Factoring a 4th degree polynomial

[mbrmbrg]

I have the equation

$$f''(t) = 0 =\frac{\ -6t-4t^4+2t^5}{(1+2t)^4}$$

which I factored to

$$0 =\frac{ (2t)(-3-2t^3+t^4)}{(1+2t)^4}$$

and set

$$2t = 0$$ or $$t^4-2t^3-3 = 0$$

How am I supposed to factor the second possibility for t? Using guess-and-check () with 0, 1, and -1, I found that t is probably a fraction between -1 and 1. And now I'm stuck for real, because no way am I guess-and-checking random fractions.

Help, please!

 

[mbrmbrg]

sorry, having latex trouble.

My second step should read 0=[(2t)(-3 - 2t^3 +t^4)]/(1+t^2)^4
but I don't think it does, and I can't figure out how to fix the code.
I'm blaming my computer display

 

[shmoe]

All rational roots can be found by the rational root theorem, you don't have to check many:

http://planetmath.org/encyclopedia/RationalRootTheorem.html

In any case, when you tried to put in t=-1, you should have gotten 0, no? So you can reduce it to a cubic at least.

hmm, $$0 =\frac{\ (2t)(-3-2t^3+t^4)}{(1+2t)^4}$$

odd, it's a cut and paste from yours.

 

[0rthodontist]

The latex doesn't work because you have an extra \ after \frac{

 

[mbrmbrg]

Ouch. Another good reason not to guess-and-check: my algebra tends to be better than my arithmetic. Oh, ouch.

Ummm, I looked at the rational root theorem, but I don't know if I'm using it right.
Does it say I have some root p/q, where my first coefficient (here 1) is divisible by p and my last coefficient (here -3) is divisible by q? So my answer is -1/3??
OK, apparently not, because that doesn't give me zero.

In either case, it looks like any root will require t to be negative, and in this case, t is time...

But now I'm curious how this rational root theorem works.

 

[Office_Shredder]

The rational root theorem:

Any rational root of a polynomial will be in the form p/q, where p is a factor of the constant term, and q is a factor of the leading coefficient (so in this case, p is +/-3, 1, and q is +/-1).

So your best bets would be to check 3, -3, 1, and -1

 

[shmoe]

The latex doesn't work because you have an extra \ after \frac{

That should just put a space in though? I don't think I changed anything in my cut and paste. Not important though.



mbrmbrg, you have the rational root theorem reversed, if p/q is a root then q divides your lead coefficient, here a 1, and p divides your constant coefficient, here -3. So you must have q=+ or -1 (so any rational root must also be an integer), and p=+/- 1 or 3. So to find all rational roots, just stick -1, 1, -3, and 3 into your equation. These are not guaranteed to be roots, but any rational root *must* be one of these.

You must have a positive real roots though. You polynomial is -3 at t=0 and greater than 0 when t is large and positive, so there's at least one root in there somewhere.

 

[mbrmbrg]

OK, got it now--thanks people!