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sara1234
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
I don't understand what it's asking for is it asking to find positive negitive and complex?
@sara1234 It's asking for no. of positive or negative zeros.!!
Yeah, Using Descartes' Rule of Signs we can determine that. I'll explain you
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?
isnt there 5 sign changes?
oh never miind i got it
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?
Now can you write f(-x), it's used to find no. of negative zeros
-3(-x)^5-8(-x)^4+25^3-8(-x)^2+(-x)-19
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?
Now count the no. of sign changes here
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?
Yeah thankssss!! Now we find complex?
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
why do we have to add the ) positive?
Because 0 could be due to 4 complex roots
theres 4 or 2 positive real zeros but why do we have to add the 0
There could be a double quadratic, with 4 roots complex. therefore we won't have any positive zero
thanks for all your help @ash2326 you really helped me :)
you're welcome. did you understand all the things?