Question

What are the zeros of the function `f(x) = 3x^4-2x^3-36x^2+36x-8` ?

Answer 1

This quartic function has one rational zero `x=2/3` and three irrational real zeros of the form:

`x = 4cos(1/3 cos^(-1)(-1/4)+(2npi)/3)" "n = 0,1,2`

Explanation:

Given:

`f(x) = 3x^4-2x^3-36x^2+36x-8`

By the rational roots theorem, the only possible rational zeros of `f(x)` are:

`+-1/3, +-2/3, +-1, +-4/3, +-2, +-8/3, +-4, +-8`

We find:

`f(-4) = 168`

`f(-8/3) = -4600/27`

`f(1/3) = -1/27`

`f(1/2) = 15/16`

`f(2/3) = 0`

`f(1) = -7`

`f(8/3) = -488/9`

`f(4) = 200`

Bolzano's theorem tells us that if `f(x)` is continuous on `[a,b]` with `f(a)` and `f(b)` having opposite signs, then there is some `x in (a, b)` such that `f(x) = 0`

So we find that f(x) has zeros in:

`(-4, -8/3)`

`(1/3, 1/2)`

`{ 2/3 }`

`(8/3, 4)`

Of these zeros, only `2/3` is rational, with `(3x-2)` therefore being a factor of `f(x)`...

`3x^4-2x^3-36x^2+36x-8 = (3x-2)(x^3-12x+4)`

It would be possible to find numerical approximations to the irrational zeros by repeatedly bisecting each of the intervals and identifying the subintervals in which the sign changes. Alternatively, the cubic can be solved using a trigonometric substitution...

Let `x = 4cos theta`, then:

`0 = x^3-12x+4`

`color(white)(0) = (4cos theta)^3-12(4 cos theta)+4`

`color(white)(0) = 16(4cos^3 theta-3 cos theta)+4`

`color(white)(0) = 16 cos 3 theta+4`

So:

`cos 3 theta = -1/4`

`3 theta = +-cos^(-1) (-1/4) + 2npi" "` for any `n in ZZ`

Hence:

`x = 4 cos theta = 4cos(1/3 cos^(-1)(-1/4)+(2npi)/3)`

giving three distinct roots for `n = 0, 1, 2`.