Algebra 2 - Imaginary roots of Polynomials

Question

Question:
One zero of $P(z) = z^3 +az^2 + 3z + 9$ is purely imaginary. If $a \in \mathbb{R}$, find $a$ and hence factorize $P(z)$ into linear factors.

What I've done:
I know that the $P(z)$ is real since its coefficients are all real. The imaginary root must be $bi$ and its conjugate is $-bi$. For $bi$ and $-bi$, their sum $= 0$ and their product $= -b^2i^2 = b^2$. These two zeros also come from the polynomial $z^2 + b^2$.
How can I finish the problem with this information? Thanks $:)$

Edit: Fixed the $az^2$

Are you sure this is the question? Because $P(bi) = 9 + i(ab + 3b - b^3) \neq 0$... — Arthur, Nov 13 '13 at 21:19
@Arthur yes I made a mistake while typing it out, $az$ was supposed to be $az^2$, I fixed it in the question now :) — Spectrewiz, Nov 13 '13 at 21:54
Ah ok, that makes it much better. An alternative solution to the one given in the answer below is computing $P(bi)$ and setting it equal to zero. — Arthur, Nov 13 '13 at 21:59

Old John · Accepted Answer · 2013-11-13T21:32:56.500

2

I am assuming that the term $az$ is meant to be $az^2$ (since the problem doesn't seem to work otherwise):

Equating coefficients in the expansion of $$(z^2+b^2)(z+k)$$

with $z^3+az^2+3z+9$, gives $b=\sqrt{3}$ and $k = a = 3$, and the factorisation is:

$$z^3+az^2+3z+9 = (z-i\sqrt{3})(z+i\sqrt{3})((z+3).$$

edited Nov 13 '13 at 21:32

answered Nov 13 '13 at 21:26

Old John

19,569
3
59
113

Yes, sorry, the $az$ was supposed to be $az^2$, fixed it. I will try to solve now by the way you did – Spectrewiz Nov 13 '13 at 21:52

Algebra 2 - Imaginary roots of Polynomials

1 Answers1