80. Given the polynomial $x^{4}+3 x^{3}-15 x^{2}-19 x+30$. Factorize it, knowing that all its roots are integers.
Solution. We will look for the factorization in the form

$$
x^{4}+3 x^{3}-15 x^{2}-19 x+30=(x-a)(x-b)(x-c)(x-d)
$$

assuming that the numbers $a, b, c$, and $d$ are its roots. Expanding the brackets on the right-hand side, we have:

$$
x^{4}+3 x^{3}-15 x^{2}-19 x+30=x^{4}-(a+b+c+d) x^{3}+
$$

$$
+(a b+a c+a d+b c+b d+c d) x^{2}-(a b c+a b d+a c d+b c d) x+a b c d
$$

From which we get

$$
\left\{\begin{array}{l}
a+b+c+d=-3 \\
a b+a c+a d+b c+b d+c d=-15 \\
a b c+a b d+a c d+b c d=19 \\
a b c d=30
\end{array}\right.
$$

We will solve the system (1) by reasoning as follows. Since the roots of our polynomial are integers, we conclude from the last equation of the system that they must be divisors of the number 30. Therefore, they should be sought among the numbers $\pm 1, \pm 2, \pm 3, \pm 5, \pm 6, \pm 10$, $\pm 15, \pm 30$. By conducting trials, we will establish that the roots of our polynomial are $-2,-5,1$ and 3. Therefore,

$$
x^{4}+3 x^{3}-15 x^{2}-19 x+30=(x-1)(x-3)(x+2)(x+5) .
$$

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