SOLUTION: How many positive real solutions are there in: f(x)=x^4-4x^3+6x^2-4x+1 a)1 or 0 b)2 or 0 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 a)1 or 0 b)2 or 0 c)4 or 2 d)2 or 0      Log On


   

Question 146280: How many positive real solutions are there in:
f(x)=x^4-4x^3+6x^2-4x+1
a)1 or 0
b)2 or 0
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!

First count the sign changes of f%28x%29=x%5E4-4x%5E3%2B6x%5E2-4x%2B1

From x%5E4 to -4x%5E3, there is a sign change from positive to negative

From -4x%5E3 to 6x%5E2, there is a sign change from negative to positive

From 6x%5E2 to -4x, there is a sign change from positive to negative

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

So there are 4 sign changes for the expression f%28x%29=x%5E4-4x%5E3%2B6x%5E2-4x%2B1.

So there are 4, 2, or 0 positive zeros