The problem states that h(-(1)/(2)) = 0. According to the Factor Theorem, if h(c)=0, then (x-c) is a factor of h(x).

The problem states that $h\left(-\frac{1}{2}\right) = 0$. According to the Factor Theorem, if $h(c)=0$, then $(x-c)$ is a factor of $h(x)$.
Here, $c = -\frac{1}{2}$, so $\left(x - \left(-\frac{1}{2}\right)\right) = \left(x + \frac{1}{2}\right)$ is a factor.
To avoid fractions, we can multiply this factor by 2, which means $(2x+1)$ is also a factor.

Let's verify $h\left(-\frac{1}{2}\right)$:

$$h\left(-\frac{1}{2}\right) = 4\left(-\frac{1}{2}\right)^3 + 4\left(-\frac{1}{2}\right)^2 + 73\left(-\frac{1}{2}\right) + 36$$ $$h\left(-\frac{1}{2}\right) = 4\left(-\frac{1}{8}\right) + 4\left(\frac{1}{4}\right) - \frac{73}{2} + 36$$ $$h\left(-\frac{1}{2}\right) = -\frac{1}{2} + 1 - \frac{73}{2} + 36$$ $$h\left(-\frac{1}{2}\right) = \frac{1}{2} - \frac{73}{2} + 36$$ $$h\left(-\frac{1}{2}\right) = -\frac{72}{2} + 36$$ $$h\left(-\frac{1}{2}\right) = -36 + 36 = 0$$

This confirms that $(2x+1)$ is a factor of $h(x)$.

We can use polynomial long division:

$$\begin{array}{c|cc cc} \multicolumn{2}{r}{2x^2} & +x & +36 \\ \cline{2-5} 2x+1 & 4x^3 & +4x^2 & +73x & +36 \\ \multicolumn{2}{r}{-(4x^3} & +2x^2) \\ \cline{2-3} \multicolumn{2}{r}{0} & 2x^2 & +73x \\ \multicolumn{2}{r}{} & -(2x^2 & +x) \\ \cline{3-4} \multicolumn{2}{r}{} & 0 & 72x & +36 \\ \multicolumn{2}{r}{} & & -(72x & +36) \\ \cline{4-5} \multicolumn{2}{r}{} & & 0 & 0 \\ \end{array}$$

The quotient is $2x^2 + x + 36$.
So, $h(x) = (2x+1)(2x^2 + x + 36)$.

We check the discriminant $\Delta = b^2 - 4ac$ for the quadratic $ax^2+bx+c$.
For $2x^2 + x + 36$, we have $a=2$, $b=1$, $c=36$.

$$\Delta = (1)^2 - 4(2)(36)$$ $$\Delta = 1 - 288$$ $$\Delta = -287$$

Since the discriminant $\Delta < 0$, the quadratic factor $2x^2 + x + 36$ has no real roots and cannot be factored further into linear factors with real coefficients.

Therefore, the factorization of $h(x)$ over real numbers is:

$$(2x+1)(2x^2+x+36)$$