Example 5 Find all real roots of the equation
$$
x^{2}-x+1=\left(x^{2}+x+1\right)\left(x^{2}+2 x+4\right)
$$

All real roots. ${ }^{[4]}$
(2011, International Invitational Competition for Young Mathematicians in Cities)
【Analysis】The most basic method to solve higher-degree equations is to convert them into lower-degree equations for solving, that is, to handle them as linear or quadratic equations. Factorization is the most powerful tool to achieve such a transformation. The preferred method for factoring higher-degree polynomials is the trial root method (synthetic division). After rearranging the equation into $f(x)=0$, it is found that all coefficients are positive. If there is a rational root, it must be a negative root. The leading coefficient is 1, and the constant term is 3. If there is a rational root, it can only be -1 or -3.

Solution: The original equation can be transformed into
$$
x^{4}+3 x^{3}+6 x^{2}+7 x+3=0.
$$

It is easy to see that $x=-1$ is a solution of the equation.
$$
\begin{array}{l}
\text { Therefore, } x^{4}+3 x^{3}+6 x^{2}+7 x+3 \\
=(x+1)\left(x^{3}+2 x^{2}+4 x+3\right) \\
=(x+1)^{2}\left(x^{2}+x+3\right)=0.
\end{array}
$$

Notice that, $x^{2}+x+3=\left(x+\frac{1}{2}\right)^{2}+\frac{11}{4}>0$.
Therefore, the equation has a unique real root -1.