Question

How do you find the number of roots for `f(x) = 2x^3 – 12x^2 + 11x + 2` using the fundamental theorem of algebra?

Answer 1

The fundamental theorem of algebra tells us that this cubic has `3` zeros counting multiplicity.

We can further discover that all `3` are Real with `2` distinct positive zeros and one negative.

Explanation:

`f(x) = 2x^3-12x^2+11x+2`

The fundamental theorem of algebra (FTOA) tells us that any polynomial of degree `n > 0` has at least one Complex, possibly Real, zero.

A straightforward corollary of that, often stated as part of the FTOA is that a polynomial of degree `n > 0` will have exactly `n` Complex (possibly Real) zeros counting multiplicity.

This follows since given a polynomial of degree `n > 0` with a zero `x=a`, we can divide the polynomial by `(x-a)` to give a polynomial of degree `n-1`. If `n - 1 > 0` then it has a possibly distinct zero `b`, etc.

In our example, `f(x)` is of degree `3` so has exactly `3` Complex, possibly Real, zeros counting multiplicity.

We can find out more by examining the discriminant `Delta`, which for a cubic of the form `ax^3+bx^2+cx+d` is given by the formula:

`Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd`

In our example, `a=2`, `b=-12`, `c=11` and `d=2`, so we find:

`Delta = 17424-10648+13824-432-9504 = 10664`

Since `Delta > 0`, all three zeros of `f(x)` are Real and distinct.

Using Descartes's Rule of Signs we can also tell that `2` of these zeros are positive and one negative.