Posted by Jaclyn on Monday, September 12, 2011 at 5:27pm.
I know about Descartes Rule but I don't really get how to figure this outPlease check my answers
I don't get this one find number of possible positive and negative real roots of f(x) = x^4x^3+2x^2 + x5
I think there are 3 sign changes so there are 3 positive but I get really confused on doing the negative real zeros would I rewrite for negative f(x) = x^4 (x^3) 2x^2 (x5) so there would be 2negative real zeros

PreCalc  Reiny, Monday, September 12, 2011 at 6:03pm
f(x) = x^4  x^3 + 2x^2 + x  5
has 3 sign changes.
So there would be either 3 or 1 positive roots.
f(x) = (x)^4  (x)^3 + 2(x)^2 + (x)  5
= x^4 + x^3 + 2x^2  x  5
I see only one sign change, so there is 1 negative roots.
So there would be either 1+ root, 1 root, and 2 complex
or
there would be 3+ roots, and 1 root
remember we could have at most 4 roots
Looks this youtube clip showing the procedure.
http://www.youtube.com/watch?v=5YAmwfT3Esc
running it through "wolfram" shows the roots
http://www.wolframalpha.com/input/?i=x%5E4x%5E3%2B2x%5E2+%2B+x5+%3D0
show 1 positive, 1 negative, and 2 complex roots

PreCalc  Jaclyn, Monday, September 12, 2011 at 7:37pm
Thank youI viewed the youtubethat was really helpful and I did use Wolfram as you suggested. I even checked some of my other problems
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