{"id":605,"date":"2019-04-29T15:59:52","date_gmt":"2019-04-29T19:59:52","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/?post_type=chapter&p=605"},"modified":"2020-02-03T13:58:12","modified_gmt":"2020-02-03T18:58:12","slug":"7-6-factoring-quadratics-of-increasing-difficulty","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/chapter\/7-6-factoring-quadratics-of-increasing-difficulty\/","title":{"raw":"7.6 Factoring Quadratics of Increasing Difficulty","rendered":"7.6 Factoring Quadratics of Increasing Difficulty"},"content":{"raw":"[latexpage]\r\n\r\nFactoring equations that are more difficult involves factoring equations and then checking the answers to see if they can be factored again.\r\n
\r\n

Example 7.6.1<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nFactor \\(y^4 - 81x^4\\).\r\n\r\nThis is a standard difference of squares that can be rewritten as \\((y^2)^2 - (9x^2)^2\\), which factors to \\((y^2 - 9x^2)(y^2 + 9x^2)\\). This is not completely factored yet, since \\((y^2 - 9x^2)\\) can be factored once more to give \\((y - 3x)(y + 3x)\\).\r\n

Therefore, \\(y^4 - 81x^4 = (y^2 + 9x^2)(y - 3x)(y + 3x)\\).<\/p>\r\nThis multiple factoring of an equation is also common in mixing differences of squares with differences of cubes.\r\n\r\n<\/div>\r\n<\/div>\r\n

\r\n

Example 7.6.2<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nFactor \\(x^6 - 64y^6\\).This is a standard difference of squares that can be rewritten as \\((x^3)^2 + (8x^3)^2\\), which factors to \\((x^3 - 8y^3)(x^3 + 8x^3)\\). This is not completely factored yet, since both \\((x^3 - 8y^3)\\) and \\((x^3 + 8x^3)\\) can be factored again.\r\n

\\((x^3-8y^3)=(x-2y)(x^2+2xy+y^2)\\) and\r\n\\((x^3+8y^3)=(x+2y)(x^2-2xy+y^2)\\)<\/p>\r\nThis means that the complete factorization for this is:\r\n

\\(x^6 - 64y^6\u00a0 = (x - 2y)(x^2 + 2xy + y^2)(x + 2y)(x^2 - 2xy + y^2)\\)<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

\r\n

Example 7.6.3<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nA more challenging equation to factor looks like \\(x^6 + 64y^6\\). This is not an equation that can be put in the factorable form of a difference of squares. However, it can be put in the form of a sum of cubes.\r\n

\\(x^6 + 64y^6 = (x^2)^3 + (4y^2)^3\\)<\/p>\r\nIn this form, \\((x^2)^3+(4y^2)^3\\) factors to \\((x^2+4y^2)(x^4+4x^2y^2+64y^4)\\).\r\n

Therefore, \\(x^6 + 64y^6 = (x^2 + 4y^2)(x^4 + 4x^2y^2 + 64y^4)\\).<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

\r\n

Example 7.6.4<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nConsider encountering a sum and difference of squares question. These can be factored as follows: \\((a + b)^2 - (2a - 3b)^2\\) factors as a standard difference of squares as shown below:\r\n

\\((a+b)^2-(2a-3b)^2=[(a+b)-(2a-3b)][(a+b)+(2a-3b)]\\)<\/p>\r\nSimplifying inside the brackets yields:\r\n

\\([a + b - 2a + 3b] [a + b + 2a - 3b]\\)<\/p>\r\nWhich reduces to:\r\n

\\([-a + 4b] [3a - 2b]\\)<\/p>\r\nTherefore:\r\n

\\((a + b)^2 - (2a - 3b)^2\u00a0 =\u00a0 [-a - 4b] [3a - 2b]\\)<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

\r\n

Examples 7.6.5<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nConsider encountering the following difference of cubes question. This can be factored as follows:\r\n\r\n\\((a + b)^3 - (2a - 3b)^3\\) factors as a standard difference of squares as shown below:\r\n

\\((a+b)^3-(2a-3b)^3\\)\r\n\\(=[(a+b)-(2a+3b)][(a+b)^2+(a+b)(2a+3b)+(2a+3b)^2]\\)<\/p>\r\nSimplifying inside the brackets yields:\r\n

\\([a+b-2a-3b][a^2+2ab+b^2+2a^2+5ab+3b^2+4a^2+12ab+9b^2]\\)<\/p>\r\nSorting and combining all similar terms yields:\r\n

\\(\\begin{array}{rrl}\r\n&[\\phantom{-1}a+\\phantom{0}b]&[\\phantom{0}a^2+\\phantom{0}2ab+\\phantom{00}b^2] \\\\\r\n&[-2a-3b]&[2a^2+\\phantom{0}5ab+\\phantom{0}3b^2] \\\\\r\n+&&[4a^2+12ab+\\phantom{0}9b^2] \\\\\r\n\\midrule\r\n&[-a-2b]&[7a^2+19ab+13b^2]\r\n\\end{array}\\)<\/p>\r\nTherefore, the result is:\r\n

\\((a + b)^3 - (2a - 3b)^3\u00a0 =\u00a0 [-a - 2b] [7a^2 + 19ab + 13b^2]\\)<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

Questions<\/h1>\r\nCompletely factor the following equations.\r\n
    \r\n \t
  1. \\(x^4-16y^4\\)<\/li>\r\n \t
  2. \\(16x^4-81y^4\\)<\/li>\r\n \t
  3. \\(x^4-256y^4\\)<\/li>\r\n \t
  4. \\(625x^4-81y^4\\)<\/li>\r\n \t
  5. \\(81x^4-16y^4\\)<\/li>\r\n \t
  6. \\(x^4-81y^4\\)<\/li>\r\n \t
  7. \\(625x^4-256y^4\\)<\/li>\r\n \t
  8. \\(x^4-81y^4\\)<\/li>\r\n \t
  9. \\(x^6-y^6\\)<\/li>\r\n \t
  10. \\(x^6+y^6\\)<\/li>\r\n \t
  11. \\(x^6-64y^6\\)<\/li>\r\n \t
  12. \\(64x^6+y^6\\)<\/li>\r\n \t
  13. \\(729x^6-y^6\\)<\/li>\r\n \t
  14. \\(729x^6+y^6\\)<\/li>\r\n \t
  15. \\(729x^6+64y^6\\)<\/li>\r\n \t
  16. \\(64x^6-15625y^6\\)<\/li>\r\n \t
  17. \\((a+b)^2-(c-d)^2\\)<\/li>\r\n \t
  18. \\((a+2b)^2-(3a-4b)^2\\)<\/li>\r\n \t
  19. \\((a+3b)^2-(2c-d)^2\\)<\/li>\r\n \t
  20. \\((3a+b)^2-(a-b)^2\\)<\/li>\r\n \t
  21. \\((a+b)^3-(c-d)^3\\)<\/li>\r\n \t
  22. \\((a+3b)^3+(4a-b)^3\\)<\/li>\r\n<\/ol>\r\nAnswer Key 7.6<\/a>","rendered":"

    Factoring equations that are more difficult involves factoring equations and then checking the answers to see if they can be factored again.<\/p>\n

    \n
    \n

    Example 7.6.1<\/p>\n<\/header>\n

    \n

    Factor \"y^4 - 81x^4\".<\/p>\n

    This is a standard difference of squares that can be rewritten as \"(y^2)^2 - (9x^2)^2\", which factors to \"(y^2 - 9x^2)(y^2 + 9x^2)\". This is not completely factored yet, since \"(y^2 - 9x^2)\" can be factored once more to give \"(y - 3x)(y + 3x)\".<\/p>\n

    Therefore, \"y^4 - 81x^4 = (y^2 + 9x^2)(y - 3x)(y + 3x)\".<\/p>\n

    This multiple factoring of an equation is also common in mixing differences of squares with differences of cubes.<\/p>\n<\/div>\n<\/div>\n

    \n
    \n

    Example 7.6.2<\/p>\n<\/header>\n

    \n

    Factor \"x^6 - 64y^6\".This is a standard difference of squares that can be rewritten as \"(x^3)^2 + (8x^3)^2\", which factors to \"(x^3 - 8y^3)(x^3 + 8x^3)\". This is not completely factored yet, since both \"(x^3 - 8y^3)\" and \"(x^3 + 8x^3)\" can be factored again.<\/p>\n

    \"(x^3-8y^3)=(x-2y)(x^2+2xy+y^2)\" and
    \n\"(x^3+8y^3)=(x+2y)(x^2-2xy+y^2)\"<\/p>\n

    This means that the complete factorization for this is:<\/p>\n

    \"x^6 - 64y^6 = (x - 2y)(x^2 + 2xy + y^2)(x + 2y)(x^2 - 2xy + y^2)\"<\/p>\n<\/div>\n<\/div>\n

    \n
    \n

    Example 7.6.3<\/p>\n<\/header>\n

    \n

    A more challenging equation to factor looks like \"x^6 + 64y^6\". This is not an equation that can be put in the factorable form of a difference of squares. However, it can be put in the form of a sum of cubes.<\/p>\n

    \"x^6 + 64y^6 = (x^2)^3 + (4y^2)^3\"<\/p>\n

    In this form, \"(x^2)^3+(4y^2)^3\" factors to \"(x^2+4y^2)(x^4+4x^2y^2+64y^4)\".<\/p>\n

    Therefore, \"x^6 + 64y^6 = (x^2 + 4y^2)(x^4 + 4x^2y^2 + 64y^4)\".<\/p>\n<\/div>\n<\/div>\n

    \n
    \n

    Example 7.6.4<\/p>\n<\/header>\n

    \n

    Consider encountering a sum and difference of squares question. These can be factored as follows: \"(a + b)^2 - (2a - 3b)^2\" factors as a standard difference of squares as shown below:<\/p>\n

    \"(a+b)^2-(2a-3b)^2=[(a+b)-(2a-3b)][(a+b)+(2a-3b)]\"<\/p>\n

    Simplifying inside the brackets yields:<\/p>\n

    \"[a + b - 2a + 3b] [a + b + 2a - 3b]\"<\/p>\n

    Which reduces to:<\/p>\n

    \"[-a + 4b] [3a - 2b]\"<\/p>\n

    Therefore:<\/p>\n

    \"(a + b)^2 - (2a - 3b)^2 = [-a - 4b] [3a - 2b]\"<\/p>\n<\/div>\n<\/div>\n

    \n
    \n

    Examples 7.6.5<\/p>\n<\/header>\n

    \n

    Consider encountering the following difference of cubes question. This can be factored as follows:<\/p>\n

    \"(a + b)^3 - (2a - 3b)^3\" factors as a standard difference of squares as shown below:<\/p>\n

    \"(a+b)^3-(2a-3b)^3\"
    \n\"=[(a+b)-(2a+3b)][(a+b)^2+(a+b)(2a+3b)+(2a+3b)^2]\"<\/p>\n

    Simplifying inside the brackets yields:<\/p>\n

    \"[a+b-2a-3b][a^2+2ab+b^2+2a^2+5ab+3b^2+4a^2+12ab+9b^2]\"<\/p>\n

    Sorting and combining all similar terms yields:<\/p>\n

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

    Therefore, the result is:<\/p>\n

    \"(a + b)^3 - (2a - 3b)^3 = [-a - 2b] [7a^2 + 19ab + 13b^2]\"<\/p>\n<\/div>\n<\/div>\n

    Questions<\/h1>\n

    Completely factor the following equations.<\/p>\n

      \n
    1. \"x^4-16y^4\"<\/li>\n
    2. \"16x^4-81y^4\"<\/li>\n
    3. \"x^4-256y^4\"<\/li>\n
    4. \"625x^4-81y^4\"<\/li>\n
    5. \"81x^4-16y^4\"<\/li>\n
    6. \"x^4-81y^4\"<\/li>\n
    7. \"625x^4-256y^4\"<\/li>\n
    8. \"x^4-81y^4\"<\/li>\n
    9. \"x^6-y^6\"<\/li>\n
    10. \"x^6+y^6\"<\/li>\n
    11. \"x^6-64y^6\"<\/li>\n
    12. \"64x^6+y^6\"<\/li>\n
    13. \"729x^6-y^6\"<\/li>\n
    14. \"729x^6+y^6\"<\/li>\n
    15. \"729x^6+64y^6\"<\/li>\n
    16. \"64x^6-15625y^6\"<\/li>\n
    17. \"(a+b)^2-(c-d)^2\"<\/li>\n
    18. \"(a+2b)^2-(3a-4b)^2\"<\/li>\n
    19. \"(a+3b)^2-(2c-d)^2\"<\/li>\n
    20. \"(3a+b)^2-(a-b)^2\"<\/li>\n
    21. \"(a+b)^3-(c-d)^3\"<\/li>\n
    22. \"(a+3b)^3+(4a-b)^3\"<\/li>\n<\/ol>\n

      Answer Key 7.6<\/a><\/p>\n","protected":false},"author":540,"menu_order":20,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-605","chapter","type-chapter","status-publish","hentry"],"part":379,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/605","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\/605\/revisions"}],"predecessor-version":[{"id":3788,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/605\/revisions\/3788"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/parts\/379"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/605\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/media?parent=605"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapter-type?post=605"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/contributor?post=605"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/license?post=605"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}