<s>[INST] Simplify third degree polynomial equations. Given an equation: 

$6x^3 - 13x^2 + 8x + 3 = 0$

Broken down to get one form

$(x + {3\over2})$

How can you divide the prior equation to know it will simplify to

$(x + {3\over2})(6x^2 + 4x + 2) = 0$

 [/INST] Following @MartinArgerami's discovery that there was probably a typo.
if we have a degree of three polynomial we can use the Rational Zero Theorem as such:

Ration Zero Theorem: For $f(x) = a_nx^n + a_{n-1}x^{n-1}+...+a_2x^2 + a_1x + a_0$, every rational zero of $f(x)$, $\frac{p}{q}$ has $p$ as a factor of $a_0$ and $q$ as a factor of $a_n$

We have the polynomial $f(x) = 6x^3 + 13x + 8x + 3$
p) Factors of the constant $a_0 = 3$ : $\pm 1$, $\pm3$
q) Factors of the coefficient $a_n = 6$ : $\pm1$, $\pm2$, $\pm3$, $\pm6$
$\frac{p}{q_1}$ : $\pm1$, $\pm3$ 
$\frac{p}{q_2}$ : $\pm\frac{1}{2}$, $\pm\frac{3}{2}$ 
$\frac{p}{q_3}$ : $\pm\frac{1}{3}$, $\pm\frac{3}{3}$
$\frac{p}{q_4}$ : $\pm\frac{1}{6}$, $\pm\frac{3}{6}$
We can see that some of these possible zeros are repeated so we can cross some of them out. Our final possible zero list is the following:
$x = \pm\frac{1}{6}$, $\pm\frac{1}{3}$, $\pm\frac{1}{2}$, $\pm1$, $\pm\frac{3}{2}$, $\pm3$ 
Using synthetic division we try these possible factors until we find one. Let's try $-3$.
When we try $-3$, we are saying, let us see if $(x+3)$ is a factor of $f(x)$
$\begin{array}{c|rrr}&6&13&8&3\\-3&&-18&15&-69\\\hline\\&6&-5&23&-66\\\end{array}$
We have a remainder of $-66$ so $-3$ is not a factor. We are looking for a remainder of $0$. For the sake of time, let us try $\frac{-3}{2}$ since you have found already that it is a factor.
$\begin{array}{c|rrr}&6&13&8&3\\-\frac{3}{2}&&-9&-6&-3\\\hline\\&6&4&2&0\\\end{array}$
We now have a $0$ remainder and $-\frac{3}{2}$ is a zero.
Thus, $f(x)$ can be written like this:  
$f(x) = divisor(x) \times quotient(x) + remainder(x) = d(x)q(x) + r(x) \\
= (x+\frac{3}{2})(6x^2+4x+2)$
you can further factor the quotient to find the remaining zeros of $f(x)$
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