# Thread: Solve all values for x

1. ## Solve all values for x

Solve all possible values of x

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

Could someone do this question for me please, really urgent.

2. Originally Posted by Dean
Solve all possibl;e values of x

[tex]x^3-3x^2+4x-12=0[tex]

Could someone do this question for me please, really urgent.
Hello,

use factorization:

$\displaystyle x^3-3x^2+4x-12=0~\Longrightarrow~x^2(x-3) + 4(x-3) = 0~\Longrightarrow~(x-3)(x^2+4)=0$

Since $\displaystyle x^2+4 > 0$ there is only one real value for x: $\displaystyle \boxed{x = 3}$
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =

EDIT: You asked for all possible solutions. You didn't mention a domain so maybe imaginary solutions are possible for you too:

$\displaystyle (x-3)(x^2+4)=0$ . A product of 2 factors is zero if one factor is zero:

$\displaystyle (x-3) = 0~\vee~(x^2+4)=0$

$\displaystyle x = 3~\vee~x^2= -4$

$\displaystyle x = 3~\vee~x= 2i~\vee~x = -2i$