Question

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`?

Answer 1

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]}