已知多项式 $\mathrm{f}(x)$ 为 $4x^3 - 6x^2 + kx + 3$，其中 $k$ 为常数。

1. Apply the Remainder Theorem by substituting $x = \frac{1}{2}$ into $\mathrm{f}(x)$ and equating to 14.

By the Remainder Theorem, $\mathrm{f}\left(\frac{1}{2}\right) = 14$.
$4\left(\frac{1}{2}\right)^3 - 6\left(\frac{1}{2}\right)^2 + k\left(\frac{1}{2}\right) + 3 = 14$

2. Simplify the equation and solve for $k$.

$4\left(\frac{1}{8}\right) - 6\left(\frac{1}{4}\right) + \frac{k}{2} + 3 = 14$
$\frac{1}{2} - \frac{3}{2} + \frac{k}{2} + 3 = 14$
$-1 + \frac{k}{2} + 3 = 14$
$\frac{k}{2} + 2 = 14$
$\frac{k}{2} = 12$
$k = 24$

1. Set up the polynomial identity with an unknown linear factor

Since $2x^2 + px + 3$ is a quadratic factor of the cubic polynomial $\mathrm{f}(x) = 4x^3 - 6x^2 + 2x + 3$, the other factor must be linear. Let it be $ax + b$.
$4x^3 - 6x^2 + 2x + 3 = (2x^2 + px + 3)(ax + b)$

2. Determine the coefficients of the linear factor

Comparing the coefficients of $x^3$:
$4 = 2a \implies a = 2$
Comparing the constant terms:
$3 = 3b \implies b = 1$
So, the identity is:
$4x^3 - 6x^2 + 2x + 3 = (2x^2 + px + 3)(2x + 1)$

3. Compare coefficients of $x^2$ (or $x$) to find $p$

Comparing the coefficients of $x^2$:
$-6 = 2(1) + p(2)$
$-6 = 2 + 2p$
$2p = -8$
$p = -4$

(Check by comparing coefficients of $x$: $2 = p(1) + 3(2) \implies 2 = p + 6 \implies p = -4$)