To solve the equation \((32x - 4)(5x^2 + 1) = 0\), we can use the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. Therefore, we set each factor equal to zero and solve for \(x\).

1. Set the first factor equal to zero:
\[
32x - 4 = 0
\]
Solve for \(x\):
\[
32x = 4
\]
\[
x = \frac{4}{32}
\]
\[
x = \frac{1}{8}
\]

2. Set the second factor equal to zero:
\[
5x^2 + 1 = 0
\]
Solve for \(x\):
\[
5x^2 = -1
\]
\[
x^2 = -\frac{1}{5}
\]
Since \(x^2\) cannot be a negative number in the set of real numbers, there are no real solutions for this equation. However, if we consider complex numbers, we can proceed as follows:
\[
x = \pm \sqrt{-\frac{1}{5}}
\]
\[
x = \pm \frac{\sqrt{5}i}{5}
\]
\[
x = \pm \frac{i}{\sqrt{5}}
\]

So, the solutions to the equation \((32x - 4)(5x^2 + 1) = 0\) are:
\[
x = \frac{1}{8}, \quad x = \frac{i}{\sqrt{5}}, \quad \text{and} \quad x = -\frac{i}{\sqrt{5}}
\]