anonymous
  • anonymous
Can you help me with this problem 3xy^2+6^2y
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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e.mccormick
  • e.mccormick
What are you supposed to do with it and where are you having trouble? `\(3xy^2+6^2y\)` makes: \(3xy^2+6^2y\)
anonymous
  • anonymous
I think the gcf is 3
e.mccormick
  • e.mccormick
Ah, if you are looking for a GCF, 3 is part of it... but what about your variables? Are there any that show up in both parts?

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anonymous
  • anonymous
The thing is that in high school they did not show me how to do this type of math and when II took the math test for college they put me in this type of math :'(
e.mccormick
  • e.mccormick
So in high school you never learned about terms and factoring something out?
jdoe0001
  • jdoe0001
hmm, sounds fishy
anonymous
  • anonymous
Yes but not this type of math like finding the gcf so I really need help
jdoe0001
  • jdoe0001
gcf is more like 7th grade :S
DebbieG
  • DebbieG
If you didn't learn how to do it in high school, then BE GLAD that they put in it in college so that you can learn how to do it! You'll need this before you can go on to higher levels of math.
e.mccormick
  • e.mccormick
All the GCF is, is what will factor out. If you factor everything out, that is the GCF. If there is something still in there that can factor out, you are not yet to GCF.
jdoe0001
  • jdoe0001
hmm.... well, my understanding is a bit different they put it in college because you're supposed to know it already, if you dunno then their material doesn't apply to you
anonymous
  • anonymous
Well sorry my school did not show me how to do this and I am not planning to fail this class
e.mccormick
  • e.mccormick
So, to get back to the question, you said you were able to factor out a 3. Fine. Is anything else able to factor out?
DebbieG
  • DebbieG
Well, @jdoe0001 , I teach math in college so I'm not sure what you mean. If this is a problem that she is doing in the class she is currently in, then that's the level the class is at, unless it is some kind of a review/prerequisites assignment. Either way, she is here now trying to learn it so why not encourage her and help her learn instead of put her down for the question she's asking??
anonymous
  • anonymous
Ok I think I got this 3×y(xy+2xy)
e.mccormick
  • e.mccormick
Hmmm.... where did the x in the second term come from?
anonymous
  • anonymous
So I did it wrong?
e.mccormick
  • e.mccormick
Yah. Do you have your work you can show? We can find out where this is going wrong.
jdoe0001
  • jdoe0001
I don't discourage folks, the only thing I can think of is... a school that didn't do so good either way, on the same token, I don't encourage folks to skip out either, and often times my judgment doesn't come to optimistic side that, this or that claim really happen, I do keep an open mind, but also I encourage "elbow grease" so from someone asking for trigonometry, I'm not about to cover factoring or anyting like so, no for trigonometry level for someone who explicitly asked for factoring, then I understand and so I cover it
anonymous
  • anonymous
The thing I did was 3×1=3 I get 3xy(xy Then 3×2=6 I get 3xy(xy+6x^2y)
e.mccormick
  • e.mccormick
You originally posted: \(3xy^2+6^2y\) That is nowhere near \(3xy(xy+6x^2y)\)...
e.mccormick
  • e.mccormick
So, lets see. Is \(3xy^2+6^2y\) correct? Was there any mistake in typing up that? Lets start there and make sure the question is right. =) You need the GCF of that.
anonymous
  • anonymous
So wait how do it I am trying my best :\
anonymous
  • anonymous
Yes that is the question 3xy^2+6x^2y
e.mccormick
  • e.mccormick
\(3xy^2+6x^2y\) AH! So NOT \(3xy^2+6^2y\) The x was missing! That makes a HUGE difference!
anonymous
  • anonymous
Omg I forgot to put the x Dx
e.mccormick
  • e.mccormick
=) All good. Now we are on the same problem.
e.mccormick
  • e.mccormick
OK. That means what you did makes more sense, but it is still not quite right. I have been woring on an example on the side, and will post it soon.
anonymous
  • anonymous
Ok than so I thinknthe gcf is 3
e.mccormick
  • e.mccormick
It is more than just 3.
anonymous
  • anonymous
Gcf:1
e.mccormick
  • e.mccormick
In a simple polynomial you have one or more terms. A term is a number, a variable, or a combination of numbers and variables, but multiplied. Many people think a polynomial must have more than one term because of the poly part, but they are forgetting something. 5 is a term and a polynomial... all by itself. May seem odd, but it is true. See, 5 can also mean: \(0x^7+5x^0\) Now, GCF, or Greatest Common Factor, is a hunt for the largest number of things that are the same in the different terms. Let me take an example problem and work it through: \(2x^2+10x\) First, I have two terms there. I want to look for anything I can factor out of both. When it comes to the numbers, this is easy. They are both even, so I know I can factor out at least a 2. \(2(x^2+5x)\) Everything in the original was divided by 2 because I factored a 2. So what I did was this: \(\dfrac{2x^2}{2}=x^2\) and \(\dfrac{10x}{2}=5x\) then put them back together into my \(2(x^2+5x)\). Now, I look at the variables in there. Are there nay that are shared? yes! The x. So again, I divide out. \(dfrac{x^2}{x}=x\) and \(dfrac{5x}{x}=5\) and I can put those parts back into the original and make: \(2x(x+5)\) That would make 2x the GCF.
anonymous
  • anonymous
Ok so it might be 3xy(xy+2xy)
anonymous
  • anonymous
Gcf is 3xy
e.mccormick
  • e.mccormick
In your problem, you got the number, 3, out just fine. What you also need to divide out is the varibales. And remember to cancel properly! If you take both x and y out, they need to come out! If there is only one y or one x in a term, it goes by-by. But if it is squared, then one comes out nad one stays behind.
e.mccormick
  • e.mccormick
The GCF of 3xy is correct, but \(3xy^2+6x^2y\ne 3xy(xy+2xy)\) You left some variables in that were taken out.
anonymous
  • anonymous
So the answer I gave it will be wrong because I need to take out y right
e.mccormick
  • e.mccormick
Well, one of the ys and one of the xs. =) Look again and you should be able to see which ones. OR, put the inside parts of the original into fractions, and do canceling!
e.mccormick
  • e.mccormick
\(3xy^2+6x^2y\) if you are going to factor 3xy out of that, it is the same thing as saying: \(3xy\left(\dfrac{3xy^2}{3xy}+\dfrac{6x^2y}{3xy}\right)\) Now, what would that be once you canceled things in those fractions? You know the 3 part fine. \(3xy\left(\dfrac{\cancel{3}xy^2}{\cancel{3}xy}+\dfrac{\cancel{6}2x^2y}{\cancel{3}xy}\right)\) \(3xy\left(\dfrac{xy^2}{xy}+\dfrac{2x^2y}{xy}\right)\) But what about canceling the x and y on each part?
anonymous
  • anonymous
Nop i dont get it
e.mccormick
  • e.mccormick
Well, what is \(\frac{x}{x}\)?
anonymous
  • anonymous
1
e.mccormick
  • e.mccormick
YES! And \(\frac{x^2}{x}\)?
anonymous
  • anonymous
-10
e.mccormick
  • e.mccormick
Hmmm... how did you get -10?
anonymous
  • anonymous
Wait hold up I am trying to do this in my head
e.mccormick
  • e.mccormick
=)
anonymous
  • anonymous
Wait a min is it 1^2
e.mccormick
  • e.mccormick
Close! Canced the bottom fine. But the top.... hmmm....
e.mccormick
  • e.mccormick
Let me show exactly how \(\frac{x}{x}=1\) works and you might see it. \[ \frac{x}{x}=\frac{x^1}{x^1} \implies \\ \, \\ \frac{x^1}{x^1}=x^1x^{-1} \implies \\ \, \\ x^1x^{-1}=x^{1-1} \implies \\ \, \\ x^{1-1}=x^0 \implies \\ \, \\ x^0=1 \, \\ \]
anonymous
  • anonymous
So is it 1 hagen
e.mccormick
  • e.mccormick
No. The 2 in the square would make it different.
anonymous
  • anonymous
I just did this 1^2=1
e.mccormick
  • e.mccormick
\[\frac{x^2}{x}=\frac{x^2}{x^1} \implies \\ \, \\ \frac{x^2}{x^1}=x^2x^{-1} \implies \\ \, \\ x^2x^{-1}=x^{2-1} \implies \\ \, \\ x^{2-1}=x^1 \implies \\ \, \\ x^1=x \, \\\]
e.mccormick
  • e.mccormick
Yes, \(1^2=1\) But you don't have \(1^2\). You have \(\dfrac{x^2}{x}\) Do you see my proof of how \(\dfrac{x^2}{x}=x\)?
anonymous
  • anonymous
Yeahh I kinda get it
e.mccormick
  • e.mccormick
OK. Do you get it enough to apply it this time?
anonymous
  • anonymous
Nop
e.mccormick
  • e.mccormick
Well, I already talked about how you accidentally left in an x and a y that were supposed to cancel. Go back to that point and take another look at it. You just need to remove the proper ones and your answer would be correct!
anonymous
  • anonymous
Ik this will be wrong 3xy(xy+2x)= 3xy+6^2y
anonymous
  • anonymous
I mean this is wrong. 3xy(xy+2x)= 3xy+6x^2y
e.mccormick
  • e.mccormick
Closer! You got rid of the correct \(\,y\)! See, the problem is shown if you try amd multiply it out: \(3xy(xy+2x)= 3x^2y+6x^2y\) is what you wrote. But: \(3xy(xy+2x)\implies \) \(3xy(xy)+3xy(2x)\implies \) \(3x^2y^2+6x^2y\implies \) \(3x^2y^2+6x^2y=3xy^2+6x^2y\) and you can see that is not true! Look at the x in the first term.
e.mccormick
  • e.mccormick
Bah, and I put a square in one of the wrong places.
anonymous
  • anonymous
So your saying that if I do this 3xy(xy+(2x))= 3xy+6x^2y
anonymous
  • anonymous
So your saying that if I do this 3xy(xy+(2x))= 3xy+6x^2y
e.mccormick
  • e.mccormick
Well, \(3xy(xy+(2x))= 3x^2y^2+6x^2y\) But your original problem was: \(3xy^2+6x^2y\). See the difference between those? \(\begin{array}{ccc} 3x^2y^2\!\!&\!\!+\!\!&\!\!6x^2y\\ 3xy^2\!\!&\!\!+\!\!&\!\!6x^2y \end{array}\)
anonymous
  • anonymous
Sooo is it 3xy(xy+(2x)= 3xy+6x^2y
e.mccormick
  • e.mccormick
You just canged the ( ) a bit... the problem... hmm.... lets try color! \(3xy(xy+2x)\) is wrong because it multiplies out to: \(3x^2y^2+6x^2y\) \(\begin{array}{cccl} 3x^\color{red}{2}y^2\!\!\!&\!\!\!+\!\!\!&\!\!\!6x^2y&\leftarrow\text{How it multiplies out.}\\ 3xy^2\!\!&\!\!\!+\!\!\!&\!\!\!6x^2y&\leftarrow\text{What you need!} \end{array}\)
anonymous
  • anonymous
3x(y^2+2^2y)
e.mccormick
  • e.mccormick
3x(y^2+2^2y) ?? \(3x(y^2+2^2y)\) I think you are getting frustrated at this point. And I get that! You are close to getting it right. It might help of you tried reading these: http://www.purplemath.com/modules/simpfact.htm http://www.purplemath.com/modules/simpexpo.htm The first one is on factoring, the real question you have. The second is on simplifying things with exponents, which is related to it.
anonymous
  • anonymous
\[3x(y ^2+2^2y)\]
e.mccormick
  • e.mccormick
See, that has other issues. First, you are no longer factoring out 3xy. And 3xy was correct! Second, the \(2^2\).
anonymous
  • anonymous
I am just getting all confused
e.mccormick
  • e.mccormick
Yah, that is why I think it would be better at this point for you to read what is on that site. They have lots of examples.
anonymous
  • anonymous
Uffffff all right u said this is right 3xy(xy+2x)
e.mccormick
  • e.mccormick
No. I said there is one and only one problem with that.
anonymous
  • anonymous
Ok what is the problem
e.mccormick
  • e.mccormick
\(3xy^2+6x^2y\) is your original equation, right?
anonymous
  • anonymous
Yes
e.mccormick
  • e.mccormick
Well, \(3xy(xy+2x)=3x^2y^2+2x^2y\) See how that is NOT your original when I multiply it out?
e.mccormick
  • e.mccormick
Look at the exponents....
anonymous
  • anonymous
Can you tell me the right answer so I can see what am I doing wrong
e.mccormick
  • e.mccormick
What do you have too much of in \(3x^2y^2+6x^2y\) .... and I forgot to change the 2 to a 6 above... my mistake.
anonymous
  • anonymous
Of x^2 y^2
e.mccormick
  • e.mccormick
Yes, it is where that is at. The \(x^2\) is too many x. It should only be x and NOT \(x^2\). So, if I need to remove an x from JUST that side, what would I need to do? \(3xy(xy+2x)=3x^\color{red}{2}y^2+6x^2y\) The red 2 needs to go!
e.mccormick
  • e.mccormick
And remember, you only need to change one thing in: \(3xy(xy+2x)\) Just one x in the right place needs to die and it is all good!
anonymous
  • anonymous
3xy(y+2x)
e.mccormick
  • e.mccormick
YES! There it goes! That eXtra X is gone!
anonymous
  • anonymous
Is that the right answeright answer then
e.mccormick
  • e.mccormick
\(3xy\left(\dfrac{3xy^2}{3xy}+\dfrac{6x^2y}{3xy}\right)\) \(3xy\left(\dfrac{\color{red}{\cancel{\color{black}{3x}}}y^{\color{red}{\cancel{\color{black}{2}}}}}{\color{red}{\cancel{\color{black}{3xy}}}}+\dfrac{\color{red}{\cancel{\color{black}6}}2x^{\color{red}{\cancel{\color{black}{2}}}}\color{red}{\cancel{\color{black}{y}}}}{\color{red}{\cancel{\color{black}{3xy}}}}\right)\) \(3xy(y+2x)\)
e.mccormick
  • e.mccormick
Yes. That is the right answer.
anonymous
  • anonymous
Can you please help me in my math hw please.
e.mccormick
  • e.mccormick
I think you need to do a little more review at this point. That page I linked is just one of many that go over Simplifying Expressions with Exponents and Polynomial Factoring. Ig you look into those, it should be a lot easier.

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