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
Jun 5, 2016

Answer:

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(x) = x^6-x^5-x^4+4x^3-12x^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]}