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Math

Registered User, __BANNED USERS regular
edited March 2011
Really confused by all this since it wasn't covered in class at all, we were just sort of told to do the homework.

Anyway I am doing the Rational Zeros Theorem stuff, then I have to use the zeroes to factor f over the real numbers.

That isn't a sentence that makes sense to me, but I do know what to do. I can get the zeros, I don't know if there is a faster way than just process of elimination but that is what I do. Start with the lowest then go to the highest. Anyways the book always stops on the lowest one, for example there could be 2 positive zeros and it find that x+1 is a factor it will never go further than that, anyway to know that nothing else is the factor without trying the rest?

Also once I do get the zeroes I am suppose to factor by grouping, I don't know what that means, it wasn't taught in class, and it isn't in the book. After that I set the factors to 0, which seems easy but I don't know what factor by grouping is.

Than the book says thus, I don't know what happened, they get an incredibly long answer and nothing is explained by then it lists the real zeros.

Fizban140 on
«13

Posts

• Registered User regular
edited February 2011
factor by grouping:

if it's 4 terms split them into two halves and find the Greatest Common Factor of each half. If you're dealing with something that is factorable by gouping then the two halves will half a common binomial factor that you can basically treat as another GCF.

Also are you just saying you are finding zeros by guess and check? because yes there are many ways to find them and that's what you should be learning. I would like to see the problems you're doing, but I'll make up a quadratic example just in case:

x^2 + 5x + 6...find the zeros:
factor it : (x+2)(x+3) = 0
so x+2 = 0 or x+3 = 0, which means x = -2 or x = -3.

If you look at the graph, that's where the function crosses the x-axis (that's why we call them "zeros" because y = 0 on the x-axis).

musanman on
• Registered User, __BANNED USERS regular
edited February 2011
f(x) = 2x^4 + x^3 - 7x^2 - 3x +3

That is the problem, I know this is probably the most simple polynomial for this but I can't figure this out. They magic out some shit like

Factoring by grouping gives you (2x-1)(x^2-3)

then they magic out some shit like

the real zeroes are 1/2 -1 square root of 3 and negative square root of 3 multiplicity 1. I have no idea what to do. This is so complex I have to learn all these things just to start to learn how to do the homework.

I mean we spent the whole class period counting sign changes and finding the factors of 6 and other small numbers, how the fuck do we jump straight into this as homework? I can't learn like this.

Fizban140 on
• The Engineer Columbia, MDRegistered User regular
edited February 2011
Fizban, did you get a proper math tutor yet?

BoomShake on
• Registered User, __BANNED USERS regular
edited February 2011
Define proper?

I will answer that now actually, no.

I have a tutor but it is just a free service my school offers, kids from the local university.

Fizban140 on
• Registered User regular
edited February 2011
Fizban140 wrote: »
f(x) = 2x^4 + x^3 - 7x^2 - 3x +3

That is the problem, I know this is probably the most simple polynomial for this but I can't figure this out. They magic out some shit like

Factoring by grouping gives you (2x-1)(x^2-3)

then they magic out some shit like

the real zeroes are 1/2 -1 square root of 3 and negative square root of 3 multiplicity 1. I have no idea what to do. This is so complex I have to learn all these things just to start to learn how to do the homework.

I mean we spent the whole class period counting sign changes and finding the factors of 6 and other small numbers, how the fuck do we jump straight into this as homework? I can't learn like this.

ok first of all I realize now you're the same guy from the other thread, you need to get a tutor.

That said, factoring the 5 term polynomial you just listed does not give that result.

musanman on
• The Engineer Columbia, MDRegistered User regular
edited February 2011
I will start off by saying
YOU NEED A REAL TUTOR. Someone who knows the material AND is trained in the teaching of it (including multiple approaches), not just some student who's good at math. The latter is good for people who just have a few small hangups or confusions here and there, but that's honestly not your case. You need someone who can give you real 1-on-1 teaching

Ok, so I don't think musaman quite understood the approach they're asking for here.

Rational Zeroes Theorem, or more commonly Rational Root Theorem, is a way of finding roots of a polynomial and thus factoring it. The wiki has an excellent example, although their method of testing it is probably more complex than where you're at.

The basic steps are as follows (underscore indicates subscript)
Given: f(x) = a_n*x^n + a_(0-1)*x^(n-1) … + a_0 , where all coefficients are integers
Let: Q = integer factors of a_n and P = integer factors of a_0
Potential roots are then x = (+/-) P/Q

You then test these systematically. When you find one, we'll call it B, that works and give you zero, you can then factor out (x+B) from f(x) using something like polynomial long division. Test the remaining unfactored portion with B again to see if the root has multiplicity. Continue down the list of potential roots until it is entirely factored and all roots are found.

Let's take your example:
f(x) = 2x^4 + x^3 - 7x^2 - 3x +3
P = 3,1 (since factors of 3 are 3 and 1)
Q = 2,1 (since factors of 2 are 2 and 1)
Potential x = (+/-)P/Q = (+/-){3,1}/{2,1} = +3/2, +3/2, +3, -3, +1/2, -1/2, +1, - 1

Test f(x) with first one, +3/2
f(+3/2) = 2*(3/2)^4 + (3/2)^3 - 7*(3/2)^2 - 3*(3/2) +3
f(+3/1) = -15/4
∴ +3/2 is not a root

Test f(x) with second one, -3/2
f(-3/2) = -3/2
∴ -3/2 is not a root

Test f(x) with third one, +3
f(+3) = 120
∴ +3 is not a root

Test f(x) with fourth one, -3
f(-3) = 84
∴ -3 is not a root

Test f(x) with fifth one, +1/2
f(+1/2) = 0
∴ +1/2 is a root

x=1/2
2x - 1 = 0
Factor out (2x-1) from f(x)
g(x) = f(x)/(2-1) = x^3 + x^2 - 3x + 3
∴ partially factored f(x) = (2x-1)(x^3 + x^2 - 3x + 3)

Test g(x) with fifth one again, +1/2
g(+1/2) = -33/8
∴ +1/2 is not another root (and only has multiplicity 1)

Test g(x) with sixth one, -1/2
g(-1/2) = -11/8
∴ -1/2 is not a root

Test g(x) with seventh one, +1
g(+1) = -4
∴ +1 is not a root

Test g(x) with eigth one, -1
g(-1) = 0
∴ -1 is a root

x=-1
x+1 = 0
Factor (x+1) out of g(x)
h(x) = g(x)/(x+1) = x^2-3
h(x) cannot be further factored.

FACTORED f(x) = (2x-1)(x+1)(x^2-3)

Finding the remaining zeroes (of h(x))
x^2-3=0
x^2=3
x=+sqrt(3) or x = -sqrt(3)

ZEROES/ROOTS x = { 1/2 , -1 , +sqrt(3) , -sqrt(3) }

BoomShake on
• Registered User regular
edited February 2011
yeah that's not factoring by grouping, and the "answer" he listed still isn't correct

musanman on
• The Engineer Columbia, MDRegistered User regular
edited February 2011
Yeah, I'm not sure why they'd be doing rational root theorem and factoring by grouping at the same time.
Either way, between our two posts he should be good.
Here's the Factoring by Grouping[/i] wiki, which has a nicely formatted example, for good measure.

BoomShake on
• Registered User, __BANNED USERS regular
edited February 2011
That factoring looks incredibly complex, is there any good way to do it besides trial and error?

One thing I am still having trouble with, when I use sythetic division how do I know I am done? Like I find out x-1 was a factor, then I do 1 again and that is a factor, when do I stop trying to factor out something? I just had a problem like that where x-1 was a factor twice in a row, and the book just stops after the first time and does factor by grouping.

Fizban140 on
• The Engineer Columbia, MDRegistered User regular
edited February 2011
Complex and tedious are two very different things.
What I did was tedious, but not complex.

Read the link I posted earlier to the Rational Root Theorem wiki page. In their example, they use the Horner Scheme to test the roots.

BoomShake on
• Registered User, __BANNED USERS regular
edited February 2011
This is ridiculous, I am just trying to get my homework done and I have to learn all this other stuff. Really makes me not want to this, just incredibly daunting to even start my homework.

Fizban140 on
• drinking coffee in the mountain cabinRegistered User regular
edited February 2011
It is indeed ridiculous that you are being asked to learn math from a textbook rather than from a teacher. As said above, your best hope is to get a tutor. The subject matter itself is not complicated or difficult, but the barrier to entry is high without a good teacher. Once you sort of crack the code, the task should become routine.

Powerpuppies on
• Registered User, __BANNED USERS regular
edited February 2011
I just dislike being given a task and having to top several dozen other things before I can begin the task. Or more accurately getting half way through the task and then having to do several dozen other tasks.

Fizban140 on
• Registered User regular
edited February 2011
this just in: mathematics is cumulative

musanman on
• Registered User, __BANNED USERS regular
edited February 2011
That would have been great if it was in any of my previous math classes.

Fizban140 on
• Nah Registered User regular
edited February 2011
Fizban140 wrote: »
That would have been great if it was in any of my previous math classes.

The reason you're having difficulty now is because it was

If you were having problems in Pre-Alg, hell yes you're going to have the same ones plus a whole host of new ones in Alg I

Usagi on
• Registered User, __BANNED USERS regular
edited February 2011
I got an A in pre algebra, than passed algebra 1 with a B I think, now taking algebra 1 again and I think I am going to fail it.

I guess I just need to find somewhere with a lot of factoring practice problems so I can commit this all to memory, but that feels like it would takes me a long time to do.

I am taking a break for the night, there just appears to be no systemic or logical approach to this, it is all just randomly done. That is as far as I can get after a couple hours of attempting to interpret this stuff.

Sometimes they synthetically divide once, then factor by grouping, sometimes they keep synthetically dividing, answers come out different but I don't see the signal that tells me when to do what. It appears random.

Math has evaded any logical attempts at understanding it again so I will try again tomorrow maybe.

Fizban140 on
• When the last moon is cast over the last star of morning And the future has past without even a last desperate warningRegistered User, Moderator mod
edited February 2011
BoomShake wrote: »
YOU NEED A REAL TUTOR.

This is really the advice you should be taking from this thread at this point. I don't mind your math threads at all, but I don't think you really grasp the terrible disservice you are doing yourself by not following this very simple advice.

H/A can only help you with this so far. You NEED to get proper, one-on-one help for this, or you are going to fail this and any other math-related or -centered course you take. It is entirely preventable. It is clear from your posts that you are oversimplifying some things in your head at the same time you are overcomplicating others, and the only way for you to sort out which is what is for you to get proper help with this.

Lacking the ability to sit across the table from you and really examine your thought process as you go through these problems, this forum is ill-equipped to help you get at the reason that these problems aren't making sense to you. That's just how it is. It's not your fault, you aren't terrible or stupid because it doesn't come to you right away, but it IS something that you need to take extra care with especially if you are going into the sciences, and it IS something that will require instruction from a proper tutor.

ceres on
And it seems like all is dying, and would leave the world to mourn
• Registered User, __BANNED USERS regular
edited February 2011
I am most likely changing my major to something that requires no math, like an English major so I can work in in food services.

Seriously though I will look into tutors for around here, and more importantly money for this service.

Fizban140 on
• Registered User regular
edited February 2011
Fizban140 wrote: »
I am most likely changing my major to something that requires no math, like an English major so I can work in in food services.

Seriously though I will look into tutors for around here, and more importantly money for this service.

Aren't there some office hours you can attend as well? Those help immensely, at least for me.

Demerdar on
• Registered User regular
edited February 2011
Have you encountered synthetic division (also known as synthetic substitution)? If so, that makes the testing of possible roots much quicker. Wikipedia has a decent example. I might be able to help you via Skype some time also.

Here's the process I would recommend:
1. If there is a greatest common factor, go ahead and factor that out first. It will reduce the rest of the coefficients and make the numbers easier to work with.

2. Determine how many roots you have. This is easy: it's the highest power of the variable (known as the "order" of the function). The one you listed in the OP will have 4 total roots.

2. Use the rational roots theorem to make a list of possible roots. BE SURE to include both the positive and negative for each one.

3. If you're familiar with Descartes Rule of Signs, use it to identify how many positive and negative roots you have. This isn't necessary but can help you save time later.

4. Start testing roots. I would recommend starting with the smallest roots, and whole numbers before fractions, just because they're easier to work with. Do it just like BoomShake showed you. If you know/can learn synthetic division it makes this easier.

5. IF IT WON'T PISS YOUR TEACHER OFF, when you're down to only needing to find 2 more roots, throw the damned thing in the Quadratic Formula to get the last two.

Ceres is right, there's no way to do this well except for face-to-face in real time. For now, though, think of it this way: think of something you're good at, a sport or music or game or whatever. How'd you get good at it? You repeated the steps over and over again until they became second nature. It happens with math too. It's nothing more than a series of rules (to be sure, often frustrating and confusing rules). If you follow those rules, you will get the answer. Where I see my students making mistakes is, when they get frustrated, they just do something that looks sort of like math, something that feels right. It's kind of a running joke in my classes, I tell them all the time, "Don't do random things." If you think through each step as you're doing it, and ask yourself "why am I doing this step?" you'll get it eventually.

GoodOmens on

IOS Game Center ID: Isotope-X
• Registered User, __BANNED USERS regular
edited February 2011
I am just trying to finish my homework before class so I don't fall behind, I mean I am doing better than most people in the class. Anyways I need to figure out what the fuck is going on...

I have 3x^3 - x^2 - 15x +5 = 0

Pretty simple, I find that (x-1/3) is a factor so I get (x-1/3)(3x^2-15x)

That can be factored down into 3x(x-5) so now I have 1/3 and 5 as my answers except the book didn't do it like that they factored the problem without ever trying to do all the stuff it taught me in the chapter, they just straight factored it which isn't even taught in this book. I don't know why, and I did everything properly I think so I should still get the same answer, this just feels all random. I can do the steps right but get a different answer than the book because they randomly decided to do it with a different methon, one not taught in the book.

I would probably slap the shit out of the author of this book, for being so bad at his job, if I ever saw him.

I am going to take a break from this, I already am not in the mood to deal with this. I just started one problem, did the steps to perfection, got the wrong answer, double checked all my math, same answer. I look at the solution manual and they do something that I can not reference anywhere in the book and it has no explanation. I just chalk that one up as a loss, if it ever come up on a test I will get it wrong but that is okay, can't get them all right.

I go to the next problem, same thing, except an entirely different type of bullshit procedure described above. I am done for now.

Fizban140 on
• Registered User regular
edited February 2011
Fizban140 wrote: »
I am just trying to finish my homework before class so I don't fall behind, I mean I am doing better than most people in the class. Anyways I need to figure out what the fuck is going on...

I have 3x^3 - x^2 - 15x +5 = 0

Pretty simple, I find that (x-1/3) is a factor so I get (x-1/3)(3x^2-15x)

That can be factored down into 3x(x-5) so now I have 1/3 and 5 as my answers except the book didn't do it like that they factored the problem without ever trying to do all the stuff it taught me in the chapter, they just straight factored it which isn't even taught in this book. I don't know why, and I did everything properly I think so I should still get the same answer, this just feels all random. I can do the steps right but get a different answer than the book because they randomly decided to do it with a different methon, one not taught in the book.

I would probably slap the shit out of the author of this book, for being so bad at his job, if I ever saw him.

I am going to take a break from this, I already am not in the mood to deal with this. I just started one problem, did the steps to perfection, got the wrong answer, double checked all my math, same answer. I look at the solution manual and they do something that I can not reference anywhere in the book and it has no explanation. I just chalk that one up as a loss, if it ever come up on a test I will get it wrong but that is okay, can't get them all right.

I go to the next problem, same thing, except an entirely different type of bullshit procedure described above. I am done for now.

I factor 3x^3-15x-x^2+5 = 0 and get:

(3x-1)(x^2-5) = 0

then i solve for the roots.

3x-1 = 0 and
x^2 - 5 = 0

where I get

x=1/3
x=+sqrt(5)
x=-sqrt(5)

A good check for these is to plot your function with a graphing calculator. You can physically see the roots in the graph. Compare those roots with your answers. If your answers are wrong.. well you fucked up.

Remember when you factor and simplify polynomials, make SURE that what you simplified is still the same function.

Do me a favor and expand what you simplified:

(x-1/3)(3x^2-15x)

does it equal

3x^3 - x^2 - 15x +5

?

Demerdar on
• Registered User, __BANNED USERS regular
edited February 2011
I am not good at multiplying like that but I did sythetically divide it by 1/3 and got 3 2 15 so it HAS to be that doesn't it? How else would the division work?

Fizban140 on
• Registered User regular
edited February 2011
Fizban140 wrote: »
I am not good at multiplying like that but I did sythetically divide it by 1/3 and got 3 2 15 so it HAS to be that doesn't it? How else would the division work?

Did you expand it out? You need to be able to expand your polynomials. That is fundamental, and provides a good check. If you have trouble expanding the polynomial, how are you going to be good at factoring them?

Those two expressions are not equal. You did something wrong. I don't know what you did wrong, but it was fundamentally bad.

Demerdar on
• Registered User, __BANNED USERS regular
edited February 2011
I have been doing synthetic division for so long I forgot how to do long division and multiplication, guess that is one more thing on the list.

I think I am getting ahead of myself, I am going to go back to the first set of problems and try and work those and see if I can decode the books messages.

Fizban140 on
• drinking coffee in the mountain cabinRegistered User regular
edited February 2011
Wait, Fiz, I think you may have just had an arithmetic error...
Fizban140 wrote: »
I am just trying to finish my homework before class so I don't fall behind, I mean I am doing better than most people in the class. Anyways I need to figure out what the fuck is going on...

I have 3x^3 - x^2 - 15x +5 = 0

Pretty simple, I find that (x-1/3) is a factor so I get (x-1/3)(3x^2-15x)

I get (x-1/3) * (3x^2 - 15)

3x^3 divided by x is 3x^2. So if x-1/3 is a factor, we try multiplying it by 3x^2 and get 3x^3 - x^2. Subtract that from the original polynomial and you end up with -15x + 5. -15x divided by x is -15, so we multiply x - 1/3 by -15 and get -15x + 5. Subtract that from the polynomial and you get zero. So all terms in the polynomial are accounted for and we know x-1/3 times 3x^2 (the first thing we multiplied by) + -15 (the second thing we multiplied by) is the original polynomial. We can check our work by straight multiplying (x-1/3)(3x^2-15), and we should indeed get 3x^3 - x^2 - 15x + 5.

I'm not going to do the math further unless you need me to, but the eventual answer should be (x - 1/3) (3) (x + sqrt(5)) (x - sqrt(5)). Roots of +1/3, +-sqrt(5), all multiplicity one. Same answers as Demerdar got doing it a different way. You are able to do it the way the book taught and get the same answer, you just had an arithmetic error. When you double check your math, do you do it forwards (doing the problem over again in the same way) or backwards (take your answer and do the reverse of whatever you did to it, to see if you get back to the original)? I find checking my work forwards to be of little help, because I'll make the same mistake again. Checking backwards gives me better results.

Powerpuppies on
• Registered User, __BANNED USERS regular
edited February 2011
I am going to actually restart this whole chapter and go back to the start, where I still have problems with the math.
Nevermind I got it finally.

Fizban140 on
• drinking coffee in the mountain cabinRegistered User regular
edited February 2011
x^3 + 2x^2 -5x -6
-
(x-1)(x^2) = x^3 -x^2
=
3x^2-5x-6
-
(x-1)(3x) = 3x^2 - 3x
=
-2x-6
-
(x-1)(-2) = -2x+2
=
4

I don't see x-1 being a factor of the original polynomial. To check my work backwards, I multiply (x-1)(x^2+3x-2) = x^3 + 3x^2 -2x - x^2 - 3x + 2 = x^3 + 2x^2-5x+2. Original polynomial was +6. If x-1 is a factor of (expr) + 2, it won't be a factor of (expr) +6, since it isn't a factor of 4.

Arithmetic errors are hurting you, man. Can you try to be more systematic? In the past when I have taught math arithmetic errors tended to be the result of skipping steps and going too fast.

Powerpuppies on
• drinking coffee in the mountain cabinRegistered User regular
edited February 2011
x^3 + 2x^2 -5x -6
-
(x-1)(x^2) = x^3 -x^2
=
3x^2-5x-6
-
(x-1)(3x) = 3x^2 - 3x
=
-2x-6
-
(x-1)(-2) = -2x+2
=
4

I don't see x-1 being a factor of the original polynomial. To check my work backwards, I multiply (x-1)(x^2+3x-2) = x^3 + 3x^2 -2x - x^2 - 3x + 2 = x^3 + 2x^2-5x+2. Original polynomial was +6. If x-1 is a factor of (expr) + 2, it won't be a factor of (expr) +6, since it isn't a factor of 4.

Arithmetic errors are hurting you, man. Can you try to be more systematic? In the past when I have taught math arithmetic errors tended to be the result of skipping steps and going too fast.

edit: beaten. 4 minutes too slow . I'll leave it up on the off chance it helps somehow.

Powerpuppies on
• Registered User, __BANNED USERS regular
edited February 2011
Yeah I am going too fast because I absolutely hate doing math, math is basically in exercise in seeing how horrible you can be and in how many different ways.

Alright so I am doing okay, did two problems in a row then I tried the third and I can't get it at all. I just did the math about 4 times, unfortunately it was correct every time.

x^4 + x^3 - 3x^2 -x +2

I find that x-1 is a factor

thus (this is where the magic happens)

(x-1)(x^3+2x^2-x-2) from that I get x^2(x+2) -1(x+2)

So I guess the answers are 1, -2 and square root of 1 plus or minus. That is wrong but I can't figure out how to get their answer of -1 multiplicity 2.

Fizban140 on
• drinking coffee in the mountain cabinRegistered User regular
edited February 2011
the square root of 1 is 1, so you have 1, -2, 1, -1. I think you're correct. Lemme do it on paper and I'll come back in a few minutes.

edit: every which way I figure it the roots I find are 1 multiplicity 2, -2 multiplicity 1, and -1 multiplicity 1. Are you sure you copied the problem and the book's answer down correctly?

Powerpuppies on
• Registered User, __BANNED USERS regular
edited February 2011
x^2 - 4 = 0 has no real solutions....why isn't it 2?

x^2 = 4
x= 2

Fizban140 on
• drinking coffee in the mountain cabinRegistered User regular
edited February 2011
It is.

What's the greater context of the problem?

edit: I mean, it's +- 2, not just +2, but no solution is obviously wrong

x^2+4=0 has no real solutions... sign error?

Powerpuppies on
• Registered User regular
edited February 2011
I just want to chime in here and say that you should check out PatrickJMT(dot)com, or find his videos on youtube (his own site is much easier to navigate). I would also recommend Paul's Online Math Notes. Both of these have been invaluable to me in my calculus classes. Both sites have material suitable for all levels, and Patrick even does some discrete math and physics, if that's your kind of thing.

Maxim Tomato on
• Registered User regular
edited February 2011
Fizban, I recommended as much in the other thread, but you need to be making use of your professor's office hours. They are free, and the professor maintains those hours to help students understand the concepts covered (or assumed) in class. Work the problem as far as you can and then take it to the prof and say "I don't think this is right, and these are the steps I did. Can you show me where I went wrong and what I should have done?"

Also, like I mentioned before, most math texts have example problems in each chapter that are worked and have the work shown. There are also usually practice problems with answers listed in the back (only the odd-numbered problems in my experience). These are extremely valuable. Work problems with known answers until you get the same answer and know why. If you can't figure those out either, show _those_ to your professor and ask for help.

H/A is great, but help with math really needs to be a one-on-one and in-person thing. We can't see when something clicks for you - your professor or a tutor can.

jclast on
• Registered User, __BANNED USERS regular
edited February 2011
Yeah I do go to his office hours but it doesn't help much to be honest, he just explains things in a way that assumes I have a masters in mathematical theory.

Anyways I am stuck on a problem, I got it down to

(x-1/2)(2x^3-18x^+48x-40)

I decided to break it up and factor it to reduce the amount of steps and I get

2x^2(x-9) 8(6x-5) that is wrong, but I don't know why, the book took out 2 and went from there, why do I have to do that? I don't understand why I am getting a wrong answer by doing the correct, but different from the book, steps. I get 1/2, 9 and 5 but the book gets 2 instead of 9.

Fizban140 on
• Nah Registered User regular
edited February 2011
You take out 2 because each term of the polynomial has 2 as a factor: 2, 18, 48, 40 become 1, 9, 24, 20

You can't factor out 2x because only the first three terms have 'x' in them - how can you factor 2x out of 40?

Usagi on
• Registered User regular
edited February 2011
You obviously don't know what factoring is. Factoring is like "undo-ing" the distributive property. Which means if you expand your factored form back out it should match the original problem.

simple example-

factor 3x^2 + 18x
by GCF = 3x(x+6)

If you use the distributive property on 3x(x+6) you get 3x*x + 3x*6 = 3x^2 + 18x and are back where you started.

If you expand your factored form it's not what you started with, so it's wrong.

musanman on
• Registered User, __BANNED USERS regular
edited February 2011
Usagi wrote: »
You take out 2 because each term of the polynomial has 2 as a factor: 2, 18, 48, 40 become 1, 9, 24, 20

You can't factor out 2x because only the first three terms have 'x' in them - how can you factor 2x out of 40?

I didn't take x out of 40, I took it out of the first two terms, 2x^2-18x

Is there some abstract and complicated math rule I am not seeing here? The math works, but the answer is wrong.

Fizban140 on
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