Question

How do you find the roots for `f(x) = 17x^15 + 41x^12 + 13x^3 - 10` using the fundamental theorem of algebra?

Answer 1

The FTOA does not help you find the zeros - it only tells you that this polynomial of degree `15` has exactly `15` Complex zeros counting multiplicity.

Explanation:

By roots, I will assume you mean zeros, i.e. values of `x` for which `f(x) = 0`.

The so called fundamental theorem of algebra (FTOA) is neither fundamental nor a theorem of algebra, but what it does tell you is that any non-zero polynomial in one variable with Complex coefficients has a zero in `CC`. That is, there is some Complex number `x_1` such that `f(x_1) = 0`.

A simple corollary of this - often stated as part of the FTOA - is that a polynomial in one variable of degree `n > 0` with Complex coefficients has `n` zeros counting multiplicity, all in `CC`.

In our example, `f(x)` is of degree `15`, so has `15` zeros counting multiplicity, all in `CC`.

The FTOA does not help you actually find the zeros.

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Bonus

What else can we find out about the zeros of this `f(x)`?

Note that the coefficients of `f(x)` have only one change of sign, so only one positive Real zero.

Note `f(0) = -10 < 0` and `f(1) = 17+41+13-10 = 61 > 0`

So the positive Real zero is in `(0, 1)`

`f(-x) = -17x^5+41x^12-13x^3-10`

has two changes of sign, so `f(x)` may have `0` or `2` negative Real zeros.

We find:

`f(-1) = -17+41-13-10 = 1 > 0`

So there is a Real zero in `(-1, 0)` and another Real zero in `(-oo, -1)`.

Any other zeros will occur in Complex conjugate pairs, since all of the coefficients of `f(x)` are Real.

Note that all of the degrees are multiples of `3` so the zeros are all cube roots of the zeros of:

`g(t) = 17t^5+41t^4+13t-10`