Question

Please Help


How could you use Descartes' rule and the Fundamental Theorem of Algebra to predict the number of complex roots to a polynomial as well as find the number of possible positive and negative real roots to a polynomial?

Your response must include:

A summary of Descartes' rule and the Fundamental Theorem of Algebra. This must be in your own words.
Two examples of the process
Provide two polynomials and predict the number of complex roots for each.
You must explain how you found the number of complex roots for each.

Answer 1

You can use Descarte's Rule of Signs to find the maximum number of positive and negative real zeros to a polynomial. The Fundamental Theorem of Algebra can be use to find the maximum number of total zeros. Then subtract the number of real zeros from the total zeros to obtain the complex zeros.

I dont know what else to do.

Answer 2

you could start by checking your books for both

https://www.youtube.com/watch?v=5YAmwfT3Esc <-- descartes rule of signs

http://www.mathsisfun.com/algebra/fundamental-theorem-algebra.html <-- fundamental theorem of algebra

Answer 3

You can use Descarte's Rule of Signs to find the maximum number of positive and negative real zeros to a polynomial. The Fundamental Theorem of Algebra can be use to find the maximum number of total zeros. Then subtract the number of real zeros from the total zeros to obtain the complex zeros. For example, taking the equation 7(x^5)-3(x^2)-9x+2. This equation has a total of two sign changes. This means the real postive roots are 2,0. Another example is -4(x^5)-5(x^2)+8x+2, There are 1 sign change in this getting a negative real root, and the only negative real root is 1.

Answer 4

Is this good?

Answer 5

@jdoe0001

Answer 6

I hope I answered this right...

Answer 7

looks close

Answer 8

from a polynomial, you'd get both, positive and negative roots
so you find both, positive ones and negative ones

then add up their possible combinations,a nd whatever is left, from the fundamental theorem, is a complex one

Answer 9

let us take your example... one sec

Answer 10

\(f(x)=7x^5-3x^2-9x+2\impliedby
\begin{cases}
\textit{2 changes}\\
\textit{positive ones}
\\\hline\\
2,0
\end{cases}
\\ \quad \\
f(-x)=-7x^5-3x^2+9x+2\impliedby
\begin{cases}
\textit{1 change}\\
\textit{negative ones}
\\\hline\\
1
\end{cases}\)

Answer 11

the polynomial degree is 5
so we have either
2 positive, 1 negative
but the degree is 5, so the roots are also 5, so 2 are missing
and those are our complex ones

2 positive, 1 negative, 2 complex 2 + 1 + 2

or

0 positive, 1 negative
5 roots, thus, 4 complex
0 positive, 1 negative, 4 complex 0 +1 + 4

Answer 12

keeping in mind that, complex root, involve a radical solution, and thus usually come in pairs
so, you'd end up getting 2 or 4 or 6 or 8 or so, but an "even" amount of complex

Answer 13

Okay

Answer 14

but as you see, we use the fundamental theorem, to get that there are 5 roots
subtract the positive ones and negative ones, and what's left is complex