Help? Using complete sentences, describe how you would analyze the zeros of the polynomial function f(x) = –3x^5 – 8x^4 +25x^3 – 8x^2 +x – 19 using Descartes’ Rule of Signs.

- anonymous

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- jim_thompson5910

in f(x), how many sign changes are there?

- anonymous

I don't know?

- jim_thompson5910

notice that going from -8x^4 to +25x^3, there's a sign change from negative to positive

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## More answers

- jim_thompson5910

do you see this?

- anonymous

Yes

- jim_thompson5910

ok there's another from +25x^3 to -8x^2

- jim_thompson5910

what's another one?

- anonymous

+x to -19?

- jim_thompson5910

there's one more

- jim_thompson5910

from -8x^2 to +x

- jim_thompson5910

so there are 4 sign changes total in f(x)

- jim_thompson5910

this means that there are at most 4 positive real roots

- anonymous

-19 to +x?

- jim_thompson5910

no that's the same sign change (just in reverse)

- anonymous

4 sign changes=4 positive real roots

- jim_thompson5910

at most 4 (there could be 0, 1, 2, 3, or 4 positive real roots)

- jim_thompson5910

4 is the maximum

- jim_thompson5910

now we must find f(-x)

- anonymous

I'm still really confuzzled...

- jim_thompson5910

where at?

- anonymous

All of it

- jim_thompson5910

the rule is
if f(x) has n sign changes, then there are AT MOST n positive real roots

- jim_thompson5910

if n = 4, then
if f(x) has 4 sign changes, then there are AT MOST 4 positive real roots

- anonymous

Okay

- jim_thompson5910

making more sense?

- anonymous

...

- jim_thompson5910

yes? no?

- anonymous

I don't know how to formulate an answer from you telling me this...

- jim_thompson5910

you don't need to find the actual roots
you just need to be able to count the possible number and type

- jim_thompson5910

more specifically, how many positive real roots you could have (in this case, at most 4)
you don't need to find the 4 or find out how many actual positive roots there are

- anonymous

But I need a definite answer, right?

- anonymous

I can't just say "maybe 4"?

- jim_thompson5910

well that's part of the answer, we still have to find f(-x)

- anonymous

Okay, and how would we go about that ? :)

- jim_thompson5910

start with f(x)
then replace each x with -x and simplify

- anonymous

Could you give me an example?

- jim_thompson5910

start with this
f(x) = –3x^5 – 8x^4 +25x^3 – 8x^2 +x – 19
then replace each 'x' with '-x' to get
f(-x) = –3(-x)^5 – 8(-x)^4 +25(-x)^3 – 8(-x)^2 +(-x) – 19
then simplify to get
???

- jim_thompson5910

does that help?

- anonymous

Yes, it does :)

- jim_thompson5910

ok what do you get

- jim_thompson5910

when you simplify

- anonymous

3x^5=f(-x)+x(x(x8x(+25)+8)+1)+19??

- jim_thompson5910

f(-x) = –3(-x)^5 – 8(-x)^4 +25(-x)^3 – 8(-x)^2 +(-x) – 19
would simplify to
f(-x) = 3x^5 – 8x^4 - 25x^3 – 8x^2 - x – 19

- jim_thompson5910

now count the sign changes in
f(-x) = 3x^5 – 8x^4 - 25x^3 – 8x^2 - x – 19
and tell me how many you got

- anonymous

No sign changes?

- jim_thompson5910

3x^5 is the same as +3x^5

- jim_thompson5910

so there's only one sign change from +3x^5 to -8x^4

- jim_thompson5910

so this means that there is at most 1 negative real root

- anonymous

Oh, I was just looking at it wrong.. Sorry :(

- jim_thompson5910

no worries

- jim_thompson5910

So far we found that there are at most 4 positive real roots and at most 1 negative real root

- anonymous

I see

- jim_thompson5910

so one possibility is that there are 4+1 =5 real roots (4 positive, 1 negative)
BUT
this is not the only possibility

- jim_thompson5910

we could have 3 positive real roots and 1 negative real root to have 3+1 = 4 real roots total
but the problem with this scenario is that you would only have 5-1 = 1 complex root...when complex roots should come in pairs

- jim_thompson5910

so that scenario of 3 positive real roots and 1 negative real root just isn't possible

- anonymous

Okay
So don't worry about the last scenario?

- jim_thompson5910

anyways, there are a ton of possibilities
the good news is that you can sum them all up by saying
there are at most 4 positive real roots (you could have 0, 1, 2, 3, or 4 positive real roots)
there is at most 1 negative real root (you could have 0 or 1 negative real roots)
there are at most 5 complex roots (if there are no real roots at all)

- jim_thompson5910

in short
there are at most 4 positive real roots
there is at most 1 negative real root
there are at most 5 complex roots

- anonymous

I got it! :D

- jim_thompson5910

ok great

- anonymous

Thanks!

- jim_thompson5910

we can stop there because we don't need to find the actual roots or the counts of the types or roots...just the max of all of the possible types of roots

- jim_thompson5910

sure thing

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