1. Instead of the ellipsis signs, insert such numbers so that the expression $\left(x^{2}+\ldots \times x+2\right) \times(x+3)=(x+\ldots) \times\left(x^{2}+\ldots \times x+6\right)$ becomes an identity.
Answer. $\left(x^{2}+3 x+2\right)(x+3)=(x+1)\left(x^{2}+5 x+6\right)$

Solution. Let the unknown coefficients be $a, b$, respectively:

$\left(x^{2}+a x+2\right)(x+3)=(x+b)\left(x^{2}+c x+6\right) \quad$ and bring $\quad$ the polynomials in the left and right parts to the standard form:

$$
x^{3}+(a+3) x^{2}+(3 a+2) x+6=x^{3}+(b+c) x^{2}+(b c+6) x+6 b
$$

This equality will be an identity if and only if the equalities $6 b=6 ; b c+6=3 a+2 ; b+c=a+3$ are simultaneously satisfied. Solving the corresponding system of equations, we get that $b=1 ; a=3 ; c=5$.