Q. Reducible polynomials in $\mathbb{Z}[x]$

Question

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?

Answer 1

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,

$(x^2 + (a + 1)x - 1)(x - 1) = x^3 - x^2 + (a + 1)x^2 - (a + 1)x - x + 1$ $= x^3 - x^2 + ax^2 + x^2 - ax - x - x + 1 = x^3 + ax^2 - (a + 2)x + 1$ $= x^3 + ax^2 + bx + 1. \tag{12}$

Answer 2

$x^3+ax^2+bx+1\in\mathbb{Z}[x]$ is of degree $3$.

If this polynomial is reducible, it must have a root in $\mathbb{Z}$ which means you can factor the polynomial like this:

$x^3+ax^2+bx+1=(x^2+cx+d)(x+e)$

Factor out the LHS and compare the coefficients.

Answer 3

We can factorize this polynomial in $\mathbb{Z}$ iff $1$ or $-1$ is one of it zeros, so

1. case $x_1=1$, by Vieta formula we have $$ 1+x_2+x_3=-a$$ $$x_2x_3+x_2+x_3 = b$$ $$x_2x_3 = -1$$ so $$-1+(-a-1) = b\implies a+b=-2$$

2. case $x_1=-1$, by Vieta formula we have $$ -1+x_2+x_3=-a$$ $$x_2x_3-x_2-x_3 = b$$ $$x_2x_3 = 1$$ so $$1+(-a+1) = b\implies a=b$$