PDA

View Full Version : Math



Drew2
08-27-2003, 04:32 PM
Rather than have one of the mods get all antsy and tell us to create a topic, I'm creating a topic for Artha to expand his mathematical skills in. He wants to play factoring, we'll play.

Also, I'll probably slip a few of my homework questions in here. I know how some of you people just love to prove your 'inteligents', so I figure this will be an easy way to get my work done for me.

Bobmuhthol
08-27-2003, 04:48 PM
Good. Give me all the questions and I'll solve them for you.

Artha
08-27-2003, 04:51 PM
lol.

imported_Kranar
08-27-2003, 06:53 PM
Do you know what the Factor Theorem is?

Drew2
08-27-2003, 07:50 PM
A polynomial f(x) has a factor (x-k) if and only if f(k) = 0

imported_Kranar
08-27-2003, 08:07 PM
Yes, now make use of it.

You can use factor theorem to factor EVERY single polynomial equation.

Bobmuhthol
08-27-2003, 08:12 PM
And you can use the quadratic formula to solve EVERY single quadratic equation. Just like you can use addition to solve EVERY single addition problem. I'm not seeing the glamour.

Drew2
08-27-2003, 08:26 PM
Ok.. instead of getting frustrated I'm just going to dump my difficulties on you all (Kranar). After I find the X intercepts from an equation (Let's say... y = 16 - 4x^2).. to find the y, do I plug the X's in individually? And find 2 y intercepts? Or do I make x=0 and do it like I did the x intercepts?

Bobmuhthol
08-27-2003, 08:27 PM
I wish I knew what the hell you were talking about, and why you're finding intercepts in college instead of 7th grade.

Caels
08-27-2003, 08:27 PM
Kranar makes everything = 0 first, he'll confuse you! Run!

imported_Kranar
08-27-2003, 08:34 PM
If you want to find the factors, you can do it by finding the x-intercepts, correct.

So you gave this example:

y = 16 - 4x^2

You just set y = 0 and solve for x to find the x-intercepts, which is what you said:

1) 0 = 16 - 4*x^2
2) -16 = -4*x^2
3) x^2 = 4
4) x = +/- 2

Okay, so you found those 2 intercepts, all you do now is invoke the factor theorem. Since when y = 0, x = +/- 2, then (x + 2) and (x - 2) are factors.

So 16 - 4x^2 = (x + 2)*(x - 2)

Factor theorem is very useful in this respect. Just find the intercepts (k) and one of your factors will be (x - k).

[Edited on 8-27-2003 by Kranar]

Bobmuhthol
08-27-2003, 08:36 PM
I'm pretty sure Tayre was looking for the Y intercept on a graph. But I forget everything from Algebra so I'm not sure if factoring it helps or not.

Drew2
08-27-2003, 08:37 PM
Right, I did that, so now how do I reverse that and find the Y intercepts?

Bob is half right, I need the Y intercepts but I did need the X too, but I hald already solved those.

[Edited on 8-27-2003 by Tayre]

imported_Kranar
08-27-2003, 08:40 PM
For intercepts...

x-int = set y to 0 and solve for x.
y-int = set x to 0 and solve for y.

You do not need to know the x-int to find the y-int, they are independent of one another.

1) 16 - 4x^2 = y
2) 16 - 4(0)^2 = y
3) 16 - 0 = y
4) y = 16

And you're done.

Drew2
08-27-2003, 08:41 PM
UGH! I DID That.. but I was like no.. that's too simple to be math.

Caels
08-27-2003, 08:45 PM
As much as I hate math, and chose not to learn it in school, I can't help but be fascinated by it.

Maybe I'll get a second chance in college.

imported_Kranar
08-27-2003, 08:46 PM
It is simple, but this is simply review. If a professor were to slam the real math that you'll be learning this year within the first week of classes, you and everyone would drop out.

[Edited on 8-27-2003 by Kranar]

Drew2
08-27-2003, 08:47 PM
Yeah yeah... I mean I did know all this... in 7th grade.. but the saying is true.. you forget things you don't use.

imported_Kranar
08-27-2003, 08:49 PM
Just wait until the time comes when you need to find the intercept of lines and planes in 3-space using matricies. That's what first year Algebra is really about, and you'll be glad you had this review to fill in the gaps.

Bobmuhthol
08-27-2003, 08:50 PM
<<Just wait until the time comes where you need to find the intercept of lines and planes in 3-space using matricies.>>

Please tell me they're more than what I've seen in Algebra I / SAT / Other various tests I've taken.

Drew2
08-27-2003, 08:54 PM
Originally posted by Bobmuhthol
<<Just wait until the time comes where you need to find the intercept of lines and planes in 3-space using matricies.>>

Please tell me they're more than what I've seen in Algebra I / SAT / Other various tests I've taken.

Please start a "Bobmuhthol is a genius" thread if you feel the need. Keep your self-inflating egotistical comments somewhere where I don't have to read them when I put my homework problems out for someone else to explain to me. Believe it or not, just because I am stumbling on one aspect of algebra does not declare my level of knowledge or intelligence. I know more, and will always know more, than you, Bob. It's a fact of life. You can only fit a certain amount of experience in a certain amount of time, and I have a few years on you. So please, when you've gotten over yourself, resume posting.

Edit: And by no means am I saying I know more than everyone. I highly doubt I know more than anyone else on these boards. Maybe Artha. :D But seriously... I was only making a point to Bob. No one else's integrity was harmed during the making of this post.

[Edited on 8-27-2003 by Tayre]

imported_Kranar
08-27-2003, 08:56 PM
<< So please, when you've gotten over yourself, resume posting. >>

He has 1014 posts, let the child pat himself on the back once in awhile.

Bob likes to feel... special.

[Edited on 8-27-2003 by Kranar]

Drew2
08-27-2003, 08:59 PM
That in itself only proves the amount of time he has to do meaningless and trivial things.

Luckily for us, his schooling starts next week, and we'll all be able to stomp his record to the ground.

Drew2
08-27-2003, 09:03 PM
::sighs:: It never ends.

y = 2x^3 - 4x^2

y = 2x(x^2 - 2x)

Then what? (factoring-wise)

imported_Kranar
08-27-2003, 09:09 PM
The problem is:
y = 2x^3 - 4x^2

Your first step was:
y = 2x(x^2 - 2x)

So you did manage to factor out the 2x. Now you have x^2 - 2x left.

There is an x common in both terms, so your next step would be to factor out that x.

(2x)(x - 2)(x)

And you simplify that to your final answer:

2x^2(x - 2)

Since x and -2 share no common factors, you're done.

As an aside, I would recommend that you try and extract as big of a chunk as possible first:

You originally had:

2x^3 - 4x^2

And you have a 2x^2 in common in both terms. So take that out right away and you would be done the problem in one step.

2x^2(x - 2)

[Edited on 8-28-2003 by Kranar]

Drew2
08-27-2003, 09:10 PM
Ugh. I seem to be doubting the simplicities of mathematics. Mostly because I don't trust them.

imported_Kranar
08-27-2003, 09:14 PM
Yep, but it's nothing to worry about. Once you continue practicing all this you'll look back at it laughing.

Keep it up and feel free to ask.

[Edited on 8-28-2003 by Kranar]

HarmNone
08-27-2003, 10:06 PM
::HarmNone peers at all the numbers and arcane symbols, shudders once, and swoons.::