1. From the condition $\left(\frac{p_{2}}{p_{1}}\right)^{\prime}=2 x \rightarrow p_{2}(x)=\left(x^{2}+c\right) p_{1}(x), p_{1}+p_{2}=p_{1}(x)\left(x^{2}+c+1\right)=(x-1)(x+2)\left(x^{2}-3\right)$

Since $p_{1}(x)$ is a polynomial, there are two possible cases

Case 1

$\left(x^{2}+c+1\right)=(x-1)(x+2)=x^{2}+x-2$ is impossible

Case 2

$\left(x^{2}+c+1\right)=x^{2}-3 \rightarrow c=-4$. Then $p_{1}=(x-1)(x+2)$ and $p_{2}=\left(x^{2}-4\right)(x-1)(x+2)$
Answer: $p_{1}=(x-1)(x+2), p_{2}=\left(x^{2}-4\right)(x-1)(x+2)$