[lJ.]

Hi, I was wondering if someone could explain and walk me through this problem.
The question is to solve:

b+1 b
X - X

[It says x to the exponential power of b+1 minus x to the exponential power of b.]

And I don't have to solve for the variables just factor the equation.

First for these factor problems we have to find the GCF if there is one then do other steps depending on if it is the difference of 2 squares, difference of 2 cubes, sum of 2 cubes, trinomial, combination, etc. Then keep factoring until there are no exponential powers or squares [like until the problem is fully broken down], for example until x squared is x.. ]

How do I find the GCF [greatest common factor] for this problem while working with the exponents?

Please help me solve this problem and understand it!

Thank you so much if you answer!

[morgaine300]

Can we clear up first that you mean:

[morgaine300]

A little hint: pick something for b, anything you want. (Well, positive.) What then would be b + 1?

Plug in your choices and think about how you would factor that.

See if you can use that as an example to factor this. (This is assuming I'm interpreting the expression correctly.)

[Unknown008]

Do you know that:



?

And



?

You can use these to help you. Post what you get! :)

[morgaine300]

Hmm, good point, but I didn't do it that way. LOL.

[lJ.]

Thanks for your suggestions! I'll try thinking about it by putting in numbers in place of b, and then I'll separate the b and 1 since it's subtraction and try to solve it.

[galactus]

Just factor out an x^b:

.

That's it.

[lJ.]

Could you please walk me through how you factored that? I think I am just confused because of the b PLUS 1 because I understand how to do it with multiplication.

[lJ.]

Would I find the GCF as just x or x to the b?

[Unknown008]

Hmm... take where I told you.



Meaning,



Put that back in your original question:



Now, you can factor out x to get what galactus told you :)

EDIT: I put it in a perhaps clearer way now.

[morgaine300]

Actually, what I was trying to get you to think about is how Galactus came up with that. It doesn't look like he did any actual work, since he said to "just factor out the x^b". That's what I was leading to - why "just factoring" that out works.

What Unknown008 is doing is a more technical way about it, and may be something they want you to work through for your class. (i.e. me and Galactus may be "cheating" as far as your class is concerned.)

It will depend on what it is they are trying to teach you and therefore what direction they want you to look at this from.

[lJ.]

Thanks you all for trying to help. I've only had 2 classes so far this year, and it's more review right now I think, from about 2 years ago, but I am not familiar with using letters with exponents and then having them add with numbers. So I wasn't sure how to factor it I think I just needed people to explain it to me.

[lJ.]

So even though the exponents are adding, I can distribute? Could I do this:

take the x to b+1 and distribute it to get xb +x1 [multiplying while distributing].. so there are no exponents then still subtract the second part of the problem which is x to the b? What do I do with the second part of the problem?

First I'm supposed to use a GCF and then work from there.
But can I just distribute it like I said above for the x to b+1 to get xb + x1..

Thank you all for trying to help me with this I appreciate it!

[morgaine300]

OK, it's not like me to do this, but I'm giving this one away.

If you had you could just factor out the (GCF) and be left with . You've just factored out the highest thing you have. And is not 4 just (3 + 1)? Therefore in this example b = 3 and b+1 is 4.

allows me to factor out the to have

Every time my left term is one bigger than my right, I can factor out the degree over on the right one, which leaves me with an exponent of 1 on the left one. Every time.

If the left one is (b + 1), it's one bigger than the one on the right term, so you can always factor out as many as are on the right term, leaving 1 on the left one.

And this:

take the x to b+1 and distribute it to get xb +x1

is not "distributing" and you aren't multiplying.

It would be useful if you'd at least use ^ for exponents so that we know what you mean.

Do you mean ? If so, then write it (x^b) - (x^1) to make it clear what you are doing.

But since you're calling it "distributing" and mentioning multiplying, and also saying "so there's no exponents," I have to wonder if you mean exactly like it looks. And no, you can't do that. It isn't so that you can distribute. The (b+1) is an exponent.

Think backwards:

. That's a rule you have.

Cause for example:
means , and since I can remove all those parenthesis, essentially I have x multiplied times itself 7 total times.

Hence:

= =
- you can just add the exponents cause that's how many times total you're going to multiply x times itself.

Therefore, I can also do it backwards:

= .
But I'm not "distributing" it, like multiplying the x times the 3 and then times to 4, cause I'm not multiplying. They're exponents.

You can do the same thing with the as what I did with the 3 and 4 above.

Once that is done, you still have to carry down the "- x" that's on the end of it -- you can't just drop that off.

Once you have , it can be rewritten without the * in there, and without the 1 exponent, which is understood: . Tack on that "- x" and start factoring.

I personally think this is just the hard way about it, but you may have to show this work.

[Unknown008]

If you still have some problems, you may look at it this way.



Right?

Okay, now, we know that 5 = 2+3

Let's put it like that:



Now, assuming that you 'distribute' it, you'll have:



Which is not true!

You have to know the product rule in index form.

That is, if , you can add the powers.

Take the previous example again.



This time, it works! :)

[lJ.]

Thank you all so much for helping me and trying to help me understand the problem. I'll ask my teacher tomorrow. One problem is x to the third minus 8. Would I split it up 3 ways or just two?

[lJ.]

Never mind about my last post, because I found in my notes we acutally did that exact problem in class.
The problem x to the 3rd power minus 8. We created a binomial of (x-2) [because 2 comes from the root of 2 cubed which is 8] and keeping the minus in that part. Then next to it a trinomial of x squared plus 2x plus 4. She said that you multiply the 2 parts of the binomial which is x and negative 2 to get negative 2x and then take the opposite of the sign... which is just switching the negative to positive so you get 2x in the trinomial. How do I then break the trinomial of (x squared plus 2x plus 4) down? I understand the rest of the problem, I just need help with that last part with the trinomial.

Thank you so much!

[lJ.]

Sorry! I kept looking at these notes.. and I found you CAN'T break down and solve the trinomial any further so the final answer is (x-2)(xsquared+2x+4)
Sorry for the really long question now it makes sense that I can't solve it any further.

[morgaine300]

OK, I'm never minding.

[Unknown008]

Lol, you should learn to use the LaTeX format to help us understand too what you are doing.

for example, your expression (x-2)(xsquared+2x+4) is written as:



Here's what you type to render this:
[math](x-2)(x^2 + 2x +4)[/màth]

You put the tags [math] [/màth] replacing the à by a.

Here are other examples:



Code:

[math]\frac{2x}{4x^2 - 9} > \int^1_0 2x^3 + x^2 dx[/màth]

\frac is for fractions
\int is for the integral sign
^ is for powers
_ is for subscript