+421 Use the RATIONAL ROOT THEOREM to find solutions for these. factors of the lead constant=96/factors of the lead coefficient=1= 1 x 96, 2 x 48, 3 x 32, 4 x 24, 6 x 16, or 8 x 12. The possible roots are 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96. 2 is a solution, so we can factor out an x+2. (x-2)*{x^4+6x^3-12x^2-56x+96}/{x-2}, and that equals (x^3+8x^2+4x-48)(x+2). Now simplifying this actually \(=x^3+\frac{8x^3-12x^2-56x+96}{x-2}\), and keep dividing this way until you get \(x^3+8x^2+4x-48\) . So now this is easily factorable, \(=\left(x-2\right)^2\left(x+4\right)\left(x+6\right)\)I am pretty sure D is right but look out for a graph that intersects x in 3 places: 2, -4, and -6!
-Pangolin14 :D
