Question

Can you help me with this problem 3xy^2+6^2y

Answer 1

What are you supposed to do with it and where are you having trouble?

`\(3xy^2+6^2y\)` makes: \(3xy^2+6^2y\)

Answer 2

I think the gcf is 3

Answer 3

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?

Answer 4

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 :'(

Answer 5

So in high school you never learned about terms and factoring something out?

Answer 6

hmm, sounds fishy

Answer 7

Yes but not this type of math like finding the gcf so I really need help

Answer 8

gcf is more like 7th grade :S

Answer 9

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.

Answer 10

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.

Answer 11

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

Answer 12

Well sorry my school did not show me how to do this and I am not planning to fail this class

Answer 13

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?

Answer 14

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??

Answer 15

Ok I think I got this 3×y(xy+2xy)

Answer 16

Hmmm.... where did the x in the second term come from?

Answer 17

So I did it wrong?

Answer 18

Yah. Do you have your work you can show? We can find out where this is going wrong.

Answer 19

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

Answer 20

The thing I did was 3×1=3 I get 3xy(xy
Then 3×2=6 I get 3xy(xy+6x^2y)

Answer 21

You originally posted:
\(3xy^2+6^2y\)

That is nowhere near \(3xy(xy+6x^2y)\)...

Answer 22

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.

Answer 23

So wait how do it I am trying my best :\

Answer 24

Yes that is the question 3xy^2+6x^2y

Answer 25

\(3xy^2+6x^2y\) AH! So NOT \(3xy^2+6^2y\) The x was missing! That makes a HUGE difference!

Answer 26

Omg I forgot to put the x Dx

Answer 27

=) All good. Now we are on the same problem.

Answer 28

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.

Answer 29

Ok than so I thinknthe gcf is 3

Answer 30

It is more than just 3.

Answer 31

Gcf:1

Answer 32

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.

Answer 33

Ok so it might be 3xy(xy+2xy)

Answer 34

Gcf is 3xy

Answer 35

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.

Answer 36

The GCF of 3xy is correct, but \(3xy^2+6x^2y\ne 3xy(xy+2xy)\) You left some variables in that were taken out.

Answer 37

So the answer I gave it will be wrong because I need to take out y right

Answer 38

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!

Answer 39

\(3xy^2+6x^2y\) 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?

Answer 40

Nop i dont get it

Answer 41

Well, what is \(\frac{x}{x}\)?

Answer 42

1

Answer 43

YES! And \(\frac{x^2}{x}\)?

Answer 44

-10

Answer 45

Hmmm... how did you get -10?

Answer 46

Wait hold up I am trying to do this in my head

Answer 47

=)

Answer 48

Wait a min is it 1^2

Answer 49

Close! Canced the bottom fine. But the top.... hmmm....

Answer 50

Let me show exactly how \(\frac{x}{x}=1\) works and you might see it.

\[
\frac{x}{x}=\frac{x^1}{x^1} \implies \\ \, \\
\frac{x^1}{x^1}=x^1x^{-1} \implies \\ \, \\
x^1x^{-1}=x^{1-1} \implies \\ \, \\
x^{1-1}=x^0 \implies \\ \, \\
x^0=1 \, \\
\]

Answer 51

So is it 1 hagen

Answer 52

No. The 2 in the square would make it different.

Answer 53

I just did this 1^2=1

Answer 54

\[\frac{x^2}{x}=\frac{x^2}{x^1} \implies \\ \, \\
\frac{x^2}{x^1}=x^2x^{-1} \implies \\ \, \\
x^2x^{-1}=x^{2-1} \implies \\ \, \\
x^{2-1}=x^1 \implies \\ \, \\
x^1=x \, \\\]

Answer 55

Yes, \(1^2=1\) But you don't have \(1^2\). You have \(\dfrac{x^2}{x}\)

Do you see my proof of how \(\dfrac{x^2}{x}=x\)?

Answer 56

Yeahh I kinda get it

Answer 57

OK. Do you get it enough to apply it this time?

Answer 58

Nop

Answer 59

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!

Answer 60

Ik this will be wrong 3xy(xy+2x)= 3xy+6^2y

Answer 61

I mean this is wrong. 3xy(xy+2x)= 3xy+6x^2y

Answer 62

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.

Answer 63

Bah, and I put a square in one of the wrong places.

Answer 64

So your saying that if I do this 3xy(xy+(2x))= 3xy+6x^2y

Answer 65

So your saying that if I do this 3xy(xy+(2x))= 3xy+6x^2y

Answer 66

Well, \(3xy(xy+(2x))= 3x^2y^2+6x^2y\) But your original problem was: \(3xy^2+6x^2y\). See the difference between those?

\(\begin{array}{ccc}
3x^2y^2\!\!&\!\!+\!\!&\!\!6x^2y\\
3xy^2\!\!&\!\!+\!\!&\!\!6x^2y
\end{array}\)

Answer 67

Sooo is it 3xy(xy+(2x)= 3xy+6x^2y

Answer 68

You just canged the ( ) a bit... the problem... hmm.... lets try color!

\(3xy(xy+2x)\) is wrong because it multiplies out to: \(3x^2y^2+6x^2y\)

\(\begin{array}{cccl}
3x^\color{red}{2}y^2\!\!\!&\!\!\!+\!\!\!&\!\!\!6x^2y&\leftarrow\text{How it multiplies out.}\\
3xy^2\!\!&\!\!\!+\!\!\!&\!\!\!6x^2y&\leftarrow\text{What you need!}
\end{array}\)

Answer 69

3x(y^2+2^2y)

Answer 70

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.

Answer 71

\[3x(y ^2+2^2y)\]

Answer 72

See, that has other issues. First, you are no longer factoring out 3xy. And 3xy was correct! Second, the \(2^2\).

Answer 73

I am just getting all confused

Answer 74

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.

Answer 75

Uffffff all right u said this is right 3xy(xy+2x)

Answer 76

No. I said there is one and only one problem with that.

Answer 77

Ok what is the problem

Answer 78

\(3xy^2+6x^2y\) is your original equation, right?

Answer 79

Yes

Answer 80

Well, \(3xy(xy+2x)=3x^2y^2+2x^2y\)

See how that is NOT your original when I multiply it out?

Answer 81

Look at the exponents....

Answer 82

Can you tell me the right answer so I can see what am I doing wrong

Answer 83

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.

Answer 84

Of x^2 y^2

Answer 85

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!

Answer 86

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!

Answer 87

3xy(y+2x)

Answer 88

YES! There it goes! That eXtra X is gone!

Answer 89

Is that the right answeright answer then

Answer 90

\(3xy\left(\dfrac{3xy^2}{3xy}+\dfrac{6x^2y}{3xy}\right)\)

\(3xy\left(\dfrac{\color{red}{\cancel{\color{black}{3x}}}y^{\color{red}{\cancel{\color{black}{2}}}}}{\color{red}{\cancel{\color{black}{3xy}}}}+\dfrac{\color{red}{\cancel{\color{black}6}}2x^{\color{red}{\cancel{\color{black}{2}}}}\color{red}{\cancel{\color{black}{y}}}}{\color{red}{\cancel{\color{black}{3xy}}}}\right)\)

\(3xy(y+2x)\)

Answer 91

Yes. That is the right answer.

Answer 92

Can you please help me in my math hw please.

Answer 93

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.