The answer is: 3,-3,-6
We need to find the values where the function tends to 0, so we need to make the equation, equal to 0:
[tex]x^3+6x^2-9x-54=0\\x^{2} (x+6)-9(x+6)=0\\\(x^{2} -9)(x+6)=0\\(x+3)(x-3)(x+6)=0[/tex]
Then, we have three factors multiplying each other, so we have to five three numbers which make at least one of those factors 0
So,
Trying with 3:
[tex](x+3)(x-3)(x+6)=0\\(3+3)(3-3)(3+6)=0\\(3+3)(0)(3+6)=0\\0=0[/tex]
Trying with -3:
[tex](x+3)(x-3)(x+6)=(-3+3)(-3-3)(-3+6)=0\\(0)(-6)(3)=0\\\ 0=0[/tex]
Trying with -6:
[tex](x+3)(x-3)(x+6)=(-6+3)(-6-3)(-6+6)=0\\(-3)(-9)(0)=0\\\ 0=0[/tex]
Have a nice day!
