Question

What are the possible number of positive, negative, and complex zeros of f(x) = x6 – x5– x4 + 4x3 – 12x2 + 12 ?

Answer 1

This is a polynomial with more than 2 as a degree. Using Descartes Rule of Signs: 
f(x) = x⁶ + x⁵ + x⁴ + 4x³ − 12x² + 12 
Signs: + + + + − + 2 sign changes ----> 2 or 0 positive roots 
f(−x) = (−x)⁶ + (−x)⁵ + (−x)⁴ + 4(−x)³ − 12(−x)² + 12 f(−x) = x⁶ − x⁵ + x⁴ − 4x³ − 12x² + 12 
Signs: + − + − − + 4 sign changes ----> 4 or 2 or 0 negative roots 
Complex roots = 0, 2, 4, or 6 

Answer 2

Descarte's Rule of Sign is useful for finding the zeroes of a polynomial. The rule will tell you how many roots you can expect and of which type not where the polynomial's zeroes are. This rule is given as follows:

$For \ a \ polynomial \ f(x)=a_{n}x^n+a_{n-1}x^{n-1}+ \ldots a_{2}x^2+a_{1}x+a_{0} \\ \\ with \ real \ coefficients \ and \ a_{0} \neq 0$

$\bullet \ The \ number \ of \ \mathbf{positive \ real \ zeros} \ of \ f \ is \ either \ equal \ to \\ the \ number \ of \ variations \ in \ sign \ of \ f(x) \ or \ less \ than \ that \\ number \ by \ an \ even \ integer. \\ \\ \bullet \ The \ number \ of \ \mathbf{negative \ real \ zeros} \ of \ f \ is \ either \ equal \ to \\ the \ number \ of \ variations \ in \ signs \ of \ f(x) \ or \ less \ than \ that \\ number \ by \ an \ even \ integer.$

That is, the function:

$f(x)=x^6-x^5-x^4+4x^3-12x^2+12 \\ \\ \\ + \ - \ - \ + \ - \ + \\ \\ Has \ four \ changes \ in \ sign$

4, 2, or 0 positive roots

On the other hand, the function:

$f(-x)=(-x)^6-(-x)^5-(-x)^4+4(-x)^3-12(-x)^2+12 \\ \\ f(-x)=x^6+x^5-x^4-4x^3-12x^2+12 \\ \\ \\ + \ + \ - \ - \ - \ + \\ \\ Has \ 2 \ changes \ in \ sign$

2, or 0 negative roots

Finally:

Complex roots: 0, 2, 4, or 6