According to Vieta's theorem, the coefficients of the new quadratic polynomial are $-(p+q)$ and $pq$ respectively. Note that the second coefficient (of $x$) of none of the written quadratic polynomials can be zero. Indeed, in this case, the constant term of all subsequent quadratic polynomials would be zero, which means that the original polynomial could not be among them. Further, for each subsequent quadratic polynomial, the absolute value of the constant term is not less than that of the previous one. Therefore, the absolute values of all constant terms are equal, which means that all second coefficients are $\pm 1$. Thus, $-(p+q) = \pm 1, p = \pm 1$. Therefore, for the first quadratic polynomial, there are only two possible options: $x^{2}+x-2$ and $x^{2}-x+2$. It is easy to check that in the first case, the quadratic polynomial does not change, while in the second case, it first becomes the quadratic polynomial $x^{2}-x-2$, and then the quadratic polynomial $x^{2}+3x+2$, the absolute value of the second coefficient of which is not equal to one.

## Answer

$x^{2}+x-2$

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