Question

How do you find the roots for `f(x) = 6x^4 – 2x – 15` using the fundamental theorem of algebra?

Answer 1

The FTOA only tells us that `f(x)` has `4` zeros counting multiplicity.

Using Descartes' Rule of Signs we can determine that it has two real zeros and a pair of complex zeros.

Explanation:

The Fundamental Theorem of Algebra (FTOA) only tells you how many roots a polynomial has - not how to find them.

Specifically the FTOA tells us that any non-constant polynomial in a single variable with Complex (possibly Real) coefficients has a Complex (possibly Real) zero.

A straightforward corollary of that, often stated as part of the FTOA is that a single variable polynomial of degree `n > 0` has exactly `n` zeros counting multiplicity.

In our example:

`f(x) = 6x^4-2x-15`

is a quartic polynomial - i.e. is of degree `4`.

So the FTOA tells us that it has exactly `4` zeros counting multiplicity.

We can use Descartes' Rule of Signs to find that `f(x)` has exactly one positive real zero and one negative zero. So it also has a complex conjugate pair of non-real Complex zeros.

All of these zeros are irrational and have a quite messy radical form.

`color(white)()`
Durand-Kerner

We can find numerical approximations for the roots using a numerical method such as the Durand-Kerner method.

In the given example, we can use a program such as this one (written in C++):

and hence find approximate zeros:

`x_1 ~~ 1.30904`

`x_2 ~~ -1.20363`

`x_(3,4) ~~ -0.0527043+-1.25854i`

See https://socratic.org/s/aEgiJEsx for more information.

`color(white)()`
Algebraic method

Since the given quartic has no term in `x^3` it has a factorisation of the form:

`6x^4-2x-15 = 6(x^2-ax+b)(x^2+ax+c)`

`color(white)(6x^4-2x-15) = 6x^4+6(b+c-a^2)x^2+6a(b-c)x+6bc`

Hence, equating the coefficients of `x^2`, `x` and the constant term we find:

`{ (b+c = a^2), (b-c = -1/(3a)), (bc = -5/2) :}`

Then:

`(a^2)^2 = (b+c)^2 = (b-c)^2+4bc = 1/(9(a^2))-10`

Hence we get a cubic in `a^2` to solve.

From the resulting values for `a` we can derive expressions for `b, c` and hence a couple of quadratics to solve.

Messy, but works.

Answer 2

See below.

Explanation:

As Gauss did at his time, taking

`f(x) = 6 x^4 - 2 x - 15`

and making

`f(u + i v)=- 2 u + 6 u^4 - 36 u^2 v^2 + 6 v^4 -15+ i (-2 v + 24 u^3 v - 24 u v^3)`

and representing the curves

`- 2 u + 6 u^4 - 36 u^2 v^2 + 6 v^4 -15=0`
`-2 v + 24 u^3 v - 24 u v^3`

the roots can be observed at the intersection points for red's and green's