Question

Following the example below, find the:

a. Possible values of the real root
b. Possible combinations for the nature of the roots
For:
P(x) = 3x5 - x3 + x2 + 2x - 1

Answer 1

please help!

Answer 2

1 real root and 4 complex
real root between 0 and 1

Answer 3

coefficient of 1st term ---------- 3
its factors ----------------1,3

coefficients of last term --------------------- 1
its factors ------------------ 1

+/- (1,3)/(1)

=+/- (1)/(1) ,+/- (3)/(1)
=1,-1,3,-3

possible roots

Answer 4

possible real roots

Answer 5

actually, not to interrupt, but those are the possible "rational roots" not real ones

Answer 6

you have a polynomial of odd degree, therefore you are guaranteed one real root. my guess is you are supposed to use descartes' rule of sign to determine the number of possible positive and negative real roots.

Answer 7

you have 3 "changes is sign" for
\[P(x) = 3x^5 - x^3 + x^2 + 2x - 1\] is it clear what that means?

Answer 8

Yes that how it look like !!!..

Answer 9

the coefficients are : 3, -1 , 1 , 2 , -1
and the change of sign are from 3 to -1 is one change, from -1 to 1 is one change, and from 2 to -1 is one change of sign. for a total of 3 changes of sign. so you can have either 3 positive zeros, or 1 positive zero

Answer 10

then look at the number of changes of sign of
\[P(-x) = 3(-x)^5 -(- x)^3 +(- x)^2 + 2(-x) - 1=-3x^5+x^3+x^2-2x-1\]

Answer 11

the coefficients are : -3, 1, 1, -2, -1 so you have two changes of sign
from -3 to 1 is one change and from 1 to -2 is another. therefore the possible number of negative zeros are : 2 negative zeros or no negative zeros

Answer 12

in summary,
number of POSSIBLE positive zeros are 3 or 1,
number of POSSIBLE negative zeros are 2 or 0

there are 5 zeros all together so you have lots of possibilities.
3 positive, 2 negative
3 positive, 0 negative, 2 complex
1 positive 2 negative, 2 complex
1 positive 0 negative, 4 complex

Answer 13

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