Chromatic polynomial for a bipartite graph I need to get the chromatic polynomial for the complete bipartite graph: $K_{2,3}$
Im using the Fundamental Reduction Theorem, and the picture below shows mi attempt to it.

I omitted vertex names because is just a small graph.
With this procedure i get: $P(K_{2,3},x)=\frac{C_4.K_2}{K_1}-\frac{K_3.K_3}{K_2}$ 
Here $C_4$ is a cycle lenght 4 joined to a complete graph lenght 2  just by one vertex. And is well known that: $P(C_4,x)=x(x-1)(x^2-3x+3)$.
I think im doing well, but the final result is: $x (-3 x^3+12 x^2-16 x+7)$ and is not correct.
The correct result is supposed to be: $x (x-1) (x^3-5 x^2+10 x-7)$
I think im close, but evidently im failing at some point.
 A: The correct answer is $x (x-1) (x^3-5 x^2+10 x-7)$. 
Here is another argument to prove it:
In any coloring of $K_{2,3}$ the two vertices in the first part will have either the same color or two different colors. So, in the first case, there are $x$ colors to choose the same color and $x-1$ colors to choose from to color the remaining three vertices. Thus there are $x(x-1)^3$ colorings of this first kind. 
In the second case, there $x(x-1)$ ways to color the two vertices with different colors and we are left with $x-2$ colors to choose from to color the remaining three vertices. Thus, there are $x(x-1)(x-2)^3$ colorings of this second kind. 
So, in total we have: $x(x-1)^3 + x(x-1)(x-2)^3$ colorings. After simplification you get your answer.
** Edit:
Also your method wields the same result:
So $P(K_{2,3},x)=\frac{P(C_4, x).P(K_2, x)}{P(K_1, x)}-\frac{P(K_3, x).P(K_3, x)}{P(K_2, x)}$ 
and we have:
$P(C_4,x)=x(x-1)(x^2-3x+3)$,
$P(K_3, x)= x(x-1)(x-2)$, 
$P(K_2, x)= x(x-1)$ 
and $P(K_1, x)= x$
So, $P(K_{2,3},x)=\frac{x(x-1)(x^2-3x+3).x(x-1)}{x}-\frac{(x(x-1)(x-2))^2}{x(x-1)}$ 
$P(K_{2,3},x)=x(x-1)^2(x^2-3x+3)- x(x-1)(x-2)^2$ 
$P(K_{2,3},x)=x(x-1)[(x-1)(x^2-3x+3)- (x-2)^2 ]$
$P(K_{2,3},x)=x(x-1)[(x-1)(x^2-3x+3)- (x-2)^2 ]$
This will give you the correct answer. 
