Question

What are all the possible rational zeros for `f(x)=4x^4+19x^2-63` and how do you find all zeros?

Answer 1

`x=+-3/2` or `x=+-sqrt(7)i`

Explanation:

`f(x) = 4x^4+19x^2-63`

By the rational roots theorem, any rational zeros of `f(x)` are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-63` 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, +-3/4, +-1, +-3/2, +-7/4, +-9/4, +-3, +-7/2, +-9/2, +-21/4, +-7, +-9, +-21/2, +-63/4, +-21, +-63/2, +-63`

We could try each of these in turn, but there are easier ways to find the zeros of `f(x)`.

Note that `f(x)` is quadratic in `x^2`, so we can use the quadratic formula to find:

`x^2 = (-19+-sqrt(19^2-4(4)(-63)))/(2*4)`

`=(-19+-sqrt(361+1008))/8`

`=(-19+-sqrt(1369))/8`

`=(-19+-37)/8`

i.e. `x^2 = 9/4` or `x^2 = -7`

Since the result is rational, we could have found this using an AC method instead, but at least the quadratic formula gives it to us directly.

Hence `x=+-3/2` or `x=+-sqrt(7)i`