{"id":1381,"date":"2021-12-02T19:37:44","date_gmt":"2021-12-03T00:37:44","guid":{"rendered":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/chapter\/7-6-factoring-quadratics-of-increasing-difficulty\/"},"modified":"2023-08-30T18:32:56","modified_gmt":"2023-08-30T22:32:56","slug":"factoring-quadratics-of-increasing-difficulty","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/chapter\/factoring-quadratics-of-increasing-difficulty\/","title":{"raw":"7.6 Factoring Quadratics of Increasing Difficulty","rendered":"7.6 Factoring Quadratics of Increasing Difficulty"},"content":{"raw":"Factoring 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 [latex]y^4 - 81x^4[\/latex].\r\n\r\nThis is a standard difference of squares that can be rewritten as [latex](y^2)^2 - (9x^2)^2[\/latex], which factors to [latex](y^2 - 9x^2)(y^2 + 9x^2)[\/latex]. This is not completely factored yet, since [latex](y^2 - 9x^2)[\/latex] can be factored once more to give [latex](y - 3x)(y + 3x)[\/latex].\r\n

Therefore, [latex]y^4 - 81x^4 = (y^2 + 9x^2)(y - 3x)(y + 3x)[\/latex].<\/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 [latex]x^6 - 64y^6[\/latex].This is a standard difference of squares that can be rewritten as [latex](x^3)^2 + (8x^3)^2[\/latex], which factors to [latex](x^3 - 8y^3)(x^3 + 8x^3)[\/latex]. This is not completely factored yet, since both [latex](x^3 - 8y^3)[\/latex] and [latex](x^3 + 8x^3)[\/latex] can be factored again.\r\n

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

[latex]x^6 - 64y^6 = (x - 2y)(x^2 + 2xy + y^2)(x + 2y)(x^2 - 2xy + y^2)[\/latex]<\/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 [latex]x^6 + 64y^6[\/latex]. 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

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

Therefore, [latex]x^6 + 64y^6 = (x^2 + 4y^2)(x^4 + 4x^2y^2 + 64y^4)[\/latex].<\/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: [latex](a + b)^2 - (2a - 3b)^2[\/latex] factors as a standard difference of squares as shown below:\r\n

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

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

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

[latex](a + b)^2 - (2a - 3b)^2 = [-a - 4b] [3a - 2b][\/latex]<\/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[latex](a + b)^3 - (2a - 3b)^3[\/latex] factors as a standard difference of squares as shown below:\r\n

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

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

[latex]\\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\\hline\r\n&[-a-2b]&[7a^2+19ab+13b^2]\r\n\\end{array}[\/latex]<\/p>\r\nTherefore, the result is:\r\n

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

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

    Therefore, [latex]y^4 - 81x^4 = (y^2 + 9x^2)(y - 3x)(y + 3x)[\/latex].<\/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 [latex]x^6 - 64y^6[\/latex].This is a standard difference of squares that can be rewritten as [latex](x^3)^2 + (8x^3)^2[\/latex], which factors to [latex](x^3 - 8y^3)(x^3 + 8x^3)[\/latex]. This is not completely factored yet, since both [latex](x^3 - 8y^3)[\/latex] and [latex](x^3 + 8x^3)[\/latex] can be factored again.<\/p>\n

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

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

    [latex]x^6 - 64y^6 = (x - 2y)(x^2 + 2xy + y^2)(x + 2y)(x^2 - 2xy + y^2)[\/latex]<\/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 [latex]x^6 + 64y^6[\/latex]. 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

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

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

    Therefore, [latex]x^6 + 64y^6 = (x^2 + 4y^2)(x^4 + 4x^2y^2 + 64y^4)[\/latex].<\/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: [latex](a + b)^2 - (2a - 3b)^2[\/latex] factors as a standard difference of squares as shown below:<\/p>\n

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

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

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

    Which reduces to:<\/p>\n

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

    Therefore:<\/p>\n

    [latex](a + b)^2 - (2a - 3b)^2 = [-a - 4b] [3a - 2b][\/latex]<\/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

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

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

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

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

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

    [latex]\\begin{array}{rrl} &[\\phantom{-1}a+\\phantom{0}b]&[\\phantom{0}a^2+\\phantom{0}2ab+\\phantom{00}b^2] \\\\ &[-2a-3b]&[2a^2+\\phantom{0}5ab+\\phantom{0}3b^2] \\\\ +&&[4a^2+12ab+\\phantom{0}9b^2] \\\\ \\hline &[-a-2b]&[7a^2+19ab+13b^2] \\end{array}[\/latex]<\/p>\n

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

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

    Questions<\/h1>\n

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

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

      Answer Key 7.6<\/a><\/p>\n","protected":false},"author":90,"menu_order":6,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":"cc-by-nc-sa"},"chapter-type":[],"contributor":[],"license":[56],"class_list":["post-1381","chapter","type-chapter","status-publish","hentry","license-cc-by-nc-sa"],"part":1369,"_links":{"self":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1381","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/users\/90"}],"version-history":[{"count":2,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1381\/revisions"}],"predecessor-version":[{"id":2127,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1381\/revisions\/2127"}],"part":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/parts\/1369"}],"metadata":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1381\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/media?parent=1381"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapter-type?post=1381"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/contributor?post=1381"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/license?post=1381"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}