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

- anonymous

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- anonymous

I don't understand what it's asking for is it asking to find positive negitive and complex?

- ash2326

@sara1234 It's asking for no. of positive or negative zeros.!!

- anonymous

huh? for no of?

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## More answers

- ash2326

Yeah, Using Descartes' Rule of Signs we can determine that. I'll explain you

- 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

isnt there 5 sign changes?

- anonymous

oh never miind i got it

- 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

yes

- ash2326

Now can you write f(-x), it's used to find no. of negative zeros

- anonymous

-3(-x)^5-8(-x)^4+25^3-8(-x)^2+(-x)-19

- 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

Yes i do

- ash2326

Now count the no. of sign changes here

- anonymous

only one

- 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

Yeah thankssss!! Now we find complex?

- 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

why do we have to add the ) positive?

- anonymous

0

- ash2326

Because 0 could be due to 4 complex roots

- anonymous

theres 4 or 2 positive real zeros but why do we have to add the 0

- ash2326

There could be a double quadratic, with 4 roots complex. therefore we won't have any positive zero

- anonymous

thanks for all your help @ash2326 you really helped me :)

- ash2326

you're welcome. did you understand all the things?

- anonymous

Yessss :D

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