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

Jun 25, 2017

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

#### 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

${z}^{4} + 5 {z}^{3} + 2 {z}^{2} + 7 z - 15$.

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 $\setminus \pm 1 , \setminus \pm 3 , \setminus \pm 5 , \setminus \pm 15$.