# Algebra

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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

And:

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
B)2,4,0
c)exactly 1
d)exactly 3
is it B?

• Algebra -

I think it's B. Good luck!

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