Question

What are the possible integral zeros of `P(z)=z^4+5z^3+2z^2+7z-15`?

Answer 1

The possible integer roots that should be tried are `\pm 1, \pm 3, \pm 5,\pm 15`.

Explanation:

Let us imagine that some other integer could be a root. We pick `2`. This is wrong. We are about to see why.

The polynomial is

`z^4+5z^3+2z^2+7z-15`.

If `z=2` then all the terms are even because they are multiples of `z`, but then the last term has to be even to make the whole sum equal to zero ... and `-15` isn't even. So `z=2` fails because the divisibility does not work out.

To get the divisibility to work out right an integer root for `z` has to be something that divides evenly into the constant term, which here is `-15`. Remembering that integers can be positive, negative or zero the candidates are `\pm 1, \pm 3, \pm 5,\pm 15`.