{"id":1373,"date":"2021-12-02T19:37:42","date_gmt":"2021-12-03T00:37:42","guid":{"rendered":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/chapter\/7-2-factoring-by-grouping\/"},"modified":"2023-08-30T18:18:50","modified_gmt":"2023-08-30T22:18:50","slug":"factoring-by-grouping","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/chapter\/factoring-by-grouping\/","title":{"raw":"7.2 Factoring by Grouping","rendered":"7.2 Factoring by Grouping"},"content":{"raw":"First thing to do when factoring is to factor out the GCF. This GCF is often a monomial, like in the problem [latex]5xy + 10xz[\/latex] where the GCF is the monomial [latex]5x[\/latex], so you would have [latex]5x(y + 2z)[\/latex]. 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 [latex]3ax - 7bx[\/latex].\r\n\r\nBy observation, one can see that both have [latex]x[\/latex] in common.\r\n\r\nThis means that [latex]3ax - 7bx = x(3a - 7b)[\/latex].\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 [latex]3a(2a + 5b) - 7b(2a + 5b)[\/latex].\r\n\r\nBoth have [latex](2a + 5b)[\/latex] as a common factor.\r\n\r\nThis means that if you factor out [latex](2a + 5b)[\/latex], you are left with [latex]3a - 7b[\/latex].\r\n\r\nThe factored polynomial is written as [latex](2a + 5b)(3a - 7b)[\/latex].\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 [latex]10ab + 15b^2 + 4a + 6b[\/latex].\r\n\r\nTo do this, first split the polynomial into two binomials.\r\n

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

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

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

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

Questions<\/h1>\r\nFactor the following polynomials.\r\n
    \r\n \t
  1. [latex]40r^3-8r^2-25r+5[\/latex]<\/li>\r\n \t
  2. [latex]35x^3-10x^2-56x+16[\/latex]<\/li>\r\n \t
  3. [latex]3n^3-2n^2-9n+6[\/latex]<\/li>\r\n \t
  4. [latex]14v^3+10v^2-7v-5[\/latex]<\/li>\r\n \t
  5. [latex]15b^3+21b^2-35b-49[\/latex]<\/li>\r\n \t
  6. [latex]6x^3-48x^2+5x-40[\/latex]<\/li>\r\n \t
  7. [latex]35x^3-28x^2-20x+16[\/latex]<\/li>\r\n \t
  8. [latex]7n^3+21n^2-5n-15[\/latex]<\/li>\r\n \t
  9. [latex]7xy-49x+5y-35[\/latex]<\/li>\r\n \t
  10. [latex]42r^3-49r^2+18r-21[\/latex]<\/li>\r\n \t
  11. [latex]16xy-56x+2y-7[\/latex]<\/li>\r\n \t
  12. [latex]3mn-8m+15n-40[\/latex]<\/li>\r\n \t
  13. [latex]2xy-8x^2+7y^3-28y^2x[\/latex]<\/li>\r\n \t
  14. [latex]5mn+2m-25n-10[\/latex]<\/li>\r\n \t
  15. [latex]40xy+35x-8y^2-7y[\/latex]<\/li>\r\n \t
  16. [latex]8xy+56x-y-7[\/latex]<\/li>\r\n \t
  17. [latex]10xy+30+25x+12y[\/latex]<\/li>\r\n \t
  18. [latex]24xy+25y^2-20x-30y^3[\/latex]<\/li>\r\n \t
  19. [latex]3uv+14u-6u^2-7v[\/latex]<\/li>\r\n \t
  20. [latex]56ab+14-49a-16b[\/latex]<\/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 [latex]5xy + 10xz[\/latex] where the GCF is the monomial [latex]5x[\/latex], so you would have [latex]5x(y + 2z)[\/latex]. 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 [latex]3ax - 7bx[\/latex].<\/p>\n

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

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

    \n
    \n

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

    \n

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

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

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

    The factored polynomial is written as [latex](2a + 5b)(3a - 7b)[\/latex].<\/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 [latex]10ab + 15b^2 + 4a + 6b[\/latex].<\/p>\n

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

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

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

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

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

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

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

    Questions<\/h1>\n

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

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

      Answer Key 7.2<\/a><\/p>\n","protected":false},"author":90,"menu_order":2,"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-1373","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\/1373","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\/1373\/revisions"}],"predecessor-version":[{"id":2123,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1373\/revisions\/2123"}],"part":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/parts\/1369"}],"metadata":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1373\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/media?parent=1373"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapter-type?post=1373"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/contributor?post=1373"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/license?post=1373"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}