Question

Some sort of algebra having imaginary roots. -___- Can someone help?

Answer 1

\[m^6 + 9m^4 +24m +16 = 0\]

Answer 2

How can this be factored out? I mean, synthetic division is not applicable here now.

Answer 3

I'm not 100% positive, but i think its not factorable. the most u can do is m^4(m^2+9)+8(3m+2)

Answer 4

you can factor it by plugging numbers in to get a sign change on y
if you plug in 3 and get 2
and you plug in 4 and get -2 you know there is a zero between 3 and 4
then I would plug in 3.5 since they're inverse signs and see what I got getting closer and closer to zero until I got out enough decimels of the x value to get zero

Answer 5

interesting means described by @lonnie455rich (a variant of the bisection method I'm assuming), but I think we need to analyze this problem a bit more.

Answer 6

Honestly, I don't have idea of what you are talking about. Come leggit I wanna solve this!

Answer 7

Can you give us the context of this problem?

Answer 8

graph (3,2) and (4,-2) . there is a zero between these two points on the graph. so it would suggest that between 3 and 4 on the x axis. there is a zero on the y axis. does that make it any more clear? and as far as the other way to examine this, he will have to explain. im curious as well

Answer 9

@lonnie455rich , are you talking to me for another way?

Answer 10

why do we need to graph (3,2) and (4, -2) ??

Answer 11

and @Yttrium , please give the context of the problem and the current math class (just hte name) you are enrolled in...

Answer 12

Differential equations.

Answer 13

yes. I would like to know as well.
because you're plugging 3 in to get 2 and 4 in to get -2. so since there is a sign change there must be a zero between 3 and 4. 3.xxxxxx will be a zero

Answer 14

Okay. It's originally a differential equation of y having D as its differential factor.. It's equation is \[(D^6 + 9D^4 + 24D^2 +16)y=0\]
And we need to find its solution.

Answer 15

even wolfram alpha gives no closed form solution, so I am assuming either you are expected to have the tools to find the roots, or you have made a mistake somewhere.

Answer 16

It's not like the sixth and fifth degree polynomials have general closed form solutions without extensions of radical definitions...

Answer 17

I believe I haven't had a mistake so, far. All I know about this is that it will give me an imaginary solution.

Answer 18

You get a few more than just one imaginary solution.

Answer 19

@ganeshie8 what can you say about this?

Answer 20

I mean *imaginary solutions

Answer 21

@Yttrium was this the original problem?

Answer 22

http://www.wolframalpha.com/input/?i=zeros+of+m%5E6%2B9m%5E4%2B24m%2B16%3D0
this is what I got on wolfram completely useless for finding the answer to your question though
I don't know how to get the imaginary solutions without the quadratic formula unless there is a way to make this into 3 quadratics to solve for them

Answer 23

@lonnie455rich the abel ruffini theorem says there aren't general closed form solutions for polynomials of degree 5 or higher. I don't know if this is a class of sixth degree polynomial that is solvable, but I think something is not right here.

Answer 24

well I guess I could foil those zeros that wolfram got and see if I get the original polynomial. be back. so.. many.. decimels.. lol

Answer 25

Oh I see. Among all other samples, it is the only one with a power of six. Why don't we try to form quadratics from this equation?

Answer 26

@lonnie455rich , not only are those roots irrational, they probably aren't expressible as a finite operation of nested nth degree roots.

@Yttrium , I think reposting the problem with the ORIGINAL question would help a lot more. To give you reference to why solving polynomials with closed form solutions proves to be so difficult, read up on http://en.wikipedia.org/wiki/Abel%E2%80%93Ruffini_theorem . The equation you have given us is not expressible as a simple quadratic, and I don't think working with the sum of quadratics would get you very far.

Answer 27

I have already given the original question.
It's just:
Find the solution of the following differential equations:
\[(D^6+9D^4+24D^2+16)y = 0\]

Answer 28

got it off google

Answer 29

try factor D^2+1 with polynomial long division; should yield conclusive results

Answer 30

WOW! I GOT THIS. The roots then are m
\[m = \pm i, \pm 2i, \pm 2i\]

Answer 31

But that was not taught in our algebra subject. :/

Answer 32

@Yttrium , it's one of those knowledge gaps. I myself don't recall learning this material

Answer 33

I want to know more about those knowledge gaps. :(