What are the solutions to the equation x^3+3x^2+4x+12=0 ?

The solutions to the equation x^3+3x^2+4x+12=0 are x=-3,x=2i,x=-2i

Find the zeros of x^3+3x^2+4x+12=0

The zeros of x^3+3x^2+4x+12=0 are x=-3,x=2i,x=-2i

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

x = - 3, x = 2 i, x = - 2 i

Solution steps

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

Factor

x^{3} + 3 x^{2} + 4 x + 12 : (x + 3) (x^{2} + 4)

(x + 3) (x^{2} + 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 3 = 0 or x^{2} + 4 = 0

Solve

x + 3 = 0 : x = - 3

Solve

x^{2} + 4 = 0 : x = 2 i, x = - 2 i

The solutions are

x = - 3, x = 2 i, x = - 2 i

z^{8} - 16 i = 0

x^{4} - 2937 x^{2} + 100 = 0

x^{2} (x - 3) = 0

4 x^{3} - 12 x^{2} = 9 x - 27

- x^{3} + 6 x^{2} + 15 x + 8 = 0

What are the solutions to the equation x^3+3x^2+4x+12=0 ?
The solutions to the equation x^3+3x^2+4x+12=0 are x=-3,x=2i,x=-2i
Find the zeros of x^3+3x^2+4x+12=0
The zeros of x^3+3x^2+4x+12=0 are x=-3,x=2i,x=-2i