# 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#?

##### 1 Answer

#### Answer:

Use Descartes' rule of signs to find that there may be

#### Explanation:

By the FTOA this has a total of

Use Decartes' rule of signs...

The signs of the coefficients have the pattern:

#+ - - + - +#

With

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

#+ + - - - +#

With

The number of Complex non-Real zeros may be

In practice there are

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