li>\n
    26. \"\dfrac{\sqrt{2}}{3\sqrt{5}}\"<\/li>\n
    27. \"\dfrac{5x^2}{4\sqrt{3x^3y^3}}\"<\/li>\n
    28. \"\dfrac{4}{5\sqrt{3xy^4}}\"<\/li>\n
    29. \"\dfrac{\sqrt{2p^2}}{\sqrt{3p}}\"<\/li>\n
    30. \"\dfrac{\sqrt{8n^2}}{\sqrt{10n}}\"<\/li>\n<\/ol>\n

      Answer Key 9.4<\/a><\/p>\n","protected":false},"author":540,"menu_order":21,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-685","chapter","type-chapter","status-publish","hentry"],"part":389,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/685","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":11,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/685\/revisions"}],"predecessor-version":[{"id":3782,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/685\/revisions\/3782"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/parts\/389"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/685\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/media?parent=685"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapter-type?post=685"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/contributor?post=685"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/license?post=685"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}