college algebra plz help

we can factorize x^3-3x^2-4x+12. in fact: 

x^3-3x^2-4x+12 == x^2(x-3)-4(x-3) == (x^2-4)(x-3)== 
== (x+2) (x-2) (x-3) 

now 
x^3-3x^2-4x+12==0 iff (x+2) (x-2) (x-3)==0 
and 
(x+2) (x-2) (x-3)==0 iff 
(x+2)==0 or (x-2)==0 or (x-3)==0 

then 
the solution set is: 
S=={x1== -2, x2==2, x3==-3} 

hope this helps 
ciao ciao 

thank you, but one quick question why did you switch the signs for +3x to -3x and -12 to +12

What are the zeros of f(x)=x^3+3x^2-4x-12?

$\begin{array}{lrrl} x^3+3x^2-4x-12\\ & & 12 = 2\cdot 2 \cdot 3\\\\ (x-x_1)\cdot (x-x_2)\cdot (x-x_3) & \Rightarrow & -12 = (-x_1)\cdot(-x_2)\cdot(-x_3)& \\ & \Rightarrow & \boxed{ \begin{array}{rcll}12 = \underbrace{x_1}_{=2}\cdot \underbrace{x_2}_{=2} \cdot \underbrace{x_3}_{=3}\end{array} } & (\text{sign} \ ?)\\\\ (x-x_1)\cdot (x-x_2)\cdot (x-x_3) & & = x^2\cdot(-x_1)+x^2\cdot(-x_2)+x^2\cdot(-x_3) \\ & \Rightarrow & 3x^2= x^2\cdot(-x_1-x_2-x_3) \\ & & 3 = -(x_1+x_2+x_3) \\ & & -3 = x_1+x_2+x_3 \\\\ & & & \text{sum} \\ & & 2+2+3 & =7~ (\text{no})\\ & & 2+2-3 & =1~ (\text{no})\\ & & 2-2+3 & =3~ (\text{no})\\ & & 2-2-3 & =-3~ (\text{yes})\\ & & -2+2+3 & =3~ (\text{no})\\ & & -2+2-3 & =-3~ (\text{yes})\\ & & -2-2+3 & =-1~ (\text{no})\\ & & -2-2-3 & =-7~ (\text{no}) \end{array}$

$\boxed{ \begin{array}{lrrl} x_1 = 2\\ x_2 = -2\\ x_3 = -3\\ x^3+3x^2-4x-12 = (x-2)\cdot(x+2)\cdot(x+3) \end{array} }$