First, consider the expansion of \(\sqrt{1 + 4x} = (1 + 4x)^{1/2}\).

Using the binomial expansion formula, the general term is given by:

\(\binom{1/2}{n} (4x)^n = \frac{1}{2} \cdot \frac{1}{2} - 1 \cdot \frac{1}{2} - 2 \cdots \frac{1}{2} - (n-1) \cdot \frac{(4x)^n}{n!}\)

For \(x^3\), we need the term where \(n = 3\):

\(\frac{1}{2} \cdot \frac{-1}{2} \cdot \frac{-3}{2} \cdot \frac{(4x)^3}{6} = \frac{1}{2} \cdot \frac{-1}{2} \cdot \frac{-3}{2} \cdot \frac{64x^3}{6} = 4x^3\)

Next, consider the term in \(x^2\) for \((1 + 4x)^{1/2}\):

\(\frac{1}{2} \cdot \frac{-1}{2} \cdot \frac{(4x)^2}{2} = \frac{1}{2} \cdot \frac{-1}{2} \cdot 16x^2 = -2x^2\)

Now, multiply \((3 + x)\) by these terms:

\((3 + x)(4x^3) = 12x^3 + 4x^4\)

\((3 + x)(-2x^2) = -6x^2 - 2x^3\)

Combine the \(x^3\) terms:

\(12x^3 - 2x^3 = 10x^3\)

Thus, the coefficient of \(x^3\) is 10.