Transform the following expression into the product of two polynomials:

$$
x^{4}+x^{3}+x^{2}+x+1
$$
I. Solution. We are looking for a factorization where both factor polynomials are quadratic and the coefficient of $x^{2}$ in both is 1. Our task is to determine the further coefficients $p, q$, and $r, s$ such that

$$
\begin{aligned}
\left(x^{2}+p x+q\right) \cdot\left(x^{2}+r x+s\right) & =x^{4}+(p+r) x^{3}+(q+p r+s) x^{2}+(q r+p s) x+q s= \\
& =x^{4}+x^{3}+x^{2}+x+1
\end{aligned}
$$

is an identity. This holds if the coefficients of $x^{3}, x^{2}, x$, and the constant term are equal on both sides, that is, if

$$
\begin{array}{r}
p+r=1 \\
q r+p s=1
\end{array}
$$

$$
\begin{array}{r}
q+p r+s=1 \\
q s=1
\end{array}
$$

Multiplying (3) by $q$ and using (1) and (4), we can eliminate $r$ and $s$. From (1), $r=1-p$, so

$$
\begin{gathered}
q^{2} r+q p s=q^{2}(1-p)+p=q \\
q^{2}-q-p\left(q^{2}-1\right)=(q-1)(q-p q-p)=0
\end{gathered}
$$

This is always true if $q=1$, in which case (4) gives $s=1$, (2) gives $p r=-1$, and (1) gives $p$ and $r$ as the roots of the equation

$$
(z-p)(z-r)=z^{2}-(p+r) z+p r=z^{2}-z-1=0
$$

which are $\frac{1+\sqrt{5}}{2}$ and $\frac{1-\sqrt{5}}{2}$. With these, the identity

$$
P(x)=x^{4}+x^{3}+x^{2}+x+1=\left(x^{2}+\frac{1+\sqrt{5}}{2} x+1\right)\left(x^{2}-\frac{\sqrt{5}-1}{2} x+1\right)
$$

holds. This gives us a factorization of the desired form.

Remarks. 1. A factorization into a linear and a cubic polynomial with real coefficients does not exist, because otherwise $P(x)$ would have a real root, and one of the factors in the found factorization would also have a root. However, the discriminant of both factors is negative, so neither has a real root.

2. It is clear that, with the choice $q=1$, the only other factorization into two quadratic polynomials with real coefficients that can be obtained is one where one factor is multiplied by a non-zero constant $c$ and the other by $1/c$. However, if we start from the equation $q-p q-p=0$, then (2) gives a quartic equation in $p$, which can be factored into quadratics by completing the square, but one of these quadratics has no real roots. Thus, there is no essentially different factorization into quadratic polynomials with real coefficients.

II. Solution. The polynomial can be rearranged in terms of powers of $x^{2}+1$:

$$
P(x)=x^{4}+1+x\left(x^{2}+1\right)+x^{2}=\left(x^{2}+1\right)^{2}+x\left(x^{2}+1\right)-x^{2}
$$

Complete the square with the first two terms to get a difference of squares, which can then be factored:

$$
P(x)=\left(x^{2}+1+\frac{1}{2} x\right)^{2}-\frac{5}{4} x^{2}=\left(x^{2}+\frac{1+\sqrt{5}}{2} x+1\right) \cdot\left(x^{2}+\frac{1-\sqrt{5}}{2} x+1\right)
$$

Remarks. 1. This method is generally applicable to polynomials of the form $x^{4}+a x^{3}+b x^{2}+a x+1$, known as symmetric polynomials, and does not differ significantly from the usual method, which involves expressing the polynomial as $\frac{1}{x^{2}}$ times a new variable $y=x+\frac{1}{x}$.

Following this path, we always arrive at a factorization into two quadratic polynomials with real coefficients if $a^{2}-4 b+8 \geqq 0$.

2. The thought process of the first solution can also be applied to the more general case mentioned, and it leads to the goal regardless of the values of the coefficients.