Question

How do you factor `(x^3 - 7x^2 + 14x - 8)`?

Answer 1

There doesn't seem to be a grouping that will work.

The rational zero theorem tells us that tlhe possible rational zeros are `+-1, +-2, +-4, +-8`.

We can see that `(1)^3-7(1)^2+14(1)-8=1-7+14-8=0`, so 1 is a zero.

The factor theorem then tells us that `x-1` is a factor.
Do the division:

`(x^3-7x^2+14x-8)/(x-1) = x^2-6x+8`

So
`(x^3-7x^2+14x-8) =(x-1) (x^2-6x+8)=(x-1)(x-2)(x-4)`