# What are the possible number of positive, negative, and complex zeros of f(x) = x^6 – x^5– x^4 + 4x^3 – 12x^2 + 12?

Jun 5, 2016

Use Descartes' rule of signs to find that there may be $0$ or $2$ positive zeros, $0 , 2$ or $4$ negative zeros and $0 , 2 , 4$ or $6$ Complex non-Real zeros.

#### Explanation:

$f \left(x\right) = {x}^{6} - {x}^{5} - {x}^{4} + 4 {x}^{3} - 12 {x}^{2} + 12$

By the FTOA this has a total of $6$ zeros counting multiplicity, since it is of degree $6$.

Use Decartes' rule of signs...

The signs of the coefficients have the pattern:

$+ - - + - +$

With $4$ changes of sign, there may be $0 , 2$ or $4$ positive Real zeros.

Reversing the signs on the terms of odd degree we get the pattern:

$+ + - - - +$

With $2$ changes of sign, there may be $0$ or $2$ negative Real zeros.

The number of Complex non-Real zeros may be $0 , 2 , 4$ or $6$.

In practice there are $2$ positive, $2$ negative and $2$ non-Real Complex zeros.

graph{x^6-x^5-x^4+4x^3-12x^2+12 [-40, 40, -20, 20]}