What's the easiest way to solve the cubic $3x^3-13x^2+3x-13$ 
Solve $3x^3-13x^2+3x-13=0$.

I've never solved a cubic that didn't have three nice (whole number) roots. I see online that there are numerous ways to solve such an equation, but they all look very long-winded. I'd like to know which method is the simplest before attempting any of the methods.
Plotting this cubic shows that it has one real root somewhere around $4.335$ and two complex roots.
 A: Hint:
$$
3x^3-13x^2+3x-13=x^2(3x-13)+(3x-13)=(3x-13)(x^2+1)
$$
A: $$3x^3-13x^2+3x-13=3x^3+3x-13x^2-13\\=3(x^3+x)-13(x^2+1)=3x(x^2+1)-13(x^2+1)\\
=(3x-13)(x^2+1)$$
A: The rational root theorem says that the possible rational real roots are $\pm 13/3$. Noting that $13/3$ is a root, we may divide to get that your polynomial is 
$$(3x-13)(x^2+1)$$
A: \begin{align*}
3x^3-13x^2+3x-13&=0\\
\Rightarrow\left(3x^3-13x^2\right)+(3x-13)&=0\\
\Rightarrow x^2(3x-13)+(3x-13)&=0\\
\Rightarrow (3x-13)\left(x^2+1\right)&=0
\end{align*}
So, either $(3x-13)=0\quad$ or, $\quad\left(x^2+1\right)=0$
Therefore, $x=\dfrac{13}{3}\quad$ or, $\quad x=\pm\sqrt{-1}=\pm i$.
A: 
Question: What's the easiest way to solve the cubic$$3x^3-13x^2+3x-13=0\tag{1}$$

Answer: By factoring. More specifically, you can factor this cubic via grouping. If you look at the coefficients of the cubic closely, you'll notice that the first two terms have a coefficient of $3$ and $-13$. While the last two terms also have $3$ and $-13$.
Using this, it's possible to group the terms together like so
$$\begin{align*}3x^3-13x^2+3x-13 & =x^2\color{blue}{(3x-13)}+\color{blue}{(3x-13)}\tag2\\ & =(3x-13)(x^2+1)\end{align*}$$
Now, set the factors equal to zero and you'll get the roots$$x=\tfrac {13}3\qquad x=\pm\sqrt{-1}$$


Extra:

Sometimes, the cubic will have coefficients that aren't consistent (i.e not of the form $ax^3+bx^2+ax+b$). In that case, you will have to separate a term if you intend to factor by grouping.
Take, for example, the cubic $x^3-4x^2+6x-4$. The coefficients go $1$ and $-4$ for the first two terms, and $6$ and $-4$ for the last two.
Instead, let's split up the $6x$ term such that the first three terms are factorable (quadratic). Testing out values, we find that $4x+2x$ works perfectly fine.$$\begin{align*}x^3-4x^2\color{red}{+6x}-4 & =x^3-4x^2+\color{red}{4x+2x}-4\\ & =x(x-2)^2+2(x-2)\\ & =(x-2)(x^2-2x+2)\end{align*}$$
