+74 Cphill Thank You, I told my friends I will marry you one day. xD. Heres another one...
Write a polynomial function in standard form that has zeros 3 and 2i (Hint: Are there other zeros?). Thank you, and will you please add in the steps? They help ALOT.
+129899 If 2i is a root.....so is its conjugate, - 2i.....so......the polynomial, P(x), is
P(x) = (x - 2i) (x + 2i)(x -3) simplify the first part
(x - 2i)(x + 2i) = x^2 - 4i^2 note.... [i^2 = -1] .........so we have, for this part
x^2 - (4)(-1) =
x^2 + 4 .......now......multiply this by the other factor, (x -3), and we get
(x^2 + 4) (x - 3) =
x^3 - 3x^2 + 4x - 12 and that's the polynomial, P(x)
This will verify the results :
http://www.wolframalpha.com/input/?i=x^3+-+3x^2+%2B+4x+-+12++%3D+0
