posted by Marissa .
State the possible number of imaginary zeros of g(x)= x^4+3x^3+7x^2-6x-13
this is how far Ive gotten:there are 4 zeros,3 positive and 1 negative
If you've found four real zeroes, then there are no imaginary zeroes. You can also see that there are no imaginary zeroes directly:
Put x = i y in the equation:
y^4 - 3iy^3 - 7y^2- 6iy - 13 = 0
Equate real and imaginary parts to zero:
y^4 - 7 y^2 - 13 = 0
y^3 + 2 y = 0
The last equation has only one solution:
y = 0
But this does not satisfy the first equation.
so what is the answer? my choices are A)3 or 1
is it B?
Yes, the answer is B
I think it's B. Good luck!