Answer
$(x-3)(x+2)(2x-1)$.
Work Step by Step
The given dividend $=2x^3-3x^2-11x+6$.
Divisor $=x-3$
$\begin{matrix}
& 2x^2 & +3x &-2 & & \leftarrow &Quotient\\
&-- &-- &--&--& \\
x-3) &2x^3&-3x^2&-11x&+6 & \\
& 2x^3 & -6x^2 & & & \leftarrow &2x^2(x-3) \\
& -- & -- & & & \leftarrow &subtract \\
& 0 & 3x^2 & -11x & & \\
& & 3x^2 & -9x & & \leftarrow & 3x(x-3) \\
& & -- & -- & & \leftarrow & subtract \\
& & 0&-2x &+6 & \\
& & & -2x& +6 & \leftarrow & -2(x-3) \\
& & & -- & -- & \leftarrow & subtract \\
& & & 0 & 0 & \leftarrow & Remainder
\end{matrix}$
The Quotient is $2x^2+3x-2$. We can say that $(x-3)$ is a factor of $2x^3-3x^2-11x+6$ because the remainder is $0$.
Now factor $2x^2+3x-2$.
Rewrite the middle term $3x$ as $4x-x$.
$\Rightarrow 2x^2+4x-x-2$
Group terms.
$\Rightarrow (2x^2+4x)+(-x-2)$
Factor each group.
$\Rightarrow 2x(x+2)-1(x+2)$
Factor out $x+2$.
$\Rightarrow (x+2)(2x-1)$
The complete factors are:
$=(x-3)(x+2)(2x-1)$.
