What is a good technique for solving polynomials? Say for example:
$6x^{3}-17x^{2}-4x+3=0$
I sort of look at it and don't know where to start, other than just guessing what the first one would be and trying to do from there. Is there a good technique for approaching such polynomials?
 A: First, to minimize the lead coefficient, we reverse the polynomial (which $\rm\color{brown}{reciprocates}$ the roots) and then we scale it to make the leading coefficient $1,\,$ i.e. we apply the AC-method, which yields
$$\begin{eqnarray}\rm f(x) \,&=&\,\rm 3\,x^3 - 4\,x^2-17\,x+ 6 \\
\rm \Rightarrow\ \ 9\,f(x)\, &=&\,\rm (3x)^3 - 4\,(3x)^2 - 51\,(3x) + 54 \\
\rm &=&\rm\ \ z^3 - \color{blue}4\, z^2 - 51\,z + \color{#0A0}{54},\ \ \ z = 3x
\end{eqnarray}$$ 
Note $\rm\:z=\color{#C00}1\:$ is a root. By Vieta, other roots have product $\color{#0A0}{-54/\color{#C00}1},\,$ sum $\color{blue}4\!-\!\color{#C00}1 = 3,\:$ so are $\:9,\,-6.\:$
Hence $\rm\:z\, =\, 1,\,9,\,-6\,\:$ so $\rm\:x\, =\, z/3\, =\, 1/3,\, 3,\, -2,\,$ which $\rm\color{brown}{reciprocated}$ yields $\:3,\,1/3,\, -1/2.$
A: $f(x)= 6x^3 -17x^2 -4x +3$
Always observe the polynomials behavior for different values of x
$f(-1)=-16$
$f(0)=3$
$f(1)=-12$
Here we see that there are two roots of $f(x)=0$ between $-1$ and $1$
Now the constant part of the term $6x^3$ is $6=(1)(2)(3)$
Check for $x=\frac{-1}{3}, \frac{1}{3},\frac{-1}{2}, \frac{1}{2}$
$f(\frac{1}{3})=0$
$f(x)$ is not an even function so discard $x=\frac{-1}{3}$
$f(\frac{-1}{2})=0$
$f(x)$ is not an even function so discard $x=\frac{1}{2}$
we see that two of the roots are $x=\frac{-1}{2}$ and $x=\frac{1}{3}$
Now $f(x)= (x +\frac{1}{2})(6x^2 -20x^2 +6)$
i.e. $f(x)= (x +\frac{1}{2})(x -\frac{1}{3})(6x -18)$
i.e. $f(x)= (6)(x +\frac{1}{2})(x -\frac{1}{3})(x -3)$
i.e. $f(x)= (1)(2)(3)(x +\frac{1}{2})(x -\frac{1}{3})(x -3)$
the solution set for $f(x)=0$ is $(\frac{-1}{2}, \frac{1}{3}, 3)$
A: You can ask Wolfram Alpha  You can use the rational root theorem which works well for class problems (less well for real world problems).  You can do numeric solutions-see any numerical analysis text.  In this case (you didn't say where it came from) the rational root theorem works very well.
