I'm having a hard time factoring polynomials, especially 3rd degree.

-x^3+2x^2+4x-8

-x^3-5x^2+20x+12

I don't know how to begin. I thought long division, but what would I divide by?

First find one root by trial and error. Call it a. Then divide (x-a) into the cubic using long division to get a quadratic equation. Get the roots of that by the usual method.

For the equation
-x^3+2x^2+4x-8,
one of the roots is x=2. Divide (x-2) into -x^3+2x^2+4x-8, and you get
-x^2 + 4 = (4-2)(x+2)
That completes the factoring.

-x^3+2x^2+4x-8 = (x-2)(x+2)(2-x)

To factor a 3rd degree polynomial, you can try using the Rational Root Theorem and synthetic division. Here's how you can approach factoring these two polynomials:

1. For the polynomial -x^3 + 2x^2 + 4x - 8:
- Begin by listing all possible rational roots. The Rational Root Theorem states that any rational root of a polynomial must be in the form of p/q, where p is a factor of the constant term (-8 in this case) and q is a factor of the leading coefficient (-1 in this case).
- For -8, the factors are ±1, ±2, ±4, and ±8. For -1, the factors are ±1.
- Therefore, the possible rational roots are ±1, ±2, ±4, ±8.

Let's try them one by one using synthetic division:

For x = 1, perform synthetic division as follows:

1 | -1 2 4 -8
| -1 1 5
|---------------
-1 1 5 -3

The remainder is -3, which means that x = 1 is not a root.

Continue testing the other potential rational roots using synthetic division until you find one that gives a remainder of 0. In this case, x = -2 is a root:

-2 | -1 2 4 -8
| 2 -8 8
|---------------
-1 4 -4 0

The remainder is 0, which means that x = -2 is a root. This means that (x + 2) is a factor. Now we have a quadratic polynomial (-1x^2 + 4x - 4) left, which can be factored further.

Therefore, -x^3 + 2x^2 + 4x - 8 can be factored as -(x + 2)(x^2 - 4x + 4).

2. For the polynomial -x^3 - 5x^2 + 20x + 12:
- Follow the same steps as above to find the possible rational roots.
- The possible rational roots are ±1, ±2, ±3, ±4, ±6, ±12.

Start with x = 1:

1 | -1 -5 20 12
| -1 -6 14
|-----------------
-1 -6 14 26

The remainder is 26, which means that x = 1 is not a root.

Continue testing the other potential rational roots until you find one that gives a remainder of 0. In this case, x = -2 is a root:

-2 | -1 -5 20 12
| 2 6 -32
|------------------
-1 -3 26 -20

The remainder is 0, which means that x = -2 is a root. This means that (x + 2) is a factor. Now we have a quadratic polynomial (-1x^2 - 3x + 26) left, which can be factored further.

Therefore, -x^3 - 5x^2 + 20x + 12 can be factored as -(x + 2)(x^2 - 3x + 26).

Keep in mind that this method works for finding rational roots, so it might not be successful in every case. In some cases, you might need to resort to other methods like factoring by grouping, completing the square, or using the quadratic formula.