What is the solution of the equation x^3+3x^2+4x+2=0, if it is integer?

We'll re-write the equation, keeping the first 3 terms to the
left side. For this teason, we'll move 2 to the right
side:

x^3+3x^2+4x = -2

We'll factorize
by x to the left side:

x(x^2 + 3x + 4) =
-2

If the solution of the equation is an integer number,x, it has to
divide -2, as well.

We'll identify the divisors of
-2:

D = -2 ; -1 ; 1 ; 2

We'll
substitute the divisor into equation.

For x =
-1

-1*[(-1)^2 + 3(-1) + 4] =
-2

-1*(4-2) = -2

-2 =
-2

x = -1 is the solution of the
equation.

We'll put x = -2:

-2*[(-2)^2
+ 3(-2) + 4] = -2

-2(4 - 6 + 4) =
-2

-2*2 = -2

-4 = -2
impossible!

x = -2 is not a solution for the
equation!

We'll put x = 1

1*[(1)^2 +
3*(1) + 4] = -2

8 = -2 impossible

x =
2

2*[(2)^2 + 3*(2) + 4] = -2

28 = -2
impossible

We notice that for positive values of x, the expression
has values bigger than -2.

The only integer solution
for the equation is x = -1.