Question

How do you find the number of roots for `f(x) = 4x^5 – 2x – 15` using the fundamental theorem of algebra?

Answer 1

`f(x)` has exactly `5`, possibly Complex, zeros counting multiplicity.

Explanation:

The Fundamental Theorem of Algebra tells us that any non-constant polynomial equation with Complex (possibly Real) coefficients has at least one Complex (possibly Real) zero.

A simple corollary of this is that a polynomial of degree `n > 0` with Complex coefficients has exactly `n` zeros in `CC` counting multiplicity.

To show the corollary: Suppose `p(x)` is a polynomial of degree `n > 0`. Then `p(x)` has some zero `r_1 in CC`. Then `(x-r_1)` is a factor of `p(x)`. So `p(x) = (x-r_1)q(x)` for some polynomial `q(x)` of degree `n-1`. If `n-1 > 0` then `q(x)` has a zero `r_2 in CC` and `q(x)` is divisible by `(x-r_2)`. Repeat until we have all `n` zeros `r_1, r_2,...,r_n in CC`.

So in the case of `f(x) = 4x^5-2x-15` we can deduce that `f(x)` has exactly `5` (possibly Complex) zeros counting multiplicity.

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Bonus

What else can we find out about these zeros?

First note that the sequence of coefficients of `f(x)` has one change of sign, so `f(x)` has at most one positive Real zero.

In fact, we can see that `f(1) = 4-2-15 = -13 < 0` and `f(2) = 128-4-15 = 109 > 0`, so `f(x)` has a zero in `(1, 2)`.

If we reverse the signs of the coefficients of the terms of odd degree, then all are negative. So `f(x)` has no negative Real roots.

Since the coefficients of `f(x)` are all Real, any non-Real Complex zeros must occur in conjugate pairs.

So we must have two pairs of Complex conjugate zeros and one positive Real zero.