Question

Which correctly describes the root of the following cubic equation? x^3-3x^2+4x-12=0

Answer 1

I know that the factors of 12 are 1, 2, 3, 4, 6, and 12
Therefore, the possible roots are 1, 2, 3, 4, 6, and 12

Answer 2

Answer:

The equation has one real root and two complex roots.

Step-by-step explanation:

The given equation is

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

The above equation is true form x=3, therefore (x-3) is a factor of above equation.

Use long division or synthetic division method to divide the equation by (x-3).

$(x-3)(x^2+4)=0$

Equate each factor equal to zero.

$x-3=0$

$x=3$

Therefore 3 is a real root of the equation.

$x^2+4=0$

$x^2=-4$

$x=\sqrt{-4}$

$x=\pm 2i$

2i and -2i are complex roots of the equation.

Therefore the equation has one real root and two complex roots.