Question

HELP!!!! descartes rule
f(x)=x^2+3x+5

Answer 1

@.Sam. @ajprincess @Eyad @RadEn

Answer 2

remind me of descartes rule

Answer 3

It was determined, by the Fundamental Theorem of Algebra, that there are five values of x that make f(x) = x5 + 2x4 – 4x3 – 4x2 – 5x – 6 equal to zero. These values for x are also known as the zeros of the function. Using the Rational Root Theorem and synthetic division, 2, –1 and –3 were found to be three of the five zeros of the function. Notice that these three zeros, which are also known as roots, are the x-intercepts of the graph of this function.

Answer 4

@perl

Answer 5

ok thanks , one moment

Answer 6

@perl, your avatar is in extremely poor taste.

Answer 7

dont like it?

Answer 8

Not even a little.

Answer 9

sammy, what are the directions to this problem?

Answer 10

using the descartes rule, solve

Answer 11

if you dont understand this one, im also stuck with this one...
find possible ration root
f(x)=2x^3+3x^2-x+4

Answer 12

ok one sec , i gotta update my pic

Answer 13

ok

Answer 14

the Rational Root Theorem gives you information about how many rational roots there could be.

Descartes' Rule of Signs gives you information about how many positive or negative roots there could be (which may or may not be rational).

for example, if p(x) = x^2 + 1, the rational root test tells us that the only rational roots (if they exist) will be -1, and 1.

applying the rule of signs, we find that there are no positive roots (because there are no sign changes for p(x)), and there are no negative roots (because there are no sign changes for p(-x), which happens to equal p(x) in this case).

so as you can see, the two tests can yield different results.

Answer 15

f(x) = x5 + 2x4 – 4x3 – 4x2 – 5x – 6

Answer 16

you have 1 sign change, so there is exactly only 1 positive real root

Answer 17

so how do i do my problem? @perl

Answer 18

then plug -x
f(-x) = (-x)^5 + 2(-x)^4 - 4(-x)^3 - 5(-x) - 6

Answer 19

f(-x) = (-x)^5 + 2(-x)^4 - 4(-x)^3 - 5(-x) - 6 = -x^5 + 2x^4 +4x^3 + 5x - 6

Answer 20

now there are 2 sign changes, so there are 2 or zero negative roots. since you found 2 negative roots using synthetic division, so we have a total of 3 roots (and 2 non real roots)

Answer 21

which problem are you solving for??? @perl

Answer 22

oh your 2nd example

Answer 23

and thats just the final answer? @perl

Answer 24

f(x) has no positive real roots,
but f(-x) = (-x)^2 +3(-x)+5 = x^2 -3x + 5 has either two negative roots or 0 negative roots

Answer 25

what about the original problem i posted? @perl

Answer 26

well descartes rule says there are either 0 negative real roots or 2 negative real roots

Answer 27

and i just checked usin quadratic formula, there are no real roots

Answer 28

descartes sign rule says there are no positive real roots, and there are either two or zero negative real roots