Question

how you could use Descartes' rule and the Fundamental Theorem to predict the number of complex roots to a polynomial. Please include at least two examples to help clarify the explanation.

Answer 1

descartes law of signs, I assume you should have read it by now
the so-called fundamental theorem, only says,
the amount of roots you'd get, is the same as the degree of a function
so
\(x^2\) +.... degree is 2, 2 roots
\(x^7\)+..... degree is 7, 7 roots
and so on

Answer 2

and then you use Descartes Law of Signs, as we did in the previous one, to find
how many "real positive" ones
how many "real negative" ones
and the complex ones, pick up any slack left from the real ones

so if you end up with say
a 7 degree function, so 7 roots
you have say 3 real positive, and 2 negative, so the slack will be 2, so 2 complex ones

Answer 3

So can I begin by stating "You can use Descarte's Rule of Signs to find the maximum number of positive and negative real zeros to a polynomial. Then, you can use Fundamental Theorem of Algebra to find the maximum number of total zeros (real and complex). Then, just subtract the number of real zeros from the total zeros to obtain the complex zeros."

Answer 4

yes

Answer 5

is that basically Descarte's Rule?

Answer 6

what are some examples that i can use to show that this is true?

Answer 7

http://www.youtube.com/watch?v=MVX2vYlc24k

Answer 8

f (x) = x^5 + x^4 + 4x^3 + 3x^2 + x + 1.

Answer 9

x^5 + x^4 + 4x^3 + 3x^2 + x + 1.
+ + + + + +
no no no no no

there was no change when hopping from one sign to the next
that means, no real positive roots
now to test for negative ones

f(-x)
- x^5 + x^4 - 4x^3 + 3x^2 - x + 1.
- + - + - +
yes yes yes yes yes

so, there were 5 changes, when hopping from sign to sign in f(-x)
that means, there are
5
5-2
5-2-2
real negative roots

Answer 10

that is
5
OR
5-2
OR
5-2-2