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August 21st, 2010, 04:09 PM  #1 
Member Joined: Jul 2010 Posts: 69 Thanks: 0  Factor by Grouping Part 2
Ok I have another problem, I figured out the last one. Here we have: (3ac2bd)+(6bcad) I figured that the factor they all have in common is abcd, so I divided this and all I get is (abcd)(3261). But the answer in the book is not even near that, the answer in the book is (a+2b)(3cd). What is the problem here? Can someone break the problem down and rebuild it? I would really appreciate it. I am trying to fit into a math placement test by Monday. 
August 21st, 2010, 04:31 PM  #2 
Member Joined: Jul 2010 Posts: 69 Thanks: 0  Re: Factor by Grouping Part 2
Ok so, intuitively I am playing with the problem and also rereading instructions from this selfhelp book. It looks like I place the terms that have some common ground together. So far I rearranged the problem so it appears as (2bdad) + (3ac +6b), can someone tell me what I am doing wrong here?

August 21st, 2010, 04:43 PM  #3 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs  Re: Factor by Grouping Part 2
Ok, you have 3ac  2bd + 6bc  ad There are two ways to approach this. (a) Two of the terms have a common a and two of the terms have a common 2b, so they could be grouped as (3ac  ad) + (6bc  2bd) In the first group, factor out the common a, giving you a(3c  d), and in the second group factor out the common 2b giving you 2b(3c  d), which allows you to write the original expression as a(3c  d) + 2b(3c  d) Now what do these two terms have in common? (b) Two of the terms have a common 3c and two of the terms have a common d, so they could be grouped as (3ac + 6bc) + (ad  2bd) In the first group, factor out the common 3c, giving you 3c(a + 2b), and in the second group factor out the common d giving you d(a + 2b), which allows you to write the original expression as 3c(a + 2b)  d(a + 2b) Now what do these two terms have in common? 
August 21st, 2010, 11:07 PM  #4  
Member Joined: Jul 2010 Posts: 69 Thanks: 0  Re: Factor by Grouping Part 2 Quote:
 
August 22nd, 2010, 12:02 AM  #5 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs  Re: Factor by Grouping Part 2
I enjoy helping, like most here. I respect anyone who will ask questions, as that shows motivation. Now, the two terms 3c(a + 2b) and  d(a + 2b) have the factor (a + 2b) in common as it appears in both. So you take that and put it in front, then write what's left in the two terms, in this case, the 3c and d as the other factor, thus 3c(a + 2b) and  d(a + 2b) = (a + 2b)(3c  d) When you are factoring, look for what factors the terms involved have in common. For instance, let's dispose of variables for now and look at turning addition of numbers into multiplication. Suppose you have 12 + 6 = ? Now disregard that you immediately know this is 18 and factor the two terms 12 and 6 and rewrite it as 2·2·3 + 2·3 You are looking for every factor that both terms have within them. Notice that both terms have 2·3 contained within their prime factorization. Pull this 2·3 out front and write what's left after you remove them within parentheses as follows: 2·3(2 + 1) = 6·3 = 18 Notice that if one of the terms is actually what both have in common, you write a 1 in its place since 1·x = x, this is the multiplicative identity, ie. any number times 1 is unchanged. As another example let's look at 30 + 45 = ? 30 = 2·3·5 and 45 = 3·3·5 both terms have a 3 and both have a 5 as a factor, so we can write 3·5(2 + 3) = 15·5 = 75 Let's now look at 72 + 108 = 2·2·2·3·3 + 2·2·3·3·3 What is it that is common to both terms? Look at both and ask what is the maximum number of 2's and 3's present in both (the greatest common divisor)? Both have at least 2 2's and 2 3's, so we can write 2·2·3·3(2 + 3) = 36·5 = 180 Now, the same holds true for terms containing variables, such as x³y² + x²y³ = x·x·x·y·y + x·x·y·y·y = x·x·y·y(x + y) = x²y²(x + y) Now, when factoring by grouping, you want to choose terms that have factors in common. Suppose you are given 2ac  ad + 4bc  2bd Look first at the first term 2ac. It's factors are 2, a, and c. Now look at the other terms to see if any of them have a 2, an a or a c as factors. The second term ad has an a, and the third term has a 2 and a c, and the fourth term has a 2. Since the third terms has the most in common with the first term, let's group the first and third in the first set of parentheses, and put the second and fourth in the second set as follows: (2ac + 4bc) + (ad  2bd) Now the first term (2ac + 4bc) = (2·a·c + 2·2·b·c) has 2·c in both, so we can write it as 2·c(a + 2·b) = 2c(a + 2b) Next, the second term (ad  2bd) = (1·a·d + 1·2·b·d) has 1·d in both , so we can write it as 1·d(a + 2·b) = d(a + 2b) Putting these two terms, now factored, back together to represent the original expression, we have (2ac + 4bc) + (ad  2bd) = 2c(a + 2b) + d(a + 2b) = 2c(a + 2b)  d(a + 2b) Recall, x +  y = x  y, while x  y = x + y. Now in the expression 2c(a + 2b)  d(a + 2b) notice that both have (a + 2b) in common. Thus, we take it out front, and write what's left in the second factor as follows (a + 2b)(2c  d) It takes practice to get the hang of it. Hang in there, you are motivated, and you will master it. 
August 22nd, 2010, 06:45 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 18,140 Thanks: 1415 
I suspect you meant "maximum" rather than "minimum", since you are seeking a greatest common divisor.

August 22nd, 2010, 08:40 AM  #7 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs  Re: Factor by Grouping Part 2
Indeed I did! I think I was thinking in terms of exponents where you use the smaller value. Thank you for taking the time to review my rather longwinded post!

August 22nd, 2010, 05:54 PM  #8 
Global Moderator Joined: Dec 2006 Posts: 18,140 Thanks: 1415 
Some confusion has arisen. Whether the word "minimum" or "maximum" is used depends on how the sentence is worded. Once you know the counts, their minimum is also the maximum common to every term.

August 22nd, 2010, 06:22 PM  #9 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs  Re: Factor by Grouping Part 2
Exactly what caused my error in wording. That's my story and I'm sticking to it! For a simple example x^3 + x^2 = x^2(x + 1) The minimum power is 2, which you factor out, but 2 is the maximum number of x's that are common to both as factors. 

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