We need to find the roots of this polynomial. We can do this by inspection:
[tex]\begin{gathered} P(x)=x^3+3x^2-4x-12 \\ P(0)=-12 \\ P(1)=1+3-4-12=-12 \\ P(2)=8+3\cdot4-4\cdot2-12=0 \\ P(-1)=-1+3-4\cdot(-1)-12=-6 \\ P(-2)=-8+3\cdot4-4\cdot(-2)-12=0 \end{gathered}[/tex]Above we found two roots at 2 and -2. Now we can write:
[tex]\begin{gathered} x^3+3x^2-4x-12=(x-b)(x-2)(x+2) \\ x^3+3x^2-4x-12=(x-b)(x^2-4)=x^3-4x-bx^2+4b \\ So, \\ 3x^2=-bx^2\Rightarrow b=-3 \\ -12=4b\Rightarrow b=-\frac{12}{4}=-3 \end{gathered}[/tex]So we can write the polynomial as:
[tex]\begin{gathered} x^3+3x^2-4x-12=(x+3)(x+2)(x-2)=0 \\ \text{The solutions are:} \\ x=2 \\ x=-2 \\ x=-3 \end{gathered}[/tex]
