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September 18, 2014

September 18, 2014

Posted by **Marissa** on Friday, August 10, 2007 at 5:13pm.

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?

Yes, the answer is B

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