Sorry for such a terribly worded question, that is just how I came to verbalize it. The real question and it's answer is provided below:
My problem comes with understanding the solution to 4bi. and 4biii. Why does setting the polynomial less than zero then determining it's positive regions ensure that the polynomial will have no zeros (with the whole polynomial having one real zero because of the x-1)? Why does doing the opposite ensure 2 real zeroes?
$1$ is certainly a root to the cubic equation. The remaining roots come from the quadratic part and we have proven that $1$ is not a root for the quadratic part.
Hence if the discriminant is negative, then the quadratic part has $0$ real root, hence the cubic equation will have $1$ real root.
If the quadratic part has $r$ real root, then the cubic equation has $r+1$ roots.