# Finding a cubic equation from it's roots

• December 3rd 2012, 07:41 PM
Tutu
Finding a cubic equation from it's roots
For the cubic equation z^3+az^2+3z+9 where one zero is purely imaginary.
I factorized into (z-a)(z-bi)(z+bi)

And on to expand and equate coefficients to get a=3, and b = +/-sqrt(3)
Now to write back the cubic equation as a product of linear factors, how do I write b?
(z-a)(z+sqrt(3)i)(z-sqrt(3)i)
But why?
Isn't it
(z-a)(z+sqrt(3))(z-sqrt(3)i) since one of the roots is a positive sqrt(3)?

Thank you so much!
• December 3rd 2012, 07:53 PM
Prove It
Re: Finding a cubic equation from it's roots
Quote:

Originally Posted by Tutu
For the cubic equation z^3+az^2+3z+9 where one zero is purely imaginary.
I factorized into (z-a)(z-bi)(z+bi)

And on to expand and equate coefficients to get a=3, and b = +/-sqrt(3)
Now to write back the cubic equation as a product of linear factors, how do I write b?
Nonreal roots always appear as complex conjugates. That means if \displaystyle \begin{align*} z - \sqrt{3}\,i \end{align*} is a factor, so is \displaystyle \begin{align*} z + \sqrt{3}\,i \end{align*}.