Question

How do you find the number of roots for `0=x^3-4x-48` using the fundamental theorem of algebra?

Answer 1

The degree of this polynomial is `3` so by the fundamental theorem of algebra it has `3` roots (counting multiplicity).

Explanation:

A non-zero polynomial equation in one variable always has as many roots as its degree. They may be repeated and/or Complex, but it will have that many roots.

The highest degree term in `0 = x^3-4x-48` is the `x^3` one, so the polynomial is of degree `3`.

Let `f(x) = x^3-4x-48`.

Notice that `f(4) = 64-16-48 = 0`, so `x=4` is one of the roots and `(x-4)` is a factor of `f(x)`.

`x^3-4x-48 = (x-4)(x^2+4x+12)`

The remaining quadratic factor `(x^2+4x+12)` is of the form `ax^2+bx+c`, with `a=1`, `b=4` and `c=12`. It has zeros given by the quadratic formula:

`x=(-b+-sqrt(b^2-4ac))/(2a) = (-4+-sqrt((-4)^2-(4xx1xx12)))/(2xx1)`

`=(-4+-sqrt(-32))/2 = -2+-2sqrt(2) i`

..that is a conjugate pair of Complex roots.