Use the remainder theorem to factorize the expression $\;$ $2x^3 + x^2 - 13x + 6$
Let $\;$ $f \left(x\right) = 2x^3 + x^2 - 13x + 6$
For $x = 2$, the value of the given expression is:
$f \left(2\right) = 2 \times 2^3 + 2^2 - 13 \times 2 + 6 = 16 + 4 - 26 + 6 = 0$
$\implies$ $\left(x - 2\right)$ is a factor of $f \left(x\right)$.
$\begin{array}{lllll}
x - 2 & ) & 2x^3 + x^2 - 13x + 6 & ( & 2x^2 + 5x - 3 \\
& & 2x^3 - 4x^2 & & \\
& & - - - - - - - - - - - - & & \\
& & \ \ \ \ \ \ \ \ 5x^2 - 13x & & \\
& & \ \ \ \ \ \ \ \ 5x^2 - 10x & & \\
& & - - - - - - - - - - - - & & \\
& & \ \ \ \ \ \ \ \ \ \ \ \ \ \ -3x + 6 & & \\
& & \ \ \ \ \ \ \ \ \ \ \ \ \ \ -3x + 6 & & \\
& & - - - - - - - - - - - - & & \\
& & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0
\end{array}$
$\begin{aligned}
\therefore \; 2x^3 + x^2 - 13x + 6 & = \left(x - 2\right) \left[2x^2 + 5x - 3\right] \\\\
& = \left(x - 2\right) \left[2x^2 + 6x - x - 3\right] \\\\
& = \left(x - 2\right) \left[2x \left(x + 3\right) - 1 \left(x + 3\right)\right] \\\\
& = \left(x - 2\right) \left(2x - 1\right) \left(x + 3\right)
\end{aligned}$