How do I factor a third power polynomial?

Question

-x³+8x²-20x+16=0

Answer 1

Because signs alternate, we know that we only have positive roots. Because this is a cubic, we are guaranteed at least one real root (this is true whenever the power of the leading coefficient is odd). From the rational root theorem, as we know all roots are positive, we only need to consider factors of 16, 1, 2, 4, 8, and 16.

We could use trial and error, including synthetic division. We also could just try 1, 2, 4, 8, and 16, and see if any of these values make the polynomial 0. We would see that 2 and 4 are roots. Graph the polynomial. We would see 0s at 2 and 4, with a double root at 2 (the curve goes up on both sides). Thus, the polynomial is -(x-2)^2 (x-4). Alternatively, using synthetic division with either 2 or 4 would leave us with an easily factored binomial, which yields the same result.

Note that the constant coefficient is -16, which equals -2*2*4, which tells us that our answer makes sense..