# factoring a polynominal

Consider the following polynomial of x $$x^2+(a+b\,g)x+c\,g^2=0$$ where $a\in \mathbb{R}$, $b\in \mathbb{R}$, and $c\in \mathbb{R}$, whereas g can be complex. $c\neq 0$.

$c$ is a free variable and can be chosen to be ANY function of $a$ and $b$, for example $c=a+b$, $c=ab$, or $c=a^2$.

When $a$ is zero and $c=b^2$, we can factor this polynomial as $$(x+(\frac{1}{2}+\frac{\sqrt{3}}{2}i)b\,g)(x+(\frac{1}{2}-\frac{\sqrt{3}}{2}i)b\,g)=0$$ What is the factorization $(x+x_1)(x+x_2)=0$ when $a\neq 0$ and $b\neq0$, $c\neq 0$?

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The quadratic formula works the same for polynomials with either real or complex coefficients. – user7530 Jan 22 '13 at 18:21
I know you can use the solution of the quadratic equations to obtain $x_1$ and $x_2$, but I was wondering if there is a clean representation like the example I gave above for a=0. – sara Jan 22 '13 at 18:28

\begin{align*}&\quad x^2 + (a+bg)x + cg^2\\ &= \left(x + \frac{a+bg}{2} + \frac{\sqrt{a^2+2abg + (b^2-4c)g^2}}{2}\right)\left(x + \frac{a+bg}{2} - \frac{\sqrt{a^2+2abg + (b^2-4c)g^2}}{2}\right), \end{align*} where $\sqrt{z}$ is the principal square root.
There is no significantly simpler formulation unless some special relationship exists between $a,b,c$ and $g$.
EDIT: I assume you're looking for special values of $a,b,c$ so that the discriminant "factors nicely":
$$\sqrt{a^2+2abg + (b^2-4c)g^2} = dg+e.$$ Squaring both sides $$a^2 + 2abg + (b^2-4c)g^2 = e^2 + 2deg + d^2g^2$$ and equating coefficients of $g$ gives $a=\pm e$, from which it follows that either $a=0$ or $b=\pm d\Rightarrow c =0$. There are no other "simple cases."