Question

How do you find the number of roots for `0=x^3+450x^2-1350` using the fundamental theorem of algebra?

Answer 1

This polynomial equation is of degree `3`, so has exactly `3` Complex roots counting multiplicity, some of which may be Real.

Actually all three roots are Real and distinct.

Explanation:

The fundamental theorem of algebra tells you that a polynomial equation in one variable of degree `n > 0` has at least one Complex root.

A simple corollary of this is that a polynomial equation in one variable of degree `n > 0` has `n` possibly repeated Complex roots. This is sometimes included as part of the statement of the fundamental theorem of algebra.

Here's a sketch of the proof of the corollary:

If a polynomial `f(x)` has one Complex zero `x_1`, then `(x-x_1)` is a factor of `f(x)` and you can write `f(x) = (x-x_1)g(x)` where `g(x)` is of degree `n-1`. Then if `n-1 > 0`, `g(x) = 0` has at least one Complex root `x_2`, which might be the same as `x_1`, etc. So counting multiplicity, a polynomial of degree `n` has exactly `n` Complex roots.

Note that the Complex roots referred to in all of this may be Real or non-Real.

The equation `0 = x^3+450x^2-1350` is of degree `3`, hence has `3` Complex roots counting multiplicity. That's all the fundamental theorem of algebra will tell you.

Beyond this, since this polynomial has Real coefficients, any non-Real Complex roots will occur in conjugate pairs. So we can deduce that our polynomial will have at least one Real root.

It so happens that all of these `3` roots are Real in this example, being approximately `-450`, `-1.74` and `1.73`

To show that all of the roots are Real, let `f(x) = x^3+450x^2-1350` and evaluate `f(x)` for a few values:

`f(-450) = -450^3+450^3-1350 = -1350 < 0`

`f(-2) = -8+450*4-1350 = -8+1800-1350 = 442 > 0`

`f(0) = 0 + 0 - 1350 = -1350 < 0`

`f(2) = 8+450*4-1350 = 8+1800-1350 = 458 > 0`

So `f(x)` crosses the `x` axis `3` times and `f(x) = 0` has `3` distinct Real roots.