Question

Descartes' Rule isn't working here


I am so confused on this question. I graphed it and found my answer, yes, but Descartes' Rule should have given it to me. I have the function
f(x) = x^(3) + 5x^(2) + x + 5
and I am supposed to find how many imaginary and how many real zeros this function has. Descartes' Rule would give me a possibility of 2 or 0 real negative zeros (if I did this correctly, which I hope I did) because when I solved for f(-x) it goes from "- x^(3)" to "+ 5x^(2)", which is one sign change, and then it goes from "+ 5x^(2)" to "- x", which is the second sign change.

So there is a possibility of 2 or 0 negative real roots. BUT NO. Desmos says (https://www.desmos.com/calculator) that the graph has only 1 (ONE!) negative real root. I am so baffled. I did synthetic division and got "- 5", which is the point on the graph as well. Since there is apparently only 1 real root, that would mean that there are 2 imaginary roots, according to the Fundamental Theorem.

WHAT IS GOING ON PLEASE HELP

Answer 1

I am attaching the graph here

Answer 2

f(-x)=-x^3+5x^2-x+5
so there are 3 changes of sign, no. of negative roots =1 or 3

Answer 3

@nikoleann

Answer 4

@jango_IN_DTOWN
OHHHH I was changing "5" to "- 5" for some reason. Thank you SO much!!!

Answer 5

n.p.