Question

Please help me!!

Answer 1

@Michele_Laino if your not busy could you help me?

Answer 2

hint:
part a)
please you have to look at the degree of the polynomial

Answer 3

the degree of the polynomial is \(3\), am I right?

Answer 4

yes thats exactly what i was going to say

Answer 5

so we have three roots

Answer 6

okay

Answer 7

part b)
here you have to replace \(x\) with \(-x\), namely:
\[f\left( { - x} \right) = 5 \cdot {\left( { - x} \right)^3} + 8 \cdot {\left( { - x} \right)^2} - 4 \cdot \left( { - x} \right) + 3 = ...?\]
please simplify

Answer 8

5*(-x)^5+8+4+3

Answer 9

hint:
we can write this:
\[\Large \begin{gathered}
{\left( { - x} \right)^3} = \left( { - x} \right) \cdot \left( { - x} \right) \cdot \left( { - x} \right) = - {x^3} \hfill \\
\hfill \\
{\left( { - x} \right)^2} = \left( { - x} \right) \cdot \left( { - x} \right) = {x^2} \hfill \\
\end{gathered} \]

Answer 10

oh okay

Answer 11

so, after a substitution, we get:
\[\Large \begin{gathered}
f\left( { - x} \right) = 5 \cdot {\left( { - x} \right)^3} + 8 \cdot {\left( { - x} \right)^2} - 4 \cdot \left( { - x} \right) + 3 = \hfill \\
\hfill \\
= - 5{x^3} + 8{x^2} + 4x + 3 \hfill \\
\end{gathered} \]
am I right?

Answer 12

oh okay i see what you did, yes.

Answer 13

so thats the answer for b?

Answer 14

yes!

Answer 15

okay lets do C

Answer 16

in order to apply the rule of cartesio, we have to establish the numbers of variation and of permanence of the sign which occur when we go from one term and the subsequent term

Answer 17

okay

Answer 18

for example:

Answer 19

ok

Answer 20

going from the first term to the second term the sign is unchanged, so we have a permanence
whereas going from the second term to the third term the sign changes from + to -
so we have one variation
similarly going to the third term to the fourth term we have another sign change: from - to +, so we have another variation.
Summarizing, we got 1 permanence and 2 variations

Answer 21

now, the Descartes rule, says that we can have one positive root for each variation and one negative root for each permanence

Answer 22

oh

Answer 23

so, we have one permanence which means one negative root, and we have two variation, which means two positive roots, at maximum

Answer 24

okay wait im confused how do i fill out this chart?

Answer 25

please wait a moment, I'm thinking...

Answer 26

okay (:

Answer 27

we have to consider the number of variation of f(x).
Such number is \(n=2\), so, using the Descartes rule, we can say that the number of positive roots is \(0,\) or \(2\)

Answer 28

okay

Answer 29

now, we have to consider the polynomial \(f(-x)\), which is:
\(\Large f(-x)=-5x^3-8x^2+4x-3\)
again the number of variations is \(2\), that means we have \(0\) or \(2\) negative roots

Answer 30

so do i put 2 in the negative real zeros?

Answer 31

please wait, I have made an error, here is the right formula:
\[\Large f\left( { - x} \right) = - 5{x^3} + 8{x^2} + 4x + 3\]
so we have only one variation, which means that we have only one negative root

Answer 32

okay
oh

Answer 33

so we have the subsequent cases:
-first case: \(0\) positive roots, \(1\) negative roots, and \(2\) imaginary roots

Answer 34

second case:
\(2\) positive roots, \(0\) negative roots, and \(1\) imaginary roots

Answer 35

third case:
\(2\) positive roots, \(1\) negative root, and \(0\) imaginary roots

Answer 36

ok

Answer 37

here is how to fill the table:

please continue