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December 20, 2014

December 20, 2014

Posted by **hawra** on Saturday, March 13, 2010 at 2:19pm.

degree 4; zeros: i, 3+i

- pre-calculus -
**Reiny**, Saturday, March 13, 2010 at 3:53pmcomplex roots always come in conjugate pairs

so if i is a root so is -i, and if 3+i is a root, so is 3-i

so f(x) = (x^2+1)(x - (3+i))(x - (3-i)

= (x^2 + 1)((x^2 + 6x + 10)

There are no real zeros.

- pre-calculus -
**hawra**, Saturday, March 13, 2010 at 4:28pmam sorry i wrote in incoreectly forgive me, it was suppose to say: information is given about the polynomial f(x) whose coefficients are real numbers. find the remaining zeros of f: degree 4; zeros: i, 3+i, sorry again and thank you for taking the time to explain to me and solve the problem, god bless you always:)

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