# What is the solution of x^3+3x^2+4x+12=0

*print*Print*list*Cite

### 3 Answers

The solution of the equation x^3+3x^2+4x+12=0 has to be determined.

x^3+3x^2+4x+12=0

=> x^2(x + 3) + 4(x + 3) = 0

=> (x^2 + 4)(x + 3) = 0

x^2 + 4 = 0

=> x = 2i and x = -2i

x + 3 = 0

=> x = -3

**The solution set of x^3+3x^2+4x+12=0 is {-2i, 2i, -3}**

x^3+3x^2+4x+12=0

x^2*(x+3)+4*(x+3)=0

(x+3)*(4+x^2)=0

x+3=0

x=-3

4+x^2=0 x=-3,2i,-2i

x^2=-4

x1,2=+-2i

given equation : x^3 +3x^2 +4x +12 = 0

=> x^2 (x+3) +4(x+3)= 0

=> (x+3)(x^2 +4)= 0

If (x+3) = 0 , then x= -3 ----(1)

If x^2 +4= 0 then,

x^2= -4

x= sq.root(-4)

x= +2i , and x= -2i

[since square root of negative number is an imaginary number denoted by 'i' ]

**x= -3, 2i, -2i <--- Answer**