Question

What are the possible integral zeros of `P(y)=y^4-5y^3-7y^2+21y+4`?

Answer 1

The "possible" integral zeros are `+-1`, `+-2`, `+-4`

None of these work, so `P(y)` has no integral zeros.

Explanation:

`P(y) = y^4-5y^3-7y^2+21y+4`

By the rational root theorem, any rational zeros of `P(x)` are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `4` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are the possible integer zeros:

`+-1, +-2, +-4`

Trying each of these, we find:

`P(1) = 1-5-7+21+4 = 14`

`P(-1) = 1+5-7-21+4 = -18`

`P(2) = 16-40-28+42+4 = -6`

`P(-2) = 16+40-28-42+4 = -10`

`P(4) = 256-320-112+84+4 = -88`

`P(-4) = 256+320-112-84+4 = 384`

So `P(y)` has no rational, let alone integer, zeros.