Let's solve the given equation:

\[
(x^2 + 1)(x - 2) + 2x = 4
\]

First, expand \((x^2 + 1)(x - 2)\):

\[
(x^2 + 1)(x - 2) = x^2(x) - x^2(2) + 1(x) - 1(2)
\]

\[
= x^3 - 2x^2 + x - 2
\]

So the equation becomes:

\[
x^3 - 2x^2 + x - 2 + 2x = 4
\]

Combine like terms:

\[
x^3 - 2x^2 + 3x - 2 = 4
\]

Subtract 4 from both sides:

\[
x^3 - 2x^2 + 3x - 6 = 0
\]

We need to solve this cubic equation. It looks like a factorable equation, so let's try to factor it.

To factor \(x^3 - 2x^2 + 3x - 6\), test potential rational roots using the Rational Root Theorem, which suggests that any rational root is a factor of the constant term divided by a factor of the leading coefficient.

Potential rational roots could be: \( \pm 1, \pm 2, \pm 3, \pm 6 \).

Testing these:

1. \(x = 1\):
\[
1^3 - 2(1)^2 + 3(1) - 6 = 1 - 2 + 3 - 6 = -4 \quad \text{(not a root)}
\]
2. \(x = 2\):
\[
2^3 - 2(2)^2 + 3(2) - 6 = 8 - 8 + 6 - 6 = 0 \quad \text{(root)}
\]

So, \(x = 2\) is a root. Now, we can divide \(x^3 - 2x^2 + 3x - 6\) by \(x - 2\) to find the other factors.

Use synthetic division:

\[
\begin{array}{r|rrrr}
2 & 1 & -2 & 3 & -6 \\
& & 2 & 0 & 6 \\
\hline
& 1 & 0 & 3 & 0 \\
\end{array}
\]

The quotient is \(x^2 + 3\), and the remainder is 0. Thus, the factorization is:

\[
(x - 2)(x^2 + 3) = 0
\]

Set each factor to zero and solve:

1. \(x - 2 = 0 \rightarrow x = 2\)
2. \(x^2 + 3 = 0 \rightarrow x^2 = -3 \rightarrow x = \pm \sqrt{-3} = \pm i\sqrt{3}\)

Thus, the solutions to the equation are:

\[
x = 2, \quad x = i\sqrt{3}, \quad \text{and} \quad x = -i\sqrt{3}
\]