SOLUTION: How many positive real solutions are there in f(x) = x^4 + 4x^3 + 6x^2 + 4x – 1 Apply Descartes' Rule of Signs. a. 1 b. 1 or 2 c. 4 or 2 d. 2 or 0

Algebra ->  Equations -> SOLUTION: How many positive real solutions are there in f(x) = x^4 + 4x^3 + 6x^2 + 4x – 1 Apply Descartes' Rule of Signs. a. 1 b. 1 or 2 c. 4 or 2 d. 2 or 0      Log On


   

Question 152871: How many positive real solutions are there in f(x) = x^4 + 4x^3 + 6x^2 + 4x – 1
Apply Descartes' Rule of Signs.
a. 1
b. 1 or 2
c. 4 or 2
d. 2 or 0
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

Count the sign changes of f%28x%29=x%5E4%2B4x%5E3%2B6x%5E2%2B4x-1

From x%5E4 to 4x%5E3, there is no change in sign

From 4x%5E3 to 6x%5E2, there is no change in sign

From 6x%5E2 to 4x, there is no change in sign

From 4x to -1, there is a sign change from positive to negative

So there is 1 sign change for the expression f%28x%29=x%5E4%2B4x%5E3%2B6x%5E2%2B4x-1.

So there are 1 positive zeros


So the answer is a) 1