<s>[INST] Q. Reducible polynomials in $\mathbb{Z}[x]$ Show that $x^3+ax^2+bx+1$ $\in \mathbb Z[x]$ is reducible on $\mathbb{Z}$  if and only if  $a=b$ or $a+b=-2$.
If it is reducible, then it has root in $\mathbb Z$. Be $u$ the root, so i can write it as:
$x^3+ax^2+bx+1=(x-u)(x^2+cx+d)$
That developing the right expression, I get my conditions.
To make the reverse path, can I simply assume a factorization and suppose conditions to reach my original polynomial? 
 [/INST] If $x^3 + ax^2 + bx + 1$ is reducible in $\Bbb Z[x]$, then we have
$x^3 + ax^2 + bx + 1 = (x^2 + cx + d)(x + e)$
$= x^3 + ex^2 + cx^2 + ecx + dx + ed = x^3 + (c + e)x^2 + (ec + d)x + ed; \tag 1$
we know that $x^3 + ax^2 + bx + 1$ must factor in this form since (i.) a cubic, if reducible, may always be written as the product of a linear and a quadratic polynomial; (ii.) the product of the leading coefficients of the factors is $1$.  Choosing each leading coefficient of the factors to be $1$ is possible since this case can always be obtained by reversing the signs of the factors.
Then, comparing coefficients of like terms,
$c + e = a, \tag 2$
$ec + d = b, \tag 3$
$ed = 1; \tag 4$
therefore,
$e = d = 1 \vee e = d = -1; \tag 5$
$e = d = 1 \Longrightarrow c + 1 = b, \; c + 1 = a \Longrightarrow a = b; \tag 6$
$e = d = -1 \Longrightarrow -c - 1 = b, \; c - 1 = a \Longrightarrow a + b = -2; \tag 7$
we can run the logic engine in reverse; if 
$a = b, \tag 8$
we simply choose
$e = d = 1, \; c = a - 1; \tag 9$
then it is easy to check that
$(x^2 + (a - 1)x + 1)(x + 1) = x^3 + x^2 + (a - 1)x^2 + (a - 1)x + x + 1$
$= x^3 + x^2 + ax^2 - x^2 + ax - x + x + 1 = x^3 + ax^2 + ax + 1$
$= x^3 + ax^2 + bx + 1; \tag{10}$ 
if $a + b = -2$, then we pick
$e = d = -1, \; c = a + 1 = -b - 1; \tag{11}$
thus,
$(x^2 + (a + 1)x - 1)(x - 1) = x^3 - x^2 + (a + 1)x^2 - (a + 1)x - x + 1$
$= x^3 - x^2 + ax^2 + x^2 - ax - x - x + 1 = x^3 + ax^2 - (a + 2)x + 1$
$= x^3 + ax^2 + bx + 1. \tag{12}$ 
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