Question

x³-3x+4
How do you Descartes' Rule and the Fundamental Theorem of Algebra to determine the zeros(complex and real)?

Answer 1

@amistre64

Answer 2

descartes i believe looks at sign changes for f(x) and f(-x) to determine the number of possible roots; but it doesnt say anythiing about the value of those roots

Answer 3

the fundamental thrm of algebra simply says that a poly has as many roots as its degree, but it doesnt say what the value of those roots are either

Answer 4

I know Descartes' Rule.. and f(x) sign changes tells me the maximum numbers of positive roots. f(-x) tells me the max number of negative roots.

Answer 5

But I don't understand Fund. The. of Alg. But, I have to write about how to find complex roots using those two terms.

Answer 6

and those maxs can be altered by 2s to determine complex roots

Answer 7

I don't know how to find complex roots.

Answer 8

spose you find 3 sign changes in f(x); then there are at most 3 positive roots, or at least 1 positive root if we consider complexes; 3real - 2complex = 1 real

Answer 9

Okay, so this equation will have only 2 complex roots?

Answer 10

its not a gaurentee, just a means of considering possibilities

Answer 11

Okay. How do I know what the roots are for this polynomial?

Answer 12

x^3 - 3x + 4
^ ^
2 sign changes; max of 2 positive roots, or 0 positive roots

2 real roots - 2 complex roots = 0 real roots

Answer 13

the only way to find the roots is to work thru the problem using other methods not descrbed in the context of the question

Answer 14

Oh.. well what method would I use to solve for complex roots?

Answer 15

we can use the rational root thrm to determine of it has ANY real roots, and that would allow us to factor the poly into more workable forms

Answer 16

Mk, how do you do that?

Answer 17

I'm sorry... I'm lost.

Answer 18

its a modified trial and error process :) you narrow down the option to a set of choices that have the best chance of being a real root; if it is none of those options, then it has no real roots

Answer 19

So, should I try -1,1,-4, and 4?

Answer 20

\[\pm\frac{factors~last\#}{factors~first\#}\]

Answer 21

hmm: i would try +- 1,2,4

Answer 22

Oh, I forgot two. :P and okay, after I figure out which works, how do I determine the complex roots?

Answer 23

with any luck, it reduces to a prime poly that can be manipulated

im not sure, but we might be able to include complex factors of the first and last to add to the options list; but ive never really tried to do that

Answer 24

lets try out our reals :)

Answer 25

Okay :)

Answer 26

Can use synthetic division?

Answer 27

x^3-3x+4 ; x=1

1-3+4 not = 0

x=2
8-6+4; not 0

x=4
64-12+4; not 0

no positive real roots; try the negatives

Answer 28

we could use synthetic if need be, but a calculator works fine if you can use one :)

Answer 29

None of them work.

Answer 30

The negatives.

Answer 31

so it doesnt have rational roots; but it has to have at least on real root (its an odd degree) .... so it would have to be an irrational number

Answer 32

Okay, do I just figure that out by guess & check? Or is there a way I can figure out the number by looking at the equation?

Answer 33

there is no simple way to do this by hand
http://www.wolframalpha.com/input/?i=x%5E3-3x%2B4

notice the complicated nature of the root

Answer 34

there are calculus methods that can get us closer and closer if we required a approximate value to a certain decimal accuracy

Answer 35

I see... I am allowed to make up my own polynomial.. So could you make an easier one for me to figure out the roots for?

Answer 36

its easier to manufacture it from roots, that way you know you have something workably simple :)

how many roots would you like, how many complex and how many neg/pos ?

Answer 37

Nothing that causes tooo much work to explain in my essay. Something simple that still has one complex root and it's conjugate + a few real roots. Hehe :)

Answer 38

lets use roots: -3,-1,2, 3i, -3i

\[(x+3)(x+1)(x-2)(x-3i)(x+3i)\]\[\hspace{3em} = x^5 +2x^4+4x^3+12x^2-45x-54\]

Answer 39

That was a little backwards from what I am supposed to do. Hehe. But it works! Now I have to explain how I used Descartes' Rule and Fund. Theo. of. Alg to find those zeros. I'll just restate the definition, I guess? & add other things like Synthetic Division and whatnot.

Answer 40

that would be fine :)

Answer 41

Thanks for the help :)

Answer 42

yw, good luck ;)

Answer 43

Can you give me one more example like the one you just gave me?

Answer 44

its the same process, define a few roots, and multiply them together to get the poly

Answer 45

Ohhhhh, okay. So it's really anything I want it to be?

Answer 46

yep, anything you can dream of

roots: -1,1,+- i

(x+1)(x-1)(x+i)(x-i) = x^4-1