# How do you find the roots for #f(x)=16x^4+8x^3+4x^2-2x-15# using the fundamental theorem of algebra?

##### 1 Answer

The FTOA tells us that there are

#### Explanation:

**Fundamental Theorem of Algebra**

The fundamental theorem of algebra (FTOA) tells us that any non-constant polynomial with Complex (possibly Real) coefficients will have a zero in the Complex numbers.

A straightforward corollary of that, often stated as part of the FTOA is that a polynomial of degree

Given:

#f(x) = 16x^4+8x^3+4x^2-2x-15#

Note that

That is all the FTOA tells us. It does not tell us how to find them.

**Rational Roots Theorem**

By the rational roots theorem, any *rational* zeros of

That means that the only possible *rational* zeros are:

#+-1/16, +-1/8, +-3/16, +-1/4, +-5/16, +-3/8, +-1/2, +-5/8, +-3/4, +-15/16, +-1, +-5/4, +-3/2, +-15/8, +-5/3, +-3, +-5, +-15#

That is rather a lot of possibilities to try, but we can cut it down radically as follows:

Note that:

#f(x) = 16x^4+8x^3+4x^2-2x-15=(2x)^4+(2x)^3+(2x)^2-(2x)-15#

So the only possible *rational* values of

#+-1/2, +-3/2, +-5/2, +-15/2#

None of these work. So

**Descartes' Rule of Signs**

The coefficients of

The coefficients of

**Where do we go from here?**

It is possible to solve

#x_1 ~~ -1.0194#

#x_2 ~~ 0.85875#

#x_(3,4) ~~ -0.16967+-1.02084i#