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

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We have to find the integral solution of x^3 + 3x^2 + 4x + 2=0

x^3 + 3x^2 + 4x + 2=0

=> x^3 + x^2 + 2x^2 + 2x + 2x + 2 = 0

=> x^2(x + 1) + 2x(x + 1) + 2(x + 1) = 0

=> (x + 1)(x^2 + 2x + 2) = 0

x + 1 = 0

=> x = -1

For x^2 + 2x + 2 = 0, we see that b^2 < 4ac as 4 < 8, so the equation only has complex roots.

**The integral solution of the equation is only x = -1**

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.