4B. Factor the polynomial into prime factors

$$
P(x)=x^{4}+x^{3}-8 x^{2}+3 x+5
$$
Solution. Let's transform the form of the polynomial

$$
\begin{aligned}
P(x) & =x^{4}+2 x^{3}-5 x^{2}-\left(x^{3}+2 x^{2}-5 x\right)-\left(x^{2}+2 x-5\right) \\
& =x^{2}\left(x^{2}+2 x-5\right)-x\left(x^{2}+2 x-5\right)-\left(x^{2}+2 x-5\right) \\
& =\left(x^{2}-x-1\right)\left(x^{2}+2 x-5\right)
\end{aligned}
$$

The zeros of the polynomial are

$$
x_{1 / 2}=\frac{1 \pm \sqrt{5}}{2} \quad \text { and } \quad x_{3 / 4}=-1 \pm \sqrt{6}
$$

According to this,

$$
P(x)=\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)\left(x-x_{4}\right) \text {. }
$$