{"id":34951,"date":"2023-01-16T08:20:26","date_gmt":"2023-01-16T02:50:26","guid":{"rendered":"https:\/\/ccssanswers.com\/?p=34951"},"modified":"2023-01-17T09:47:15","modified_gmt":"2023-01-17T04:17:15","slug":"factoring-terms-by-grouping","status":"publish","type":"post","link":"https:\/\/ccssanswers.com\/factoring-terms-by-grouping\/","title":{"rendered":"Factoring Terms by Grouping | How to Factor Terms by Grouping?"},"content":{"rendered":"

Finding factors for an algebraic expression is simple when it consists of only a few terms. But when it comes to an expression that has more than two or three terms, students feel difficult to solve those problems. Students need a better process to solve algebraic expression factorization. Every student who is looking for the best method to solve algebraic expression factorization<\/strong><\/a> can follow the grouping method.<\/p>\n

Factoring Terms by Grouping is the easy and best method to solve different expressions easily. Also, the process of Factoring by Grouping The Terms is very simple compared to other methods.<\/p>\n

Procedure for Factoring Algebraic Expressions by Grouping<\/h2>\n

Follow the below steps to find the factorization of a given expression using the below steps.<\/p>\n

(i) Take out a factor from each group from the groups of the given expression.
\n(ii) Factorize each group
\n(iii) Lastly, take out the common factor.<\/p>\n

Factoring Terms by Grouping Examples<\/h3>\n

1. Factoring of algebraic expression<\/p>\n

(i) 2ma + mb + 2na + nb<\/p>\n

Solution:
\nGiven expression is 2ma + mb + 2na + nb.
\nGroup the first two terms and last two terms.
\nThe first two terms are 2ma + mb and the second two terms are 2na + nb.
\nTake m common from the first two terms.
\nm (2a + b)
\nTake n common from the second two terms.
\nn (2a + b)
\nm (2a + b) + n (2a + b)
\nThen, take (2a + b) common from the above expression.
\n(2a + b) (m + n)<\/p>\n

The final answer is (2a + b) (m + n).<\/p>\n

(ii) 3xm – ym – 3xn + yn<\/p>\n

Solution:
\nGiven expression is 3xm – ym – 3xn + yn.
\nGroup the first two terms and last two terms.
\nThe first two terms are 3xm – ym and the second two terms are – 3xn + yn.
\nTake m common from the first two terms.
\nm (3x – y)
\nTake -n common from the second two terms.
\n-n (3x – y)
\nm (3x – y) – n (3x – y)
\nThen, take (3x – y) common from the above expression.
\n(3x – y) (m – n)<\/p>\n

The final answer is (3x – y) (m – n).<\/p>\n

(iii) 12a2<\/sup> + 6ab – 4ma \u2013 2mb<\/p>\n

Solution:
\nGiven expression is 12a2<\/sup> + 6ab – 4ma \u2013 2mb.
\nGroup the first two terms and last two terms.
\nThe first two terms are 12a2<\/sup> + 6ab and the second two terms are – 4ma \u2013 2mb.
\nTake 6a common from the first two terms.
\n6a (2a + b)
\nTake -2m common from the second two terms.
\n-2m (2a + b)
\n6a (2a + b) – 2m (2a + b)
\nThen, take (2a + b) common from the above expression.
\n(2a + b) (6a – 2m)<\/p>\n

The final answer is (2a + b) (6a – 2m).<\/p>\n

(iv) am2<\/sup> – bm2<\/sup> + an2<\/sup> – bn2<\/sup> + ar2<\/sup> – br2<\/sup><\/p>\n

Solution:
\nGiven expression is am2<\/sup> – bm2<\/sup> + an2<\/sup> – bn2<\/sup> + ar2<\/sup> – br2<\/sup>.
\nGroup the first two terms, middle two terms, and last two terms.
\nThe first two terms are am2<\/sup> – bm2<\/sup>, the middle two terms are + an2<\/sup> – bn2<\/sup>, and the last two terms are + ar2<\/sup> – br2<\/sup>.
\nTake m2<\/sup> common from the first two terms.
\nm2<\/sup> (a – b)
\nTake n2<\/sup> common from the middle two terms.
\nn2<\/sup> (a – b)
\nTake r2<\/sup> common from the middle two terms.
\nr2<\/sup> (a – b)
\nm2<\/sup> (a – b) + n2<\/sup> (a – b) + r2<\/sup> (a – b)
\nThen, take (a – b) common from the above expression.
\n(a – b) (m2<\/sup> + n2<\/sup> + r2<\/sup>)<\/p>\n

The final answer is (a – b) (m2<\/sup> + n2<\/sup> + r2<\/sup>).<\/p>\n

(v) ax – ay + bx \u2013 by<\/p>\n

Solution:
\nGiven expression is ax – ay + bx \u2013 by.
\nGroup the first two terms and last two terms.
\nThe first two terms are ax – ay and the second two terms are + bx \u2013 by.
\nTake a common from the first two terms.
\na (x – y)
\nTake b common from the second two terms.
\nb (x – y)
\na (x – y) + b (x – y)
\nThen, take (x – y) common from the above expression.
\n(x – y) (a + b)<\/p>\n

The final answer is (x – y) (a + b).<\/p>\n

2. Factoring the following algebraic expression<\/p>\n

(i) 4a + 2ab + b + 2<\/p>\n

Solution:
\nGiven expression is 4a + 2ab + b + 2.
\nGroup the first two terms and last two terms.
\nThe first two terms are 4a + 2ab and the second two terms are b + 2.
\nTake 2a common from the first two terms.
\n2a (2 + b)
\nTake 1 common from the second two terms.
\n1 (b + 2)
\n2a (2 + b) + 1 (2 + b)
\nThen, take (2 + b) common from the above expression.
\n(2 + b) (2a + 1)<\/p>\n

The final answer is (2 + b) (2a + 1).<\/p>\n

(ii) 3m3<\/sup> + 5m2<\/sup> + 3m + 5<\/p>\n

Solution:
\nGiven expression is 3m3<\/sup> + 5m2<\/sup> + 3m + 5.
\nGroup the first two terms and last two terms.
\nThe first two terms are 3m3<\/sup> + 5m2<\/sup> and the second two terms are 3m + 5.
\nTake m2 <\/sup>common from the first two terms.
\nm2<\/sup> (3m + 5)
\nTake 1 common from the second two terms.
\n1 (3m + 5)
\nm2<\/sup> (3m + 5) + 1 (3m + 5)
\nThen, take (3m + 5) common from the above expression.
\n(3m + 5) (m2<\/sup> + 1)<\/p>\n

The final answer is (3m + 5) (m2<\/sup> + 1).<\/p>\n

(iii) b3<\/sup> + 3b2<\/sup> + b + 3<\/p>\n

Solution:
\nGiven expression is b3<\/sup> + 3b2<\/sup> + b + 3.
\nGroup the first two terms and last two terms.
\nThe first two terms are b3<\/sup> + 3b2<\/sup> and the second two terms are b + 3.
\nTake b2<\/sup>\u00a0<\/sup>common from the first two terms.
\nb2<\/sup> (b + 3)
\nTake 1 common from the second two terms.
\n1 (b + 3)
\nb2<\/sup> (b + 3) + 1 (b + 3)
\nThen, take (b + 3) common from the above expression.
\n(b + 3) (b2<\/sup> + 1)<\/p>\n

The final answer is (b + 3) (b2<\/sup> + 1).<\/p>\n

(iv) 1 + s + s2<\/sup>t + s3<\/sup>t<\/p>\n

Solution:
\nGiven expression is 1 + s + s2<\/sup>t + s3<\/sup>t.
\nGroup the first two terms and last two terms.
\nThe first two terms are 1 + s and the second two terms are s2<\/sup>t + s3<\/sup>t.
\nTake 1\u00a0<\/sup>common from the first two terms.
\n1 (1 + s)
\nTake s2<\/sup>t common from the second two terms.
\ns2<\/sup>t (1 + s)
\n1 (1 + s) + s2<\/sup>t (1 + s)
\nThen, take (1 + s) common from the above expression.
\n(1 + s) (1 + s2<\/sup>t)<\/p>\n

The final answer is (1 + s) (1 + s2<\/sup>t).<\/p>\n

(v) m – 1 – (m – 1)2<\/sup> + bm \u2013 b<\/p>\n

Solution:
\nGiven expression is m – 1 – (m – 1)2<\/sup> + bm \u2013 b.
\nGroup the first two terms, middle and last two terms.
\nThe first two terms are m – 1, the middle term is – (m – 1)2<\/sup>, and the last two terms is bm \u2013 b.
\nTake 1\u00a0<\/sup>common from the first two terms.
\n1 (m – 1)
\nTake 1 common from the middle term.
\n– (m – 1)2
\n<\/sup>Take b common from the last two terms.
\nb (m \u2013 1)
\n1 (m – 1) – (m – 1)2<\/sup>+ b (m \u2013 1)
\nThen, take (m \u2013 1) common from the above expression.
\n(m \u2013 1) (1 – m + 1 + b)
\n(m \u2013 1) (2 + b – m)<\/p>\n

The final answer is (m \u2013 1) (2 + b – m).<\/p>\n","protected":false},"excerpt":{"rendered":"

Finding factors for an algebraic expression is simple when it consists of only a few terms. But when it comes to an expression that has more than two or three terms, students feel difficult to solve those problems. Students need a better process to solve algebraic expression factorization. Every student who is looking for the … Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/34951"}],"collection":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/comments?post=34951"}],"version-history":[{"count":1,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/34951\/revisions"}],"predecessor-version":[{"id":138499,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/34951\/revisions\/138499"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=34951"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=34951"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=34951"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}