# Thread: Finding a cubic equation from it's roots

1. ## 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?
The answer is
(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!

2. ## Re: Finding a cubic equation from it's roots

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?
The answer is
(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!
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*}.