Okay so I have to solve for x
$\displaystyle 120 = 2(x-2)(x-3)(x+5)
$
$\displaystyle 60 = (x-2)(x-3)(x+5)$
$\displaystyle 60 = x^3 - 19x + 30$
I am stuck from here. I don't know if I should have expanded or not.
now subtract 60 from both sides
$\displaystyle x^3-19x-30=0$
Now we need to use the rational roots theorem. So we know the only roots are factors of 30.
So if we try x=5 we get
$\displaystyle 5^3-19(5)-30=125-95-30=0$
so 5 is a root so x-5 is a factor. So by long division we get
$\displaystyle (x-5)(x^2+5x+6)=0$
$\displaystyle (x-5)(x+2)(x+3)=0$
I hope this helps