anonymous
  • anonymous
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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katieb
  • katieb
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anonymous
  • anonymous
I don't understand what it's asking for is it asking to find positive negitive and complex?
ash2326
  • ash2326
@sara1234 It's asking for no. of positive or negative zeros.!!
anonymous
  • anonymous
huh? for no of?

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ash2326
  • ash2326
Yeah, Using Descartes' Rule of Signs we can determine that. I'll explain you
ash2326
  • ash2326
We have the polynomial \[f(x) = –3x^5 – 8x^4 +25x^3 – 8x^2 +x – 19\] First step it to count the no. of sign changes in f(x) I'll underline wherever there is a sign change between two consecutive terms i.e from + to - or - to + \[f(x) = –3x^5 \underline{– 8x^4 +25x^3} – 8x^2 +x – 19 \longrightarrow 1st\ change\] \[f(x) = –3x^5 – 8x^4 \underline{+25x^3 – 8x^2} +x – 19 \longrightarrow 2nd\ change\] \[f(x) = –3x^5 – 8x^4 +25x^3 \underline{– 8x^2 +x} – 19 \longrightarrow 3rd\ change\] \[f(x) = –3x^5 – 8x^4 +25x^3 – 8x^2 \underline{+x – 19} \longrightarrow 4th\ change\] so there are 4 sign changes, do you get this part?
anonymous
  • anonymous
isnt there 5 sign changes?
anonymous
  • anonymous
oh never miind i got it
ash2326
  • ash2326
So there could be 4 positive zero, However, some of the roots may be generated by the Quadratic Formula, and these pairs of roots may be complex. so we have to down count by 2. Either it has 4 or 2 or 0 positive roots. Do you get this?
anonymous
  • anonymous
yes
ash2326
  • ash2326
Now can you write f(-x), it's used to find no. of negative zeros
anonymous
  • anonymous
-3(-x)^5-8(-x)^4+25^3-8(-x)^2+(-x)-19
ash2326
  • ash2326
Good , we need to simplify it. I'll write that \[f(-x)=-3(-x)^5-8(-x)^4+25(-x)^3-8(-x)^2+(-x)-19\] \[f(-x)=3x^5-8x^4-25x^3-8x^2-x-19\] do you get this?
anonymous
  • anonymous
Yes i do
ash2326
  • ash2326
Now count the no. of sign changes here
anonymous
  • anonymous
only one
ash2326
  • ash2326
Yeah so it's having 1 negative zero, we can't down count by 2, it'll become -1. So it has 1 negative zero and 4 or 2 or 0 positive zeros. Do you get this?
anonymous
  • anonymous
Yeah thankssss!! Now we find complex?
ash2326
  • ash2326
We found that there are either 4 or 2 or 0 positive roots. But these could also be complex, so if we have have 4 positive roots there will be 0 complex if we have 2 positive there will be 2 complex or if we have 0 positive there will be 4 complex
anonymous
  • anonymous
why do we have to add the ) positive?
anonymous
  • anonymous
0
ash2326
  • ash2326
Because 0 could be due to 4 complex roots
anonymous
  • anonymous
theres 4 or 2 positive real zeros but why do we have to add the 0
ash2326
  • ash2326
There could be a double quadratic, with 4 roots complex. therefore we won't have any positive zero
anonymous
  • anonymous
thanks for all your help @ash2326 you really helped me :)
ash2326
  • ash2326
you're welcome. did you understand all the things?
anonymous
  • anonymous
Yessss :D

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