{"id":596,"date":"2019-04-29T15:57:34","date_gmt":"2019-04-29T19:57:34","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/?post_type=chapter&p=596"},"modified":"2019-12-03T16:59:48","modified_gmt":"2019-12-03T21:59:48","slug":"7-2-factoring-by-grouping","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/chapter\/7-2-factoring-by-grouping\/","title":{"raw":"7.2 Factoring by Grouping","rendered":"7.2 Factoring by Grouping"},"content":{"raw":"[latexpage]\r\n\r\nFirst thing to do when factoring is to factor out the GCF. This GCF is often a monomial, like in the problem \\(5xy + 10xz\\) where the GCF is the monomial \\(5x\\), so you would have \\(5x(y + 2z)\\). However, a GCF does not have to be a monomial; it could be a binomial. Consider the following two examples.\r\n
\r\n

Example 7.2.1<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nFind and factor out the GCF for \\(3ax - 7bx\\).\r\n\r\nBy observation, one can see that both have \\(x\\) in common.\r\n\r\nThis means that \\(3ax - 7bx =\u00a0 x(3a - 7b)\\).\r\n\r\n<\/div>\r\n<\/div>\r\n
\r\n

Example 7.2.2<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nFind and factor out the GCF for \\(3a(2a + 5b) - 7b(2a + 5b)\\).\r\n\r\nBoth have \\((2a + 5b)\\) as a common factor.\r\n\r\nThis means that if you factor out \\((2a + 5b)\\), you are left with \\(3a - 7b\\).\r\n\r\nThe factored polynomial is written as \\((2a + 5b)(3a - 7b)\\).\r\n\r\n<\/div>\r\n<\/div>\r\n

In the same way as factoring out a GCF from a binomial, there is a process known as grouping to factor out common binomials from a polynomial containing four terms.<\/span><\/p>\r\n\r\n

\r\n
\r\n\r\nFind and factor out the GCF for \\(10ab + 15b^2 + 4a + 6b\\).\r\n\r\nTo do this, first split the polynomial into two binomials.\r\n

\\(10ab + 15b^2 + 4a + 6b\\) becomes \\(10ab + 15b^2\\) and \\(4a + 6b\\).<\/p>\r\nNow find the common factor from each binomial.\r\n

\\(10ab + 15b^2\\) has a common factor of \\(5b\\) and becomes \\(5b(2a + 3b)\\).<\/p>\r\n

\\(4a + 6b\\) has a common factor of 2 and becomes \\(2(2a + 3b)\\).<\/p>\r\nThis means that \\(10ab + 15b^2 + 4a + 6b = 5b(2a + 3b) + 2(2a + 3b)\\).\r\n

\\(5b(2a + 3b) + 2(2a + 3b)\\) can be factored as \\((2a + 3b)(5b + 2)\\).<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

Questions<\/h1>\r\nFactor the following polynomials.\r\n
    \r\n \t
  1. \\(40r^3-8r^2-25r+5\\)<\/li>\r\n \t
  2. \\(35x^3-10x^2-56x+16\\)<\/li>\r\n \t
  3. \\(3n^3-2n^2-9n+6\\)<\/li>\r\n \t
  4. \\(14v^3+10v^2-7v-5\\)<\/li>\r\n \t
  5. \\(15b^3+21b^2-35b-49\\)<\/li>\r\n \t
  6. \\(6x^3-48x^2+5x-40\\)<\/li>\r\n \t
  7. \\(35x^3-28x^2-20x+16\\)<\/li>\r\n \t
  8. \\(7n^3+21n^2-5n-15\\)<\/li>\r\n \t
  9. \\(7xy-49x+5y-35\\)<\/li>\r\n \t
  10. \\(42r^3-49r^2+18r-21\\)<\/li>\r\n \t
  11. \\(16xy-56x+2y-7\\)<\/li>\r\n \t
  12. \\(3mn-8m+15n-40\\)<\/li>\r\n \t
  13. \\(2xy-8x^2+7y^3-28y^2x\\)<\/li>\r\n \t
  14. \\(5mn+2m-25n-10\\)<\/li>\r\n \t
  15. \\(40xy+35x-8y^2-7y\\)<\/li>\r\n \t
  16. \\(8xy+56x-y-7\\)<\/li>\r\n \t
  17. \\(10xy+30+25x+12y\\)<\/li>\r\n \t
  18. \\(24xy+25y^2-20x-30y^3\\)<\/li>\r\n \t
  19. \\(3uv+14u-6u^2-7v\\)<\/li>\r\n \t
  20. \\(56ab+14-49a-16b\\)<\/li>\r\n<\/ol>\r\nAnswer Key 7.2<\/a>","rendered":"

    First thing to do when factoring is to factor out the GCF. This GCF is often a monomial, like in the problem \"5xy + 10xz\" where the GCF is the monomial \"5x\", so you would have \"5x(y + 2z)\". However, a GCF does not have to be a monomial; it could be a binomial. Consider the following two examples.<\/p>\n

    \n
    \n

    Example 7.2.1<\/p>\n<\/header>\n

    \n

    Find and factor out the GCF for \"3ax - 7bx\".<\/p>\n

    By observation, one can see that both have \"x\" in common.<\/p>\n

    This means that \"3ax - 7bx = x(3a - 7b)\".<\/p>\n<\/div>\n<\/div>\n

    \n
    \n

    Example 7.2.2<\/p>\n<\/header>\n

    \n

    Find and factor out the GCF for \"3a(2a + 5b) - 7b(2a + 5b)\".<\/p>\n

    Both have \"(2a + 5b)\" as a common factor.<\/p>\n

    This means that if you factor out \"(2a + 5b)\", you are left with \"3a - 7b\".<\/p>\n

    The factored polynomial is written as \"(2a + 5b)(3a - 7b)\".<\/p>\n<\/div>\n<\/div>\n

    In the same way as factoring out a GCF from a binomial, there is a process known as grouping to factor out common binomials from a polynomial containing four terms.<\/span><\/p>\n

    \n
    \n

    Find and factor out the GCF for \"10ab + 15b^2 + 4a + 6b\".<\/p>\n

    To do this, first split the polynomial into two binomials.<\/p>\n

    \"10ab + 15b^2 + 4a + 6b\" becomes \"10ab + 15b^2\" and \"4a + 6b\".<\/p>\n

    Now find the common factor from each binomial.<\/p>\n

    \"10ab + 15b^2\" has a common factor of \"5b\" and becomes \"5b(2a + 3b)\".<\/p>\n

    \"4a + 6b\" has a common factor of 2 and becomes \"2(2a + 3b)\".<\/p>\n

    This means that \"10ab + 15b^2 + 4a + 6b = 5b(2a + 3b) + 2(2a + 3b)\".<\/p>\n

    \"5b(2a + 3b) + 2(2a + 3b)\" can be factored as \"(2a + 3b)(5b + 2)\".<\/p>\n<\/div>\n<\/div>\n

    Questions<\/h1>\n

    Factor the following polynomials.<\/p>\n

      \n
    1. \"40r^3-8r^2-25r+5\"<\/li>\n
    2. \"35x^3-10x^2-56x+16\"<\/li>\n
    3. \"3n^3-2n^2-9n+6\"<\/li>\n
    4. \"14v^3+10v^2-7v-5\"<\/li>\n
    5. \"15b^3+21b^2-35b-49\"<\/li>\n
    6. \"6x^3-48x^2+5x-40\"<\/li>\n
    7. \"35x^3-28x^2-20x+16\"<\/li>\n
    8. \"7n^3+21n^2-5n-15\"<\/li>\n
    9. \"7xy-49x+5y-35\"<\/li>\n
    10. \"42r^3-49r^2+18r-21\"<\/li>\n
    11. \"16xy-56x+2y-7\"<\/li>\n
    12. \"3mn-8m+15n-40\"<\/li>\n
    13. \"2xy-8x^2+7y^3-28y^2x\"<\/li>\n
    14. \"5mn+2m-25n-10\"<\/li>\n
    15. \"40xy+35x-8y^2-7y\"<\/li>\n
    16. \"8xy+56x-y-7\"<\/li>\n
    17. \"10xy+30+25x+12y\"<\/li>\n
    18. \"24xy+25y^2-20x-30y^3\"<\/li>\n
    19. \"3uv+14u-6u^2-7v\"<\/li>\n
    20. \"56ab+14-49a-16b\"<\/li>\n<\/ol>\n

      Answer Key 7.2<\/a><\/p>\n","protected":false},"author":540,"menu_order":16,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-596","chapter","type-chapter","status-publish","hentry"],"part":379,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/596","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\/596\/revisions"}],"predecessor-version":[{"id":3617,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapters\/596\/revisions\/3617"}],"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\/596\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/media?parent=596"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/pressbooks\/v2\/chapter-type?post=596"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/contributor?post=596"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/intermediatealgebrakpu\/wp-json\/wp\/v2\/license?post=596"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}