Question

Find the equation of a 3rd degree polynomial with zeros 2 and 3i. And f(1)=3

Answer 1

start with
\[a(x-2)(x-3i)(x+3i)\] and then multiply out. it is not too hard since
\[(x-3i)(x+3i)=x^2+9\]

Answer 2

you should get
\[a(x^3-2 x^2+9 x-18)\] and you know if x = 1, then the result is 3, so replace x by 1 and set the answer = 3 and solve for a

Answer 3

i get
\[-10a=3\] so
\[a=-\frac{3}{10}\] but you should check my work, because i am tired and my algebra is sloppy.

Answer 4

Ok thank you very much! :D