State the possible number of imaginary zeros of g(x)= x4 + 3x3 + 7x2 - 6x - 13.
I came up with the answer of 2,4 or 0.
Is this correct?
Thanks for your help.
Substitute x = i t where t is assumed to be real:
t^4 - 3it^3 - 7t^2 -6 i t - 13 = 0
Take the imaginary part of the equation:
t^3 + 2 t = 0 --->
t(t^2 + 2) = 0
Now, we can see from the original equation for t that t = 0 is not a solution (this is actually implied by the real part of the equation). Since t has to be real, the equation:
t^2 + 2 = 0
cannot be satisfied.
Therefore the equation has no imaginary solutions.
To determine the possible number of imaginary zeros for a polynomial, we can use the Fundamental Theorem of Algebra. According to this theorem, the number of possible imaginary zeros is equal to the degree of the polynomial.
In this case, the degree of the polynomial g(x) = x^4 + 3x^3 + 7x^2 - 6x - 13 is 4. Therefore, the possible number of imaginary zeros is 4.
Hence, your answer of 2, 4, or 0 is incorrect. The correct answer is 4 possible imaginary zeros.