51. Find all integers $a$ for which the polynomial

$$
y=x^{4}-3 x^{3}+a x^{2}-9 x-2
$$

can be factored into the product of two quadratic trinomials with integer coefficients.
$\triangle$ The coefficients of $x^{2}$ in both quadratic trinomials can be set to 1. As for the constant terms, since their product is -2, they are either 2 and -1, or 1 and -2.

Consider the two cases.

1) Suppose the constant terms of the quadratic trinomials are 2 and -1. We get:

$$
x^{4}-3 x^{3}+a x^{2}-9 x-2=\left(x^{2}+p x+2\right)\left(x^{2}+q x-1\right)
$$

where the numbers $p$ and $q$ are integers. Expand the brackets on the right side of this identity:

$$
x^{4}-3 x^{3}+a x^{2}-9 x-2=x^{4}+(p+q) x^{3}+(1+p q) x^{2}+(2 q-p) x-2
$$

By the theorem of the identity of two polynomials, we have:

$$
p+q=-3, \quad 1+p q=a, \quad 2 q-p=-9
$$

From the first and third equations of this system, we find $p$ and $q: p=1, q=-4$. From the second equation, we find $a: a=1-4=-3$.

2) Suppose the constant terms of the quadratic trinomials are 1 and -2. We get:

$$
\begin{gathered}
x^{4}-3 x^{3}+a x^{2}-9 x-2=\left(x^{2}+p x+1\right)\left(x^{2}+q x-2\right) \\
x^{4}-3 x^{3}+a x^{2}-9 x-2=x^{4}+(p+q) x^{3}+(p q-1) x^{2}+(q-2 p) x-2
\end{gathered}
$$

Then

$$
p+q=-3, \quad p q-1=a, \quad q-2 p=-9
$$

From this,

$$
p=2, \quad q=-5, \quad a=-11
$$

Answer: $-3,-11$.