According to the Fundamental Theorem of Algebra (p. 244-245), [tex]f(x)=x^3-2 x^2+2 x-6[/tex] will have $\square$ total zeroes because it has a degree of 3. Create a table and use the FTA and the numbers of possible positive/negative real zeroes to determine the possible number of imaginary zeroes for the function.
| Number of Positive | Number of Negative | Number of Imaginary | Total Number |
| :----------------- | :----------------- | :----------------- | :------------- |
| Real Zeroes | Real Zeroes | Zeroes | = |
| + | + | | = |
| + | + | | |
| + | + | | = |
| + | + | | = |
Answer (2)
The polynomial f ( x ) = x 3 − 2 x 2 + 2 x − 6 can have 3 positive real roots or 1 positive real root, with the remaining roots being imaginary. By analyzing the signs, there are no negative roots. This gives us a total of either 3 real roots or 1 real root and 2 imaginary roots.
;
The polynomial f ( x ) = x 3 − 2 x 2 + 2 x − 6 has degree 3, so it has 3 total roots.
By Descartes' Rule of Signs, f ( x ) has either 3 or 1 positive real roots and 0 negative real roots.
The possible combinations of roots are:
3 positive real roots, 0 negative real roots, 0 imaginary roots.
1 positive real root, 0 negative real roots, 2 imaginary roots.
The completed table summarizes these possibilities.
Explanation
Understanding the Problem We are given the polynomial f ( x ) = x 3 − 2 x 2 + 2 x − 6 . We need to use the Fundamental Theorem of Algebra and Descartes' Rule of Signs to determine the possible number of positive, negative, and imaginary zeroes of this polynomial.
Applying the Fundamental Theorem of Algebra The Fundamental Theorem of Algebra states that a polynomial of degree n has exactly n complex roots (counting multiplicities). Since f ( x ) has degree 3, it has a total of 3 complex roots.
Determining Possible Positive Real Roots Descartes' Rule of Signs helps us determine the possible number of positive and negative real roots. We examine the sign changes in the coefficients of f ( x ) . The coefficients are 1 , − 2 , 2 , − 6 . The sign changes are:
From 1 to − 2 (positive to negative)
From − 2 to 2 (negative to positive)
From 2 to − 6 (positive to negative)
There are 3 sign changes, so there can be 3 or 1 positive real roots (we subtract by even numbers until we reach 1 or 0).
Determining Possible Negative Real Roots Now we find f ( − x ) to determine the possible number of negative real roots:
f ( − x ) = ( − x ) 3 − 2 ( − x ) 2 + 2 ( − x ) − 6 = − x 3 − 2 x 2 − 2 x − 6
The coefficients of f ( − x ) are − 1 , − 2 , − 2 , − 6 . There are no sign changes, so there are 0 negative real roots.
Creating the Table Now we create a table to summarize the possible combinations of positive, negative, and imaginary roots. Since the total number of roots must be 3, we have the following possibilities:
Case 1: 3 positive real roots, 0 negative real roots, and 0 imaginary roots. Case 2: 1 positive real root, 0 negative real roots, and 2 imaginary roots.
Final Table Here's the completed table:
Number of Positive Real Zeroes
Number of Negative Real Zeroes
Number of Imaginary Zeroes
Total Number of Zeroes
3
0
0
= 3
1
0
2
= 3
Examples
Understanding the nature of polynomial roots is crucial in many engineering and physics applications. For instance, when designing a control system, the stability of the system depends on the roots of a characteristic polynomial. If the roots have positive real parts, the system is unstable. Similarly, in quantum mechanics, the energy levels of a system are determined by the roots of a polynomial equation. Knowing the possible number of real and imaginary roots helps engineers and scientists predict the behavior of these systems.
