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Given an equation:

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

Broken down to get one form

$(x + {3over2})$

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

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

It looks like you have typo. If you take the polynomial $$p(x)=6 x^3 + 13 x^2 + 8 x + 3,$$ then $-3/2$ is a root.

This guarantees that $x+3/2$ is a factor of $p$. This follows easily from the Division Algorithm: if you write $$p=(x+3/2) q(x)+r(x),$$ with $deg r<deg (x+3/2)=1$, then $r$ is constant; from $p(3/2)=0$ it follows that $r=0$. So now you want to find $q$. You can do it by long division, or simply writing $q(x)=ax^2+bx+c$, and then you have
begin{align}
6 x^3 + 13 x^2 + 8 x + 3&=(x+3/2)(ax^2+bx+c)\
&=ax^3+(3a/2+b)x^2+(c+3b/2)x+3c/2.
end{align}
So $3c/2=3$, and $c=2$. Then $2+3b/2=8$, and $b=4$; finally, $3a/2+4=13$, and $a=6$. Thus
$$
q(x)=6x^2+4x+2.
$$

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 $a_n = 6$ : $pm1$, $pm2$, $pm3$, $pm6$

$frac{p}{q_1}$ : $pm1$, $pm3$

$frac{p}{q_2}$ : $pmfrac{1}{2}$, $pmfrac{3}{2}$

$frac{p}{q_3}$ : $pmfrac{1}{3}$, $pmfrac{3}{3}$

$frac{p}{q_4}$ : $pmfrac{1}{6}$, $pmfrac{3}{6}$

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:

$x = pmfrac{1}{6}$, $pmfrac{1}{3}$, $pmfrac{1}{2}$, $pm1$, $pmfrac{3}{2}$, $pm3$

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 $f(x)$