Question

What are the possible numbers 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

Answer:

2 or 0 positive  zeros

4 or 2 or 0 negative zeros

complex zeros 0,2,4,6  

Step-by-step explanation:

f(x) = x^6 + x^5 + x^4 + 4 x^3 − 12 x^2 + 12

We use Descartes's rule of sign

LEts see the change of sign

f(x) = x^6 + x^5 + x^4 + 4 x^3 − 12 x^2 + 12

In f(x) , there is 2 sign changes

so 2 positive  zeros . then subtract 2 to get possible number of zeros

2 or 0 positive  zeros

To find negative zeros replace x with -x

f(-x) = (-x)^6 + (-x)^5 + (-x)^4 + 4 (-x)^3 − 12 (-x)^2 + 12

f(x) = x^6 - x^5 + x^4 - 4 x^3 − 12 x^2 + 12

4 sign changes

4 or 2 or 0 negative zeros

complex zeros 0,2,4,6