Figure 1: Check the form the polynomial takes before finding factors or a numerical solution.<\/p>\n
If the polynomial expression has common factors other than 1 or -1, the common factors can be factored out, before looking at the rest of the equation. This is a very important step, because the polynomial that is left may be an ax2<\/sup> +bx +c expression. Suppose the expression is 9x3<\/sup> + 27x2<\/sup> + 24x. The common factor to each element of the expression is 3x. Dividing 9x3<\/sup> by 3xis 3x2<\/sup>; 27x2<\/sup> by 3x is 9x, and 24x by 3x is 8. Therefore one factor is 3x, and the other factor is in the form ax2<\/sup> +bx +c.<\/p>\n Figure 2: An example of reviewing the polynomial for common factors.<\/p>\n The next step is to multiply the leading coefficient a by the constant c. In this example 3\u22198 is 24. In the polynomial 14x2<\/sup> + 67x + 81, 14\u221981 is 1134. In the polynomial x2<\/sup> -3x + 7, 1\u22197 is 7. In symbol terms, find integers p and q such that pq =ac. In symbol terms, integers p and q are such that pq =ac AND p + q =the coefficient b. Then the expression can be factored as (x +p) (x +q). This is also called splitting the middle term, and it is a method of trial and error. Suppose the expression is 3x2<\/sup> -10x -8. The expression is in the form ax2<\/sup> +bx +c. That a coefficient is 3, b is -10, and c is 8, so there are no common factors. In this example ac is equal to -24, so the task is to find two integers p and q, so that pq = -24 and p + q = -10. The factors that are equal to -24 are (-1, 24) with a sum of 23; (1, -24) with a sum of -23, (-2, 12) with a sum of 10, and (2,-12) with a sum of -10; (-3, 8) with a sum of 5; and (3, -8) with a sum of -5; (-4, 6) with a sum of 2, and (4, -6) with a sum of -2. The integers that follow both conditions are 2 and -12.<\/p>\n Figure 3: The process of defining AC and splitting the middle term.<\/p>\n Next, write the expression out and group the factors. The polynomial 3x2<\/sup> -10x – 8 = 3x2<\/sup> – 12x + 2x \u2013 8 = 3x(x – 4), because 3x \u2219 x is 3x2<\/sup> and 3x \u2219 (-4) is -12x. Also 2x – 8 can be factored as 2(x – 4), because the expression 2x – 8 is equal to 2(x – 4). Putting the expression together 3x(x – 4) +2(x – 4) equals (3x + 2) (x – 4), which can be checked by multiplying using FOIL, so that 3x \u2219 x = 3x2<\/sup>, 2x – 12x = -10x, and 2 \u2219 4 is 8.<\/p>\n Figure 4: An example of factoring by grouping.<\/p>\n Interested in algebra tutoring services<\/a>? Learn more about how we are assisting thousands of students each academic year.<\/p>\n SchoolTutoring Academy<\/span> is the premier educational services company for K-12 and college students. We offer tutoring programs for students in K-12, AP classes, and college. To learn more about how we help parents and students in Shiloh, PA: visit Tutoring in Shiloh, PA<\/a><\/p>\n","protected":false},"excerpt":{"rendered":" Overview The AC-method, also known as factoring by grouping, is often used to factor polynomials when the a coefficient is equal to 1. It can also be used when the a coefficient is not equal to 1, similar to the FOIL method. Questioning the Polynomial Expression The first step to defining the polynomial expression, before […]<\/p>\n","protected":false},"author":22,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"inline_featured_image":false,"footnotes":""},"categories":[2],"tags":[2622,2774,2819,2820],"class_list":["post-7435","post","type-post","status-publish","format-standard","hentry","category-algebra","tag-factoring-polynomials","tag-factoring-polynomials-by-grouping","tag-splitting-the-middle-term-of-a-polynomial","tag-using-common-factors"],"acf":[],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/posts\/7435","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/users\/22"}],"replies":[{"embeddable":true,"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/comments?post=7435"}],"version-history":[{"count":0,"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/posts\/7435\/revisions"}],"wp:attachment":[{"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/media?parent=7435"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/categories?post=7435"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/schooltutoring.com\/help\/wp-json\/wp\/v2\/tags?post=7435"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}![](\"https:\/\/schooltutoring.com\/help\/wp-content\/uploads\/sites\/2\/2014\/09\/Questioning-the-polynomial-expression.png\")
<\/a><\/p>\nReview for Common Factors<\/h3>\n
<\/a><\/p>\nDefining AC and Splitting the Middle Term<\/h3>\n
<\/a><\/p>\nFactoring by Grouping<\/h3>\n
<\/a><\/p>\n