Answer
There are 3 negative real zeros or there is 3-2=1 negative real zero.
Work Step by Step
$f(x)=x^3+7x^2+x+7$
To find possibilities for positive real zeros, we count the numer of sign changes in the equation for f(x). Because all the coefficients are positive, there are no variations in sign. Thus, there are no positive real zeros.
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:
$f(-x)=(-x)^3+7(-x)^2+(-x)+7$
$f(-x)=-x^3+7x^2-x+7$
Now we count the sign changes. There are three variations in sign. The number of negative real zeros of f is equal to the number of sign changes 3, or is less than this number by an even integrer.
There are 3 negative real zeros or there is 3-2=1 negative real zero.
