Question

A polynomial with rational coefficients having roots 3,3,and 3-i

Answer 1

How do you have "3-i" only once, I can't really think of such polynomial. Unless a polynomial has imaginary numbers in it.

Answer 2

When you solve an equation (for x intercepts of a polynomial) you should be getting something like
Plus Minus imaginary... (not just an imaginary root ONCE)

Answer 3

I think that to have RATIONAL coefficients (meaning they are real numbers, and not square roots) in this case is not possible.

Answer 4

would 3-i have to be acquainted with 3+i? would that make a difference?

Answer 5

well, if it was -3-i then it is like you have
(of roots are 3, 3+i, -3-i)

f(x)=(x-3)(x-3-i)(x+3+i)
for the "(x-3-i)(x+3+i)" positive and negative 3i cancel, and i times -i = 1 (real, rational number).

positive xi and negative xi also cancel.
then this product "(x-3-i)(x+3+i)" after the expansion of which you get only real, rationals, is multiplied times (x+3), .... so all values/coefficients will then be real

Answer 6

will be real and rational.

Answer 7

I am saying MINUS 3 minus i, because
the +- (3+i)
would give 3+i (and when minus) -3-i

(but not a 3-i. the +- somthing... they are supposed to be additive inverses of each other.)