Question

find a 3rd degree polynomial equation w/ rational coeffecients that has routes -6 and 3+i

Answer 1

A 3rd degree polynomial must have three roots--real or complex. If we have one complex root, we know its conjugate is also there. The conjugate of a + bi being a - bi. Thus, our three roots are -6, 3+i, and 3-i.
We also know that they are roots and by definition, satisfy:
(x-6)(x-(3+i))(x-(3-i)) = 0.

After the proper expansion, you should get
x^3-12 x^2+46 x-60

Answer 2

thats incorrect according to the sheet - we r tryin to figure out how they get the answer they got

Answer 3

OH I AM SO SORRY. Your root is -6. I thought it was 6. I apologize!! The correct equation should be

(x+6)(x-(3+i))(x-(3-i)) = 0. (the difference is that it is now x + 6, not x - 6

After expansion,
x^3-26 x+60