take a look at this question I did several years ago
http://www.jiskha.com/display.cgi?id=1315862859
a. 0 positive, 0 negative, 3 non real roots
b. 0 positive, 1 or 3 negative, 0 or 2 non real roots
c. 0 positive, 1 or 3 negative, 1 or 3 non real roots
d. 0 or 2 positive, 0 or 2 negative, 0 or 2 non real roots
e. 0 or 2 positive, 1 or 3 negative, 0 or 2 nonreal roots
http://www.jiskha.com/display.cgi?id=1315862859
Given the equation 4x^3+7x^2+7x+3=0, let's look at the sign changes:
1. Start by writing down the coefficients in descending order:
a_3 = 4 (coefficient of x^3)
a_2 = 7 (coefficient of x^2)
a_1 = 7 (coefficient of x)
a_0 = 3 (constant term)
2. Count the number of sign changes in the coefficients. In this case, we have two sign changes. We go from positive (4x^3) to positive (7x^2), then to negative (7x), and finally to positive (3).
3. The number of possible positive roots is either equal to the number of sign changes (2) or less than that by an even integer (0 or 2). In this case, we can conclude that there are either 0 or 2 possible positive roots.
4. The number of possible negative roots is given by the number of sign changes in the equation f(-x). We take the opposite sign of each term (change positive to negative, and negative to positive) and count the number of sign changes. In this case, f(-x) = 4(-x)^3 + 7(-x)^2 + 7(-x) + 3 = -4x^3 + 7x^2 - 7x + 3.
By checking the sign changes, we see that there is one sign change (from positive to negative). Therefore, there is either 1 or 3 possible negative roots.
5. The remaining possible roots fall into the nonreal category. Since the number of nonreal roots is equal to the number of sign changes or less than that by an even integer, we can conclude that there are either 0 or 2 nonreal roots.
Based on the above analysis, the correct option is:
e. 0 or 2 positive, 1 or 3 negative, 0 or 2 nonreal roots.