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Ok, going to try this again... I'd posted a similar Q once but figured it out (maybe I'll be lucky and figure this one out after posting)... anyway...

First off, this is NOT homework, or more specifically it's homework I've assigned myself so I can learn this.

I know how to remove greatest common denominators, and I know how to factor trinomials, and I know how to group to allow factoring...

In this problem:

f(x) = 2x^4 + 14x^3 + 25x^2 - 4x - 28

I know that I can group the first two and the last two and go from here:

2x^3(x + 7) + 25x^2 - 4(x + 7)

to here:

(2x^3 - 4)(x + 7) + 25x^2

But since the stated intent of this problem is "State all possible rational zeros" and I assume that means values for x such that the end result IS zero, I'm stumped as to how to get from my last point to THAT point...

All the things I've found about "grouping" are based on grouping 2 things together.

Understand, I'm interested in knowing HOW to do this, not the answer to this one specifically...

Thanks.

First off, this is NOT homework, or more specifically it's homework I've assigned myself so I can learn this.

I know how to remove greatest common denominators, and I know how to factor trinomials, and I know how to group to allow factoring...

In this problem:

f(x) = 2x^4 + 14x^3 + 25x^2 - 4x - 28

I know that I can group the first two and the last two and go from here:

2x^3(x + 7) + 25x^2 - 4(x + 7)

to here:

(2x^3 - 4)(x + 7) + 25x^2

But since the stated intent of this problem is "State all possible rational zeros" and I assume that means values for x such that the end result IS zero, I'm stumped as to how to get from my last point to THAT point...

All the things I've found about "grouping" are based on grouping 2 things together.

Understand, I'm interested in knowing HOW to do this, not the answer to this one specifically...

Thanks.

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Start your 7-day free trialGo through a few integers close to 0, to see if any give f(that number)=0.

I tried from -3, then -2, ...

And found that f(-2)=0. So factor out (x+2):

f(x) = (x+2)(2x³+10x²+5x-14)

Use a similar method on g(x)=2x³+10x²+5x-14, you should see again that g(-2)=0 as well. So factor that from g(x):

f(x) = (x+2)(x+2)(2x²+6x-7) = (x+2)²(2x²+6x-7)

...Can you solve it from there?

(2x^3 - 4)(x + 7) + 25x^2

namely x = -2

So, you get :

(x + 2)(2x^3 + 10x^2 + 5x - 14)

Once more, you can easily spot that x = -2 is again a zero :

(x + 2)(x + 2)(2x^2 + 6x - 7)

Then we have a quadratic which can easily be solved :

(-6 +/- sqrt(36 + 56)) / 4 = (-3 +/- sqrt(23)) / 2

So we get :

2(x + 2)(x + 2)(x + (3 + sqrt(23))/2)(x + (3 - sqrt(23))/2)

There's no real rule that always works - sometimes obvious zeros can be seen, which can help you further like in this case.

What can help you is to recognise general rules like :

a^2 - b^2 = (a - b)(a + b)

(a + b)^2 = a^2 + 2ab + b^2

(a - b)^2 = a^2 - 2ab + b^2

a^3 + b^3 = (a + b)(a^2 - ab + b^2)

a^3 - b^3 = (a - b)(a^2 + ab + b^2)

etc.

Note that a and b can be anything : x, 5, x^2, etc.

It's mostly just a question of recognising patterns ...

I've only glanced at that link so far but is it supposed to be easy to understand because it doesn't seem to be... (remember, I only glanced though)...

>InteractiveMind...

Is there a specific reason you picked the numbers to test you did or were you just trial and erroring...?

In some special cases, you can get lucky and find simple factors.

Newton-Raphson can find roots numerically

http://www.mathwords.com/r/rational_root_theorem.htm

I used trial and error. The reason for this, is that usually if you are asked to solve a polynomial f(x)=0 in an exam, for example, then they usually make the factors fairly simple, such that a quick trial and error will allow you to solve it (as I did above).

Obviously, in the real world, polynomials rarely work out so easily - in which case, ozo's general solution would be the root you'd take..

It's partly guessing, partly deducing :

f(x) = 2x^4 + 14x^3 + 25x^2 - 4x - 28

Since this is an exercise, there is most likely an easy solution, so since we can't spot a grouping at first sight, we'll start trying integer values.

Notice that positive values don't have to be checked, since there aren't enough negative terms to "counter" the positive terms. The 0 can be easily discarded too, since we have a constant term.

So, we try -1 which doesn't work. Then we try -2 ... and that one works.

It's not merely guessing, but it comes pretty close :)

As ozo said : luck often plays an important role too.

2x^4 + 14x^3 + 25x^2 - 4x - 28

the rational root theorem tells you that you only need to try factors of 28

Thanks for the help...

4x^2 + 12xy + 9y^2

Not meaning it has to be that simple but the idea in general...

I just wasn't sure whether there was something I was missing... As I said, I was reading about grouping and thought, maybe in this case, the idea was to group one group of two and one group of three...

I'm guessing that could be the way to go sometimes but clearly this wasn't one of those times. =]

I don't suppose you could explain how that link you sent would work in this case, could you...?

I've been looking at it and it's just not making any sense whatsoever...

I mean it starts with

z^4 + a[sub3]z^3 + a[sub2]z^2 + a[sub1]z + a[sub0] = 0

then the next thing is all x's and a's with subs

and then after the next formula they're using nothing but z's with subs and exponents...

Scary because it says algebra and I thought algebra was not that advanced...

=]

Indeed ... if you're lucky, you can easily spot a grouping. This time it wasn't that obvious ... but still doable if you can spot this grouping :

(x^2 + 4 x + 4)(2x^2 + 6x - 7)

Which isn't entirely impossible :)

I'm not sure we're talking about the same thing or not...

I meant for instance having:

qa^4 + rb^3 + sc^2 + tb + u

and being able to break it into:

(qa^4 + rb^3 + tb) + (sc^2 + u)

kind of thing... and having that actually be obvious for a breakdown...

>> (qa^4 + rb^3 + tb) + (sc^2 + u)

That usually doesn't bring you a lot closer to factoring the polynomial. I say usually, because in certain cases it will help you :)

I know all of this sounds vague, but a lot of exercise can help you :)

Yeah, from what you lot are saying it's a lot like just doing normal trinomials... enough practice and some of them become almost obvious at first look...

Though I love math, it SO should not involve guessing... =]

The problem is that the alternative to "guessing" in this case involves almost brute-force like operations (which isn't really better than guessing).

If you are lucky, you can see a pattern, and use that to solve the problem without guessing ... but usually, some level of guessing is needed.

Oh well, you get it :)

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http://mathworld.wolfram.com/QuarticEquation.html