# Factorization of polynomials over $\mathbb{C}$

I'm stuck I don't know how to write this complex number equation as two factors although I know one of those factors is $z - 3$. Any ideas/advice appreciated.

$$f(z) = z^3 + (-6+2j)z^2 + (12-15j)z + (-9+27j)$$ $f(3)=0$, therefore $z-3$ is a factor. Write as two factors $$f(z) = (z-3)(\cdots?\cdots)$$

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Long division. – Arturo Magidin May 28 '12 at 0:48
Thanks @ArturoMagidin, could you provide an example or a starting point? – NSDigital May 28 '12 at 0:49

You need to do long division, just like you would for polynomials with rational/real coefficients. (We can even do synthetic division!) Or you can do it on the fly:

We have $$f(z) = z^3 + (-6+2j)z^2 + (12-15j)z + (-9+27j)$$

Now, if we write $f(z) = (z-3)(\text{something})$, the something will have to be of degree $2$. So it will be of the form $az^2+bz+d$.

Now, to get the cubic term right, $az^2$ has to be $z^2$. So we will have: $$z^3 + (-6+2j)z^2 + (12-15j)z + (-9+27j) = (z-3)(z^2+\cdots).$$ Now, when you multiply $-3$ times $z^2$, you'll get $-3z^2$. What you actually want is $(-6+2j)z^2$, so that means that the linear term $bz$ should be such that when you multiply by $z$ to get $bz^2$, you'll have $bz^2-3z^2 = (-6+2j)z^2$. So $b$ should be $-3+2j$. So we have: $$z^3 + (-6+2j)z^2 + (12-15j)z + (-9+27j) = (z-3)(z^2 + (-3+2j)z + \cdots).$$ When you multiply $(-3+2j)z$ by $-3$, you'll get $(9-6j)z$; what you want to get is $(12-15j)z$. So the missing $(3-9j)z$ must come from multiplying the $z$ in $(z-3)$ by the constant term. So the constant term must be $3-9j$; that is, $$z^3 + (-6+2j)z^2 + (12-15j)z + (-9+27j) = (z-3)(z^2+(-3+2j)z + (3-9j)).$$ Now verify that we get the "right" constant term: $-3(3-9j) = -9 + 27j$. Good. So we're done.

(You can also do this to write the quotient-with-remainder, by proceeding as above, and then just adding whatever is "left over" to make the equality work.)

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Either the polynomial division algorithm or undetermined coefficients work. For the latter

$$\begin{eqnarray}\rm (z-3)\,(z^2 + ax + b)\, &=&\,\rm \ z^3 &+&\rm (\color{blue}{a\, - 3})\ z^2 &+&\ \rm (\color{brown}{b-\,3a})\ z &\color{green}{-}&\rm \color{green}{3b} \\ &=&\,\rm\ z^3 &+&\rm (\color{blue}{2j - 6})\,z^2 &+&\rm (\color{brown}{12-15j})z &+&\rm \color{green}{27j-9} \end{eqnarray}$$

Hence $\rm\: \color{blue}{a = 2j\!-\!3}\:$ and $\rm\: \color{green}{b = 3\!-\!9j}.\:$ Note that this is equivalent to using Vieta's formulas.

Note $\$ The $\rm\color{brown}{middle}$ coefficients need not be computed here. Generally they will be equal iff $\rm\, x = 3\:$ is a root of the given polynomial, and you already know that. If you didn't know that then you would need to verify that $\rm\:\color{brown}{b-3a = 12-15j}\:$ to ensure that the factorization is correct.

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Long division is not actually needed if you use Vieta's formulas.

Let the roots of $f$ be $3$, $\alpha$, and $\beta$. Then

$3+\alpha+\beta = -(−6+2j)$

$3\alpha\beta = -(−9+27j)$

So the factor you want is $(z-\alpha)(z-\beta)=(z^2-(\alpha+\beta)z-\alpha\beta)=z^2-(3-2j)z+(3-9j)$.

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