Factoring a cube

Question

Hello,

Can you show me how to factor these cubes? I can factor a quadratic but don't know if that knowledge is useful here.

-8x^3-60x^2+36x+32

x^3-7x^2-38x+240

Answer 1

For the first one start by taking out factor of -4:

-4(2x^3 +15x^2 -9x - 8)

then I would use rational root theorem to get some candidates for possible solutions, rational root theorem hints at ±8/±2, ±4/±2, ±2/±2, ±1/±2, ±8/±1, ±4/±1, ±2/±1, ±1/±1 being possible zeros to the polynomial but visibly x = 1 works as a zero for the polynomial (always check values like 1 or -1 as they are easy to do in your head). That means that (2x^3 +15x^2 -9x - 8) can be decomposed to (x-1) * q(x) for some q(x). Dividing (2x^3 +15x^2 -9x - 8) by (x-1) (I suggest synthetic division). This gives 2x^2 + 17x + 8, which you can then factor using normal rules or again trying rational root to get 2x^2+17x+8 = (2x+1)(x+8).

Thus -8x^3-60x^2+36x+32 factors to -4(x-1)(2x+1)(x+8).

For the second question similarly we can look at potential rational roots, and test the possible cases. Look at the rational root theorem for reasoning behind this but by observation/testing x=5 gives a zero of the polynomial. Then dividing x^3-7x^2-38x+240 by (x-5) we get x^2-2x-48, which then factoring the quadratic any way you want you get (x+6)(x-8).

Thus the second equation factors to be x^3-7x^2-38x+240 = (x-5)(x+6)(x-8).

To sum up the idea just check for potential rational roots using the rational root theorem and then divide them out to a simple polynomial.