Question 119560
subtracting 12 gives x^3-13x-12=0 __ substituting -1 for x satisfies the equation, so x+1 is a factor


dividing x^3-13x-12 by x+1 gives x^2-x-12, which factors into (x-4) and (x+3)


so x^3-13x-12=0 factors into (x+1)(x+3)(x-4)=0


setting each of the factors equal to zero, gives the "real" values of x __ -1, -3, 4