Answer
There are 3 positive real zeros, or there is 3-2=1 positive real zero.
Work Step by Step
$f(x)=5x^3-3x^2+3x-1$
To find possibilities for positive real zeros, we count the numer of sign changes in the equation for f(x). There are three positive real zeros. The number of positive real zeros of f is equal to the number of sign changes 3, or is less than this number by a positive real zero.
There are 3 positive real zeros, or there is 3-2=1 positive real zero.
To find possibilities for negative real zeros, we count the numer of sign changes in the equation for f(-x). We obtain this equation by replacing x with -x in the function. We have:
$f(-x)=5(-x)^3-3(-x)^2+3(-x)-1$
$f(-x)=-5x^3-3x^2-3x-1$
Now we count the sign changes. Because all the coefficients are negative, there are no variations in sign. Thus, there are no negative real zeros.
