anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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jim_thompson5910
  • jim_thompson5910
in f(x), how many sign changes are there?
anonymous
  • anonymous
I don't know?
jim_thompson5910
  • jim_thompson5910
notice that going from -8x^4 to +25x^3, there's a sign change from negative to positive

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jim_thompson5910
  • jim_thompson5910
do you see this?
anonymous
  • anonymous
Yes
jim_thompson5910
  • jim_thompson5910
ok there's another from +25x^3 to -8x^2
jim_thompson5910
  • jim_thompson5910
what's another one?
anonymous
  • anonymous
+x to -19?
jim_thompson5910
  • jim_thompson5910
there's one more
jim_thompson5910
  • jim_thompson5910
from -8x^2 to +x
jim_thompson5910
  • jim_thompson5910
so there are 4 sign changes total in f(x)
jim_thompson5910
  • jim_thompson5910
this means that there are at most 4 positive real roots
anonymous
  • anonymous
-19 to +x?
jim_thompson5910
  • jim_thompson5910
no that's the same sign change (just in reverse)
anonymous
  • anonymous
4 sign changes=4 positive real roots
jim_thompson5910
  • jim_thompson5910
at most 4 (there could be 0, 1, 2, 3, or 4 positive real roots)
jim_thompson5910
  • jim_thompson5910
4 is the maximum
jim_thompson5910
  • jim_thompson5910
now we must find f(-x)
anonymous
  • anonymous
I'm still really confuzzled...
jim_thompson5910
  • jim_thompson5910
where at?
anonymous
  • anonymous
All of it
jim_thompson5910
  • jim_thompson5910
the rule is if f(x) has n sign changes, then there are AT MOST n positive real roots
jim_thompson5910
  • jim_thompson5910
if n = 4, then if f(x) has 4 sign changes, then there are AT MOST 4 positive real roots
anonymous
  • anonymous
Okay
jim_thompson5910
  • jim_thompson5910
making more sense?
anonymous
  • anonymous
...
jim_thompson5910
  • jim_thompson5910
yes? no?
anonymous
  • anonymous
I don't know how to formulate an answer from you telling me this...
jim_thompson5910
  • 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
  • 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
  • anonymous
But I need a definite answer, right?
anonymous
  • anonymous
I can't just say "maybe 4"?
jim_thompson5910
  • jim_thompson5910
well that's part of the answer, we still have to find f(-x)
anonymous
  • anonymous
Okay, and how would we go about that ? :)
jim_thompson5910
  • jim_thompson5910
start with f(x) then replace each x with -x and simplify
anonymous
  • anonymous
Could you give me an example?
jim_thompson5910
  • 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
  • jim_thompson5910
does that help?
anonymous
  • anonymous
Yes, it does :)
jim_thompson5910
  • jim_thompson5910
ok what do you get
jim_thompson5910
  • jim_thompson5910
when you simplify
anonymous
  • anonymous
3x^5=f(-x)+x(x(x8x(+25)+8)+1)+19??
jim_thompson5910
  • 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
  • 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
  • anonymous
No sign changes?
jim_thompson5910
  • jim_thompson5910
3x^5 is the same as +3x^5
jim_thompson5910
  • jim_thompson5910
so there's only one sign change from +3x^5 to -8x^4
jim_thompson5910
  • jim_thompson5910
so this means that there is at most 1 negative real root
anonymous
  • anonymous
Oh, I was just looking at it wrong.. Sorry :(
jim_thompson5910
  • jim_thompson5910
no worries
jim_thompson5910
  • jim_thompson5910
So far we found that there are at most 4 positive real roots and at most 1 negative real root
anonymous
  • anonymous
I see
jim_thompson5910
  • 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
  • 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
  • jim_thompson5910
so that scenario of 3 positive real roots and 1 negative real root just isn't possible
anonymous
  • anonymous
Okay So don't worry about the last scenario?
jim_thompson5910
  • 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
  • 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
  • anonymous
I got it! :D
jim_thompson5910
  • jim_thompson5910
ok great
anonymous
  • anonymous
Thanks!
jim_thompson5910
  • 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
  • jim_thompson5910
sure thing

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