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

##### 2 Answers

The FTOA only tells us that

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

In our example:

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

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

So the FTOA tells us that it has exactly

We can use Descartes' Rule of Signs to find that

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

**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.

**Algebraic method**

Since the given quartic has no term in

#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

#{ (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

From the resulting values for

Messy, but works.

See below.

#### Explanation:

As Gauss did at his time, taking

and making

and representing the curves

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