Solve all possible values of x
$\displaystyle x^3-3x^2+4x-12=0$
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:
$\displaystyle (x-3) = 0~\vee~(x^2+4)=0$
$\displaystyle x = 3~\vee~x^2= -4$
$\displaystyle x = 3~\vee~x= 2i~\vee~x = -2i$