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

Apr 29, 2017

The FTOA only tells us that $f \left(x\right)$ 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:

$f \left(x\right) = 6 {x}^{4} - 2 x - 15$

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 $f \left(x\right)$ 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.

$\textcolor{w h i t e}{}$
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:

${x}_{1} \approx 1.30904$

${x}_{2} \approx - 1.20363$

${x}_{3 , 4} \approx - 0.0527043 \pm 1.25854 i$

$\textcolor{w h i t e}{}$
Algebraic method

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

$6 {x}^{4} - 2 x - 15 = 6 \left({x}^{2} - a x + b\right) \left({x}^{2} + a x + c\right)$

$\textcolor{w h i t e}{6 {x}^{4} - 2 x - 15} = 6 {x}^{4} + 6 \left(b + c - {a}^{2}\right) {x}^{2} + 6 a \left(b - c\right) x + 6 b c$

Hence, equating the coefficients of ${x}^{2}$, $x$ and the constant term we find:

$\left\{\begin{matrix}b + c = {a}^{2} \\ b - c = - \frac{1}{3 a} \\ b c = - \frac{5}{2}\end{matrix}\right.$

Then:

${\left({a}^{2}\right)}^{2} = {\left(b + c\right)}^{2} = {\left(b - c\right)}^{2} + 4 b c = \frac{1}{9 \left({a}^{2}\right)} - 10$

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.

Apr 29, 2017

See below.

#### Explanation:

As Gauss did at his time, taking

$f \left(x\right) = 6 {x}^{4} - 2 x - 15$

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 