## sara1234 3 years ago 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

1. sara1234

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

2. ash2326

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

3. sara1234

huh? for no of?

4. ash2326

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

5. 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?

6. sara1234

isnt there 5 sign changes?

7. sara1234

oh never miind i got it

8. 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?

9. sara1234

yes

10. ash2326

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

11. sara1234

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

12. 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?

13. sara1234

Yes i do

14. ash2326

Now count the no. of sign changes here

15. sara1234

only one

16. 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?

17. sara1234

Yeah thankssss!! Now we find complex?

18. 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

19. sara1234

why do we have to add the ) positive?

20. sara1234

0

21. ash2326

Because 0 could be due to 4 complex roots

22. sara1234

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

23. ash2326

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

24. sara1234

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

25. ash2326

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

26. sara1234

Yessss :D