# Pre-Calc

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I know about Descartes Rule but I don't really get how to figure this out-Please check my answers-

I don't get this one find number of possible positive and negative real roots of f(x) = x^4-x^3+2x^2 + x-5
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 -(x-5) so there would be 2negative real zeros

• Pre-Calc -

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.

running it through "wolfram" shows the roots
http://www.wolframalpha.com/input/?i=x%5E4-x%5E3%2B2x%5E2+%2B+x-5+%3D0

show 1 positive, 1 negative, and 2 complex roots

• Pre-Calc -

Thank you-I viewed the youtube-that was really helpful and I did use Wolfram as you suggested. I even checked some of my other problems