Question

What are all the possible rational zeros for `f(x)=4x^4-16x^3+12x-30` and how do you find all zeros?

Answer 1

The "possible" rational zeros are:

`+-1/2, +-1, +-3/2, +-5/2, +-3, +-5, +-15`

but this quartic has two irrational and two complex zeros.

Explanation:

Note that all of the coefficients of `f(x)` are divisible by `2`, so separate that off as a factor before applying the rational roots theorem:

`f(x) = 4x^4-16x^3+12x-30 = 2(2x^4-8x^3+6x-15)`

By the rational roots theorem, any rational zeros of `2x^4-8x^3+6x-15` are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `-15` and `q` a divisor of the coefficient `2` of the leading term.

That means that the only possible rational zeros are:

`+-1/2, +-1, +-3/2, +-5/2, +-3, +-5, +-15`

Note that:

`2(color(blue)(5)^4) = 1250 > 1045 = 8(color(blue)(5)^3)+6(color(blue)(5))+15`

Hence any zeros lie strictly inside the circle `abs(x) = 5` in the complex plane.

So we can discard `+-5` and `+-15` as possibilities.

To cut a long story short, none of the "possible" rational zeros are zeros of this quartic.

It is a typical "nasty" quartic with complicated irrational zeros.

It is possible to solve algebraically, but much simpler to find numerical approximations to the zeros using a method such as Durand-Kerner.

`x_1 ~~ -1.2905`

`x_2 ~~ 3.9293`

`x_(3,4) ~~ 0.68061+-1.00785i`

See https://socratic.org/s/aB9Ee9wQ for another example.