Pre Calculus Synthetic Division two of the roots of the equation $2x^3-3x^2+px+q=0$ are $3$ and $-2$. Find the third root of the equation.
 A: Hint:
Let $f(x) = 2x^{3} - 3x^{2} + px + q$. 
Since there are $3$ roots to a cubic equation, and $3$ and $-2$ are roots of $f(x)$, we can factor $f(x)$ as
$f(x) = (2x - k)(x-3)(x+2)$ 
where $k/2$ is the third root of $f(x)$. 
Since $f(x) = 2x^{3} - 3x^{2} + px + q$, we can expand the previous factorization of $f(x)$ to get linear equations which will allow us be able to solve for $k$. 
A: The cubic must factor as $2(x-r_1)(x-r_2)(x-r_3)$ and the coefficient of $x^2$ is $2 \cdot (-3/2).$ So the sum of the roots is $3/2$ and you know two roots are $3,-2$ with sum $1$, forcing the third root to be $1/2$.
Note that we do not have to find $p,q$ to use this method, since the sum of the roots of a monic cubic gives the negative of the coefficient of $x^2$, and the given equation may be divided by $2$ to be
$$x^3-(3/2)x^2+(p/2)x+(q/2)=0.$$
A: $\newcommand{\+}{^{\dagger}}%
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$\ds{{\rm f}\pars{x} = 2x^{3} - 3x^{2} + px + q}$.

${\rm f}\pars{x}$ is of the form
${\rm f}\pars{x} = 2\pars{x - \tilde{x}}\pars{x - 3}\bracks{x - \pars{-2}}
= 2\pars{x - \tilde{x}}\pars{x - 3}\pars{x + 2}$ where
$\tilde{x}$ is the unknown root.
$$
{\rm f}\pars{0} = 12\tilde{x}
\quad\imp\quad
\color{#0000ff}{\large\tilde{x}}
=
{{\rm f}\pars{0} \over 12} = \color{#0000ff}{\large{q \over 12}}
\quad\imp\quad
$$
Then ${\rm f}\pars{x} = 2\pars{x - q/12}\pars{x - 3}\pars{x + 2}$ 

Moreover
\begin{align}
{{\rm f}'\pars{x} \over {\rm f}\pars{x}}
&=
{1 \over x - q/12} + {1 \over x - 3} + {1 \over x + 2}
\\[3mm]
{{\rm f}''\pars{x} \over {\rm f}\pars{x}} - {{\rm f}'^{2}\pars{x}
\over
{\rm f}^{2}\pars{x}}
&=
-\,{1 \over \pars{x - q/12}^{2}} - {1 \over \pars{x - 3}^{2}}
-
{1 \over \pars{x + 2}^{2}}
\end{align}
Evaluating those formulas at $x = 0$; we get ${\rm f}\pars{0} = q$,
${\rm f}'\pars{0} = p$ and ${\rm f}''\pars{0} = -6$

\begin{align}
{p \over q} & = -\,{12 \over q} - {1 \over 3} + {1 \over 2} 
=
-\,{12 \over q} + {1 \over 6}
\\[3mm]
-\,{6 \over q} - {p^{2} \over q^{2}}
&=
-\,{144 \over q^{2}} - {1 \over 9} - {1 \over 4}=
-\,{144 \over q^{2}} - {13 \over 36}
\end{align}
Those equations are reduced to
$$
-\,{6 \over q} - \pars{-\,{12 \over q} + {1 \over 6}}^{2}
=
-\,{144 \over q^{2}} - {13 \over 36}\
\imp\
-\,{6 \over q} + {4 \over q} - {1 \over 36} 
=
-\,{13 \over 36}\
\imp\
-\,{2 \over q} = -\,{1 \over 3}
$$

Then, $q = 6$ and $p = -12 + q/6 = -11$. 
\begin{align}
&\mbox{The roots are at}\ 3, -2\ \mbox{and}\
\color{#0000ff}{\large\tilde{x} = {1 \over 2}}
\\[3mm]
&\mbox{The polinomia is given by}\ \color{#ff0000}{\large 2x^{3} - 3x^{2} - 11x + 6}
\end{align}
