Which is NOT a possible...

Which is NOT a possible classification of the zeros of f(x) = x⁴ + 2x³ - 7x² - 7x + 3 according to Descartes' rule of signs?
A. 2 positive real zeros, 2 negative real zeros, and 0 imaninary zeros
B. 0 positive real zeros, 2 negative real zeros, and 2 imaninary zeros
C. 0 positive real zeros, 0 negative real zeros, and 4 imaninary zeros
D. 1 positive real zero, 1 negative real zero, and 2 imaninary zeros

And if you could, please explain your choice. Thanks!

f(x) can be described in sign changes as:

f(x) = + + - - +

Which means there are two sign changes. It goes from positive to negative once and from negative to positive once.

Descartes’ Rule of Signs says that the number of positive real zeros of f is equal to the number of changes in sign of the coefficients of f(x) or is less than this by an even number.

That means that this function has either 2 positive real zeros, or less than that by an even number, which means zero positive real zeros.

Obviously the only choice NOT possible is D. 1 positive real zero, etc.

:)

Let roots = a, b, c, d. Then:
f(x) = (x-a) (x-b) (x-c) (x-d)
f(x) = x⁴ . . . . + (-a)(-b)(-c)(-d) = x⁴ . . . + abcd = x⁴ + 2x³ - 7x² - 7x + 3
-----> a*b*c*d = 3 > 0

Note: product of 2 imaginary zeros is always > 0
(A + Bi) (A - Bi) = A² - B²i² = A² + B² > 0

A. Product of 2 positive roots and 2 negative roots > 0 ----> ok
B. Product of 2 negative roots ans 2 imaginary roots > 0 ----> ok
C. Product of 4 imaginary roots > 0 ----> ok
D. Product of 1 positive root, 1 negative root and 2 imaginary roots < 0 ----> NOT ok

Answer: D