The equation x^6 - 2x^5 + 3x^2 - 2 has 6 complex solutions, which 3 or 1 are positive, 1 is negative, and 2 or 4 are imaginary.
The fundamental theorem of algebra states that "a polynomial function f(x) of degree n (where n > 0) has n complex solutions for the equation f(x) = 0.
The equation x^6 - 2x^5 + 3x^2 - 2 has a degree of 6 since its largest exponent is 6. The degree of the polynomial dictates how many solution it has. In this case, a polynomial of degree 6 has 6 complex solutions.
On the other hand, Descartes' rule of sign tells us that the number of positive real zeros in a polynomial function f(x) is the same or less than by an even numbers as the number of changes in the sign of the coefficients while the number of negative real zeros of the f(x) is the same as the number of changes in sign of the coefficients of the terms of f(-x) or less than this by an even number.
The coefficients of the equation x^6 - 2x^5 + 3x^2 - 2 are:
+1 -2 +3 -2
Changes of sign can be seen between 1st and 2nd, between 2nd and 3rd, and between 3rd and 4th. Therefore, the number of positive real zeros can either be 3 or 1 only.
To check the number of negative real zeros, consider f(-x) by negating the sign of each terms.
x^6 - 2x^5 + 3x^2 - 2 ⇒(-x)^6 - 2(-x)^5 + 3(-x)^2 - 2
= x^6 + 2x^5 + 3x^2 - 2
The coefficients of the equation x^6 - 2x^5 + 3x^2 - 2 are:
+1 +2 +3 -2
Changes in sign occur only once. Hence, There is only 1 negative real zeros.
If there is 1 negative real zero, and 3 or 1 positive real zero, and if the equation has 6 complex solutions, then it can have 2 or 4 imaginary solutions.
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