Question

Using Descartes' rule of signs, how many sign changes are there in the function:
h(x)=4x^4-5x^3+2x^2-x+5 ?

Answer 1

are 6 positive real roots possible? how many total roots must there be in this fourth-degree function?

Answer 2

Review this:

http://www.mathwords.com/d/descartes_rule_of_signs.htm

Answer 3

so according to that link, i got that there are 4 sign changes in this function... is that correct?

Answer 4

Yes, correct.

Answer 5

thanks :) so for "how many total roots must there be in this fourth-degree function" would that also be 4?

Answer 6

okay :) would that mean that there are 6 positive real roots that are possible?

Answer 7

Well, actually let me clarify. There must be 4 roots since the highest degree is 4.

Answer 8

But for positive zeroes, there can't be more than 4

Answer 9

oh i see, that makes sense!

Answer 10

You can have either zero, two, or four positive roots.

Answer 11

would zero, two, or four positive roots be possible real roots?

Answer 12

Yes, exactly.

Answer 13

If it isn't too much to ask, can you help me understand what this is asking me to do? This is the third and fourth part of this same question, but I am so confused!

Answer 14

Remember, when in doubt, graph the function. It will tell you how many zeroes for sure.

Answer 15

finding how many zeros there are will help me to find the number of possible negative real roots?

Answer 16

by replacing every x with (-x) would the new function be: 4(-x)^4-5(-x)^3+2(-x)^2-(-x)+5 causing there to be 1 sign change?

Answer 17

Yes. When you input f(-x) and count the sign changes, it will tell you the number of possible negative zeroes.

Answer 18

when i input that function it keeps telling me that it is an incomplete input

Answer 19

Just compute f(-x) manually (by hand-pencil and paper) then count the number of sign changes to get the number of possible negative zeroes

Answer 20

i keep getting 1, is that correct?

Answer 21

Graphing will give you the total actual number of zeroes, but not the total "possible zeroes".

Answer 22

Assuming you did it correctly:
f(-x) = 4(-x)^4-5(-x)^3+2(-x)^2-(-x)+5

Results in
4x^4 + 5x^3 + 2x^2 + x + 5

So no negative roots possible.

Answer 23

so in the new function there are also no sign changes?

Answer 24

What do you mean "also".

Answer 25

You got 4 sign changes for f(x) and 0 sign changes for f(-x)

Answer 26

the other question was "how many sign changes are there in the new function"

Answer 27

sorry, thats what i meant! there are 0 sign changes for f(-x)

Answer 28

in this new function of f(-x) would there be 0 possible complex roots since there are 0 possible roots and 0 sign changes?

Answer 29

I'm not really too pleased with your teacher calling f(-x) the "new function". f(x) is the function of interest. We're trying to figure out the possible real roots and negative roots of f(x).

Finding f(-x) is just a strategy used to find the number of negative roots of f(x).

Answer 30

So far there are 4, or 2, or 0 positive roots. Zero negative roots. Which means there are 2 or 4 or 6 complex roots.

Answer 31

By graphing, we can figure out exactly how many roots.

Answer 32

oh wow, that's not too good to hear, thats the way I have been learning!

Answer 33

Well, review this:

http://www.purplemath.com/modules/drofsign.htm

Answer 34

thank you so much for ALL of your help! I really appreciate it! :)