Question

What are the possible rational zeros of `P(x) = 4x^4+19x^3-x^2+19x-5` ?

Answer 1

The "possible" rational zeros are: `+-1/4, +-1/2, +-1, +-5/4, +-5/2, +-5`

The actual rational zeros are: `1/4` and `-5`

The other two zeros are: `+-i`

Explanation:

Given:

`P(x) = 4x^4+19x^3-x^2+19x-5`

By the rational roots 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 `-5` and `q` a divisor of the coefficient `4` of the leading term.

That means that the only possible rational zeros are:

`+-1/4, +-1/2, +-1, +-5/4, +-5/2, +-5`

We find:

`P(1/4) = 4(color(blue)(1/4))^4+19(color(blue)(1/4))^3-(color(blue)(1/4))^2+19(color(blue)(1/4))-5`

`= 1/64+19/64-4/64+304/64-320/64 = 0`

So `x=1/4` is a zero and `(4x-1)` a factor:

`4x^4+19x^3-x^2+19x-5 = (4x-1)(x^3+5x^2+x+5)`

`color(white)(4x^4+19x^3-x^2+19x-5) = (4x-1)((x^3+5x^2)+(x+5))`

`color(white)(4x^4+19x^3-x^2+19x-5) = (4x-1)(x^2(x+5)+1(x+5))`

`color(white)(4x^4+19x^3-x^2+19x-5) = (4x-1)(x^2+1)(x+5)`

`color(white)(4x^4+19x^3-x^2+19x-5) = (4x-1)(x^2-i^2)(x+5)`

`color(white)(4x^4+19x^3-x^2+19x-5) = (4x-1)(x-i)(x+i)(x+5)`

So the rational zeros are `1/4` and `-5`. The other two zeros are `+-i`.