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

justaguide | College Teacher | (Level 2) Distinguished Educator

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

Now 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

For x +1 = 0, x = -1.

The equation x^2 + 2x + 2  =0 does not have any real roots. This follows from the fact that for a quadratic equation ax^2 + bx +c = 0 to have real roots, b^2- 4ac should not be less than 0. Here we see that 2^2 - 4*2*1  = -4 which is less than 0. Therefore it only has two complex roots.

Therefore the only integral root of root x^3+3x^2+4x+2 =0 is -1.

neela | High School Teacher | (Level 3) Valedictorian

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To find the solution of  x^3+3x^2+4x+2= 0.

Given that the solution is in integer. To solve the equation.

We see that the sum of the alternate coefficients are equal. Thherefore -1 is a solution.

x+1)x^3+3x^2+4x+2( x^2

x^3+x^2

----------------------

x+1)2x^2+4x(x^2+2x

2x^2+2x

--------------------

x+1)   2x +2 ( x^2+2x+2

2x+2

---------------------

0.

Therefore x^3+3x^2+4x+2 = (x+1)(x^2+2x+2).

Therefore , the other solutions are the roots of x^2+2x+2 = 0 which are given by:

x1 = {-2+sqrt(2^2-4*2)}/2 = -1 + (sqrt2)i

x2 = -1 -(sqrt2)i.

Therefore the solutions of the given equation are -1 , (-1+ 2^(1/2)*i) and (-1 - 2^(1/2) * i).

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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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.