# Factor x^2+2xy+y^2+3x+3y!!! HELP

• 11-24-2007, 11:29 AM
Gam3freQ
Factor x^2+2xy+y^2+3x+3y!!! HELP
x^2+2xy+y^2+3x+3y
factor it completely

Explain steps.

Spoiler for Factored x^2+2xy+y^2+3x+3y:
(x+y)(x+y+3)

The reason why I'm asking is because I need to show my work and do not know how to factor it completely to get the answer above.
• 11-24-2007, 11:34 AM
J_C_103
So you cheated? I'm guessing thats for an algebra 2 course.
• 11-24-2007, 11:37 AM
Archaemic
I'm guessing it's one of those odd problems that you can look up in the back of the book, so he looked up the answer, but didn't know how to get to it, and thus he's asking for help. The immediate thing to do is to reread the chapter.
• 11-24-2007, 11:42 AM
xpack
i recommend working backwards from the answer you got, IF all you want is to get the work, but if you want to learn how to do it the go with trying it out, I tried it backwards and got most of the work but then got bored and so i stopped
• 11-24-2007, 11:43 AM
J_C_103
wouldn't you have to combine like terms? And wheres Freeplay when you need him?
-= Double Post =-
My bad. There aren't any like terms. Plus I think that would just be simplifying.
• 11-24-2007, 11:53 AM
-chw42-
I'm trying it now...I should know this, if not I need to go and stab myself.

EDIT: I got (X+Y)^2 + 3 (X+Y)...
• 11-24-2007, 11:55 AM
Gam3freQ
Quote:

Originally Posted by Archaemic
I'm guessing it's one of those odd problems that you can look up in the back of the book, so he looked up the answer, but didn't know how to get to it, and thus he's asking for help. The immediate thing to do is to reread the chapter.

No, this is actually a question I got wrong on the test, and the teacher gave me the answer, and didn't have enough time to explain how to get it, so I wrote it down. I know how to do questions of this type, but not how to do this question, as I got x(x+2y+3)+y(y+3) (WRONG). and -CHW42-, I already have the answer, I just want to know how to get it. The question is the title of the thread, and the answer is in the <spoiler> tags.
• 11-24-2007, 11:57 AM
-chw42-
Quote:

Originally Posted by Gam3freQ
No, this is actually a question I got wrong on the test, and the teacher gave me the answer, and didn't have enough time to explain how to get it, so I wrote it down. I know how to do questions of this type, but not how to do this question, as I got x(x+2y+3)+y(y+3) (WRONG).

My answer is wrong too, but what I did was separate the Y^2 with the 3X. So you have two factor-able equations. X^2 + 2XY + Y^2 goes on to be (X+Y)^2 while the 3X + 3Y is 3(X+Y).

EDIT: I got it right now, after you get (X+Y)^2 + 3(X+Y), you have common factors with (X+Y) so you take that out. You have a square with X+Y though so you leave one in. Leaving you with one factor of (X+Y) and another of (X+Y+3). There is your answer...

Now I don't need to stab myself.
• 11-24-2007, 12:05 PM
Gam3freQ
Quote:

Originally Posted by -CHW42-
My answer is wrong too, but what I did was separate the Y^2 with the 3X. So you have two factor-able equations. X^2 + 2XY + Y^2 goes on to be (X+Y)^2 while the 3X + 3Y is 3(X+Y).

EDIT: I got it right now, after you get (X+Y)^2 + 3(X+Y), you have common factors with (X+Y) so you take that out. You have a square with X+Y though so you leave one in. Leaving you with one factor of (X+Y) and another of (X+Y+3). There is your answer...

Now I don't need to stab myself.

I'm glad you don't have to stab yourself and go to the hospital and have your health insurance rates go up while you have to go to a psychiatrist and lose your job, go bankrupt, and become a loser at life and then really kill yourself.

I understand how your previous answer makes the real answer, but how do you get (x+y)^2 + 3(x+y)?
Quote:

Originally Posted by jaymes
So you cheated? I'm guessing thats for an algebra 2 course.

No, I didn't cheat, and yes it is for Algebra II Honors in high school 10th grade :)
• 11-24-2007, 12:08 PM
-chw42-
Quote:

Originally Posted by Gam3freQ
I'm glad you don't have to stab yourself and go to the hospital and have your health insurance rates go up while you have to go to a psychiatrist and lose your job, go bankrupt, and become a loser at life and then really kill yourself.

so how do you get (x+y)^2 + 3(x+y)?

Oh god...

Okay here I go, I'll show you in steps without words.

x^2 + 2xy + y^2 + 3x + 3y
x^2 + 2xy + y^2 = (x+y)^2
3x + 3y = 3(x+y)
(x+y)^2 + 3(x+y) (combined)
(x+y) (x+y) + 3 (take (x+y) out because it's a common factor of both)
(x+y) (x+y+3)

There you have it, pretty simple.
• 11-24-2007, 12:09 PM
J_C_103
Good work!!!
• 11-24-2007, 12:14 PM
Gam3freQ
Quote:

Originally Posted by -CHW42-
Oh god...

Okay here I go, I'll show you in steps without words.

x^2 + 2xy + y^2 + 3x + 3y
x^2 + 2xy + y^2 = (x+y)^2
3x + 3y = 3(x+y)
(x+y)^2 + 3(x+y) (combined)
(x+y) (x+y) + 3 (take (x+y) out because it's a common factor of both)
(x+y) (x+y+3)

There you have it, pretty simple.

thanks you are a life saver --> wasn't thinking solve by grouping when i took the test

Quote:

Originally Posted by jaymes
Good work!!!

Quote:

Originally Posted by jaymes
wouldn't you have to combine like terms? And wheres Freeplay when you need him?
My bad. There aren't any like terms. Plus I think that would just be simplifying.

Spoiler for what i was about to post:
You are SO lost lol. It is already simplified, it just needs to factored (opposite of simplifying).
lol I typed this and was about to post it but then he gave the steps on how to do it so i decided not to, but then i decided to so i posted it now lol. FreePlay would probably be like "this is the answer, *****. figure it out urself, ur in school. dumbass."
• 11-24-2007, 12:23 PM
J_C_103
I know, I pointed that out later in the post if you didn't notice.
• 11-24-2007, 12:26 PM
Gam3freQ
Quote:

Originally Posted by jaymes
I know, I pointed that out later in the post if you didn't notice.

i know thats why i lol'd i mean no offense i didn't intend any. i noticed
• 11-24-2007, 12:28 PM
J_C_103
ok. none taken, but its been about two weeks since ive done this in school. =]
• 11-24-2007, 02:29 PM
Archaemic
Quote:

Originally Posted by -CHW42-
Oh god...

Okay here I go, I'll show you in steps without words.

x^2 + 2xy + y^2 + 3x + 3y
x^2 + 2xy + y^2 = (x+y)^2
3x + 3y = 3(x+y)
(x+y)^2 + 3(x+y) (combined)
(x+y) (x+y) + 3 (take (x+y) out because it's a common factor of both)
(x+y) (x+y+3)

There you have it, pretty simple.

Surely I'm not the only one who sees a problem here.
• 11-24-2007, 02:38 PM
CtrlAltDeleteDie
It looks right to me.
• 11-24-2007, 02:44 PM
PopeOfDope
There is a set of brackets missing on his 2nd last line.
Instead of (x+y) (x+y) + 3
It should say (x+y) [(x+y) + 3]
• 11-24-2007, 04:58 PM
-chw42-
Quote:

Originally Posted by PopeOfDope
There is a set of brackets missing on his 2nd last line.
Instead of (x+y) (x+y) + 3
It should say (x+y) [(x+y) + 3]

OMG the world is going to end, come on guys, stop being like my parents...
• 11-24-2007, 04:59 PM
Chathurga
Quote:

Originally Posted by -CHW42-
OMG the world is going to end, come on guys, stop being like my parents...

:neutral:
• 11-24-2007, 05:03 PM
-chw42-
Quote:

Originally Posted by Chathurga
:neutral:

When I miss the simplest things they go Osama on me and never stop. "OMG no equal sign, that means the equation is not equal! But you got the right answer, however it is still wrong! OMG you forgot the parenthesis, but you did the problem right, OMG you got it wrong!"
• 11-24-2007, 05:06 PM
NYYFAN
Quote:

Originally Posted by -CHW42-
When I miss the simplest things they go Osama on me and never stop. "OMG no equal sign, that means the equation is not equal! But you got the right answer, however it is still wrong! OMG you forgot the parenthesis, but you did the problem right, OMG you got it wrong!"

Which is why I hate math
• 11-24-2007, 05:18 PM
PopeOfDope
Quote:

Originally Posted by -CHW42-
When I miss the simplest things they go Osama on me and never stop. "OMG no equal sign, that means the equation is not equal! But you got the right answer, however it is still wrong! OMG you forgot the parenthesis, but you did the problem right, OMG you got it wrong!"

The parenthesis make the whole thing completely different.
(x+y) (x+y) + 3 expands to x^2 + 2xy + y^2 + 3
(x+y) [(x+y) + 3] expands to x^2 + 2xy + y^2 + 3x + 3y
If you miss the parenthesis, you would be marked wrong in an exam.
• 11-24-2007, 05:49 PM
FreePlay
Quote:

Originally Posted by jaymes
And wheres Freeplay when you need him?

Jeez, can't a brother get a vacation?

(also, re: your user title... the abyss calls to the abyss? emptiness calls to emptiness? ehh??)
• 11-24-2007, 05:52 PM
J_C_103
Your being a bit too literal. In most instances it tranlates to hell.
-= Double Post =-