P.nut1996
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.
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jim_thompson5910
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in f(x), how many sign changes are there?
P.nut1996
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I don't know?
jim_thompson5910
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notice that going from -8x^4 to +25x^3, there's a sign change from negative to positive
jim_thompson5910
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do you see this?
P.nut1996
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Yes
jim_thompson5910
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ok there's another from +25x^3 to -8x^2
jim_thompson5910
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what's another one?
P.nut1996
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+x to -19?
jim_thompson5910
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there's one more
jim_thompson5910
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from -8x^2 to +x
jim_thompson5910
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so there are 4 sign changes total in f(x)
jim_thompson5910
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this means that there are at most 4 positive real roots
P.nut1996
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-19 to +x?
jim_thompson5910
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no that's the same sign change (just in reverse)
P.nut1996
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4 sign changes=4 positive real roots
jim_thompson5910
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at most 4 (there could be 0, 1, 2, 3, or 4 positive real roots)
jim_thompson5910
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4 is the maximum
jim_thompson5910
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now we must find f(-x)
P.nut1996
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I'm still really confuzzled...
jim_thompson5910
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where at?
P.nut1996
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All of it
jim_thompson5910
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the rule is
if f(x) has n sign changes, then there are AT MOST n positive real roots
jim_thompson5910
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if n = 4, then
if f(x) has 4 sign changes, then there are AT MOST 4 positive real roots
P.nut1996
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Okay
jim_thompson5910
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making more sense?
P.nut1996
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...
jim_thompson5910
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yes? no?
P.nut1996
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I don't know how to formulate an answer from you telling me this...
jim_thompson5910
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you don't need to find the actual roots
you just need to be able to count the possible number and type
jim_thompson5910
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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
P.nut1996
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But I need a definite answer, right?
P.nut1996
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I can't just say "maybe 4"?
jim_thompson5910
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well that's part of the answer, we still have to find f(-x)
P.nut1996
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Okay, and how would we go about that ? :)
jim_thompson5910
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start with f(x)
then replace each x with -x and simplify
P.nut1996
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Could you give me an example?
jim_thompson5910
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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
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does that help?
P.nut1996
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Yes, it does :)
jim_thompson5910
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ok what do you get
jim_thompson5910
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when you simplify
P.nut1996
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3x^5=f(-x)+x(x(x8x(+25)+8)+1)+19??
jim_thompson5910
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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
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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
P.nut1996
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No sign changes?
jim_thompson5910
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3x^5 is the same as +3x^5
jim_thompson5910
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so there's only one sign change from +3x^5 to -8x^4
jim_thompson5910
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so this means that there is at most 1 negative real root
P.nut1996
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Oh, I was just looking at it wrong.. Sorry :(
jim_thompson5910
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no worries
jim_thompson5910
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So far we found that there are at most 4 positive real roots and at most 1 negative real root
P.nut1996
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I see
jim_thompson5910
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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
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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
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so that scenario of 3 positive real roots and 1 negative real root just isn't possible
P.nut1996
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Okay
So don't worry about the last scenario?
jim_thompson5910
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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
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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
P.nut1996
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I got it! :D
jim_thompson5910
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ok great
P.nut1996
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Thanks!
jim_thompson5910
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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
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sure thing