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sara1234
Group Title
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
 2 years ago
 2 years ago
sara1234 Group Title
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
 2 years ago
 2 years ago

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sara1234 Group TitleBest ResponseYou've already chosen the best response.1
I don't understand what it's asking for is it asking to find positive negitive and complex?
 2 years ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.3
@sara1234 It's asking for no. of positive or negative zeros.!!
 2 years ago

sara1234 Group TitleBest ResponseYou've already chosen the best response.1
huh? for no of?
 2 years ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.3
Yeah, Using Descartes' Rule of Signs we can determine that. I'll explain you
 2 years ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.3
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?
 2 years ago

sara1234 Group TitleBest ResponseYou've already chosen the best response.1
isnt there 5 sign changes?
 2 years ago

sara1234 Group TitleBest ResponseYou've already chosen the best response.1
oh never miind i got it
 2 years ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.3
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?
 2 years ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.3
Now can you write f(x), it's used to find no. of negative zeros
 2 years ago

sara1234 Group TitleBest ResponseYou've already chosen the best response.1
3(x)^58(x)^4+25^38(x)^2+(x)19
 2 years ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.3
Good , we need to simplify it. I'll write that \[f(x)=3(x)^58(x)^4+25(x)^38(x)^2+(x)19\] \[f(x)=3x^58x^425x^38x^2x19\] do you get this?
 2 years ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.3
Now count the no. of sign changes here
 2 years ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.3
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?
 2 years ago

sara1234 Group TitleBest ResponseYou've already chosen the best response.1
Yeah thankssss!! Now we find complex?
 2 years ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.3
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
 2 years ago

sara1234 Group TitleBest ResponseYou've already chosen the best response.1
why do we have to add the ) positive?
 2 years ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.3
Because 0 could be due to 4 complex roots
 2 years ago

sara1234 Group TitleBest ResponseYou've already chosen the best response.1
theres 4 or 2 positive real zeros but why do we have to add the 0
 2 years ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.3
There could be a double quadratic, with 4 roots complex. therefore we won't have any positive zero
 2 years ago

sara1234 Group TitleBest ResponseYou've already chosen the best response.1
thanks for all your help @ash2326 you really helped me :)
 2 years ago

ash2326 Group TitleBest ResponseYou've already chosen the best response.3
you're welcome. did you understand all the things?
 2 years ago
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