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August 20th, 2010, 02:01 PM  #1 
Member Joined: Jul 2010 Posts: 69 Thanks: 0  Factor by Grouping
Hello, I am needing assistance. This doesn't appear to be too hard but some of these are tricky. I am using a self help book and it the problem I have here is: cxdx+cydy My answer was not what they go in the book. Can someone walk me through this step by step. This is the problem that I tried to break down, (cxdx) + (cydy) cd? (xx) + cd(yy) factor (xy) cd (xx) + cd (yy) cd (xx) + cd (yy)  = cd xy ????? The answer in the back of the book is: (x + y)(c  d) How am I supposed to know where to put the plus and minus signs? If someone can walk me through this step by step I would really appreciate it. Thanks. 
August 20th, 2010, 02:16 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 462 Math Focus: Calculus/ODEs  Re: Factor by Grouping
Ok, the problem given is: cx  dx + cy  dy First, group them either by c's and d's or x's and y's i.e (cx  dx) + (cy  dy) or (cx + cy)  (dx + dy) that way you can factor each group. 
August 20th, 2010, 02:30 PM  #3 
Member Joined: Jul 2010 Posts: 69 Thanks: 0  Re: Factor by Grouping
Thanks, but how am I supposed to get the answer (x + y)(c  d), why can't it be (xy)(cd), in other words, how am I supposed to know where to put the +/ signs? Thanks.

August 20th, 2010, 02:38 PM  #4  
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 462 Math Focus: Calculus/ODEs  Re: Factor by Grouping Quote:
cx  dx  cy + dy See how that differs from the problem given? Let's group them as follows: (cx  dx) + (cy  dy) If you notice, the first group has a common x and the second group has a common y. Factor them out. x(c  d) + y(c  d) Now, what is common to both terms? What can be factored out?  
August 20th, 2010, 02:52 PM  #5 
Member Joined: Jul 2010 Posts: 69 Thanks: 0  Re: Factor by Grouping
Thanks for your help, the answer should be: xcxd + ycyd? So how does the answer work when you factor by group?: This is the answer in the book: (x+y)(cd) If someone has time to walk me through this step by step that would be great, thanks. Mainly I need to know how they got the + between the x and y and the  between the c and d. 
August 20th, 2010, 03:01 PM  #6 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 462 Math Focus: Calculus/ODEs  Re: Factor by Grouping
What I'm getting at is when you have x(c  d) + y(c  d) there is a common (c  d) to both terms that can be factored out. For simplification, let u = (c  d). We can then write it as ux + uy how would you factor that? As for "how they got the + between the x and y and the  between the c and d" let's look at the original: cx  dx + cy  dy Notice that both cx and cy have a + in front of them, while dx and dy both have a  in front of them. When you have an expression of the type u  v this can be thought of as (1)u + (1)v Then, the common (1) can be factored out giving (1)(u + v) = (u + v) Thus cx  dx + cy  dy can be written cx + cy  dx  dy = (cx + cy)  (dx + dy) The first group has a common c and the second group has a common d, thus we can write c(x + y)  d(x + y) Now both terms have a common (x + y) allowing us to write (x + y)(c  d) Now, let's go back to where we had x(c  d) + y(c  d) Notice both terms have a common (c  d) allowing us to write (c  d)(x + y) 
August 20th, 2010, 03:27 PM  #7 
Member Joined: Jul 2010 Posts: 69 Thanks: 0  Re: Factor by Grouping
If you have the time and patience, can you break down the problem and build it back up? Thank you for your help!

August 20th, 2010, 03:38 PM  #8 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 462 Math Focus: Calculus/ODEs  Re: Factor by Grouping
Ok, let's begin with the problem as presented: cx  dx + cy  dy Notice that there are two ways to group these four terms into groups of two so that there will be a common factor. (cx  dx) + (cy  dy) OR (cx + cy) + (dx  dy) Let's use the one on the left, so that we will not have to factor the negative out of the second term in the one on the right. So, we have grouped them as (cx  dx) + (cy  dy) Let's start with the left group (cx  dx) What do both terms have as a common factor within the parentheses? How can this be factored? 
August 20th, 2010, 10:23 PM  #9  
Member Joined: Jul 2010 Posts: 69 Thanks: 0  Re: Factor by Grouping Quote:
 
August 20th, 2010, 11:25 PM  #10 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 462 Math Focus: Calculus/ODEs  Re: Factor by Grouping
Yes, you are right that the common factor is x, but your factoring technique is incorrect. You take that common x out in front of the parentheses as follows: (cx  dx) = x(c  d) You can check your result by multiplying it back to ensure you get what you started with. x(c  d) = xc  xd Now, by the commutative property of multiplication which says that ab = ba, we can rewrite this as cx  dx which is what we started with. Your result of x²(cd) would give you cdx² which is not what we began with. When you are presented with an expression of the form ab + ac First, observe what factor(s) the terms have in common, then take the common factors out and place them in front of the parentheses, the divide the terms by that common factor to get what remains inside, cancelling, or dividing out the common factor(s): abc + abd = ab(abc/ab + abd/ab) = ab(c + d) when you have expressions of the form ab + ac or ab  ac or ab + ac you pull the common factor out front, and keep the signs intact as follows a(b + c) or a(b  c) or a(b + c) I would recommend going back and reviewing simpler factoring problems to get the hang of it. Factoring is very important in algebra, particularly with quadratic equations, and it will impede your progress unless you become proficient at it. Do not hesitate to post your questions, there are many helpful and skillful folks here. 

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