Synthetic Division when all roots are imaginary

[questionposter]

Ok, so I'm doing synthetic division for a polynomial and I have to show all my work, but no matter what number I use from the ration root theorem, nothing works, which means all the 0s are imaginary, but then how do I find out exactly which imaginary numbers there are? Because I couldn't find a "quartic" formula and I'm not always dealing with a polynomial where there's an automatic formula for it like a quadratic formula, so is there still some quick way I can do this? Preferably using synthetic division or should I just say in an equation like that where the highest degree is in x^n, there's n "i"s?

Edited October 10, 2011 by questionposter

[DrRocket]

Ok, so I'm doing synthetic division for a polynomial and I have to show all my work, but no matter what number I use from the ration root theorem, nothing works, which means all the 0s are imaginary, but then how do I find out exactly which imaginary numbers there are? Because I couldn't find a "quartic" formula and I'm not always dealing with a polynomial where there's an automatic formula for it like a quadratic formula, so is there still some quick way I can do this? Preferably using synthetic division or should I just say in an equation like that where the highest degree is in x^n, there's n "i"s?

 

The fact that some polynomial has no rational roots does not mean that it has no real roots. E.g. [math]X^2-2[/math]

[questionposter]

The fact that some polynomial has no rational roots does not mean that it has no real roots. E.g. [math]X^2-2[/math]

 

Yet every time I've graphed an equation like the one I'm saying, they never crossed the x axis. I mean I can just do the constant and lead coefficient divided by x as separate equations to get a list of factors, and nothing seems to work, and there's only like 3 plusorminus whole numbers and they don't work, so what am I suppose to do with that? There's literally infinite possibilities for p/q and I really don't have a lot of time to test that many.

 

Here's the equation. Maybe there's some step I'm missing, but synthetic division to get roots doesn't seem to work when using the rational root theorem. Isn't the rational root theorem and synthetic division suppose to work for complex numbers too? In fact I've gotten complex numbers before, but I've never used this process when all numbers were imaginary or complex, so what am I doing wrong by using the rational root theorem and synthetic division to find the factors?

 

"x^4+0x^3+10x^2+0x+9"

 

The original equation doesn't have the 0 Xs, but you need them for any polynomial division.

 

 

And what's even stranger is I can just factor it by eye and see that its also (x^2+9)(x^2+1), yet stuff still doesn't work with the rational root theorem and synthetic division.

Edited October 10, 2011 by questionposter

[DrRocket]

 

 

And what's even stranger is I can just factor it by eye and see that its also (x^2+9)(x^2+1), yet stuff still doesn't work with the rational root theorem and synthetic division.

 

So you can see what the roots are. Since none of them are rational numbers, why would you expect any simplification from the rational root theorem ?

 

If you plug in the obvious roots, which are imaginary, you should get a simplification using synthetic division, or ordinary polynomial division.

 

Remember that synthetic division only works for factors that are first degree polynomials, so you would have to use complex numbers -- you cannot divide synthetically by x^2+9 or x^2+1.

[mathematic]

So you can see what the roots are. Since none of them are rational numbers, why would you expect any simplification from the rational root theorem ?

 

If you plug in the obvious roots, which are imaginary, you should get a simplification using synthetic division, or ordinary polynomial division.

 

Remember that synthetic division only works for factors that are first degree polynomials, so you would have to use complex numbers -- you cannot divide synthetically by x^2+9 or x^2+1.

Your last statement is not true. If 3i and -3i are roots, you can divide by x^2 + 9.

[questionposter]

So you can see what the roots are. Since none of them are rational numbers, why would you expect any simplification from the rational root theorem ?

 

If you plug in the obvious roots, which are imaginary, you should get a simplification using synthetic division, or ordinary polynomial division.

 

Remember that synthetic division only works for factors that are first degree polynomials, so you would have to use complex numbers -- you cannot divide synthetically by x^2+9 or x^2+1.

 

Yeah I suppose I'll just have to factor by grouping or use the quadratic formula if I can break it down into second degree factors

Edited October 11, 2011 by questionposter

[DrRocket]

Your last statement is not true. If 3i and -3i are roots, you can divide by x^2 + 9.

 

Wrong. Read my post again.

 

You can ndivide by x^2+9 but you cannot do it synthetically. Synthetic division applies only to linear factors. To divide by X^2+9 you must divide by x-3i and x+3i sequentially.

 

Don't tell your grandmother how to sucfk eggs.

Edited October 11, 2011 by DrRocket

[mathematic]

Wrong. Read my post again.

 

You can ndivide by x^2+9 but you cannot do it synthetically. Synthetic division applies only to linear factors. To divide by X^2+9 you must divide by x-3i and x+3i sequentially.

 

Don't tell your grandmother how to sucfk eggs.

 

Maybe you and I have a different concept of synthetic division. Any polynomial can be divided by a polynomial of equal or lower degree and get a polynomial quotient and a remainder. When the remainder = 0, then there is division.

 

P.S. I looked up the definition, via Google.

http://en.wikipedia.org/wiki/Synthetic_division

Your definition is for regular synthetic division. Mine is for expanded synthetic division.

Edited October 11, 2011 by mathematic

[questionposter]

Maybe you and I have a different concept of synthetic division. Any polynomial can be divided by a polynomial of equal or lower degree and get a polynomial quotient and a remainder. When the remainder = 0, then there is division.

 

P.S. I looked up the definition, via Google.

http://en.wikipedia....thetic_division

Your definition is for regular synthetic division. Mine is for expanded synthetic division.

 

Synthetic division doesn't really work to get real roots because you need to use the rational root theorem in order to figure out what factors there are I guess. It kinda sucks, but the formulas for polynomials can still be used when you break the polynomial down into those types of products.

[questionposter]

But wait, what do I do if I have something like

 

x^6+x+57? I can't use factoring of any kind, I can't use rational root theorem and synthetic division, and I don't know of a sextic formula, and say I don't have a graphing calculator, now what?

Say I'm stranded on the island and I know the fuel in something decreases at a rate of x^6+x+57 but I have a big helicopter with full fuel, how do I figure out how far it could take me before the fuel runs to 0? I'm only saying this in case someone's not going to bother if they think x^6 never occurs in reality and therefore I shouldn't worry about it.

Edited October 12, 2011 by questionposter

[imatfaal]

The real mathematicians will have a better answer for you - but I do sometimes get pretty vile equations in my line of work, and most of the time you can simplify or make assumptions. for f(x) = x^6+x+57 you immediately know that no +ve real can give f(x) = 0, x^6 is always +ve so any real roots must be both -ve and damn close to zero, you can differentiate to find the turning point (gives you a quintic but one that is easy to solve), if the value of f(x) is positive at the minima then you aint gonna have any real roots - and if all else fails I think you can use the Newton-raphson method to estimate the result once you are sure they do have real roots.