# How do you find the number of complex, real and rational roots of #4x^5-5x^4+x^3-2x^2+2x-6=0#?

##### 1 Answer

This quintic equation has one positive irrational Real zero and two complex conjugate pairs of non-Real Complex zeros.

#### Explanation:

Given:

#f(x) = 4x^5-5x^4+x^3-2x^2+2x-6#

**Fundamental Theorem of Algebra**

The Fundamental Theorem of Algebra (FTOA) tells us that a polynomial in one variable of degree

A straightforward corollary of the FTOA, often stated with it, is that a polynomial of degree

In our example,

**Descartes' Rule of Signs**

The signs of

The signs of

**Rational Roots Theorem**

By the rational roots theorem, any rational zeros of

In addition, from Descartes' Rule of Signs, we know that such zeros can only be positive. So the only possible rational zeros are:

#1/4, 1/2, 3/4, 1, 3/2, 2, 3, 6#

We find:

#f(1/4) = -45/8#

#f(1/2) = -89/16#

#f(3/4) = -747/128#

#f(1) = -6#

#f(3/2) = 15/16#

#f(2) = 46#

#f(3) = 576#

#f(6) = 24774#

So

**Durand-Kerner Algorithm**

We can use the Durand-Kerner algorithm to find numerical approximations to the zeros.

In our example, we find approximations:

#x_1 ~~ 1.47294#

#x_(2,3) ~~ 0.547393+-0.873706i#

#x_(4,5) ~~ -0.658864+-0.723817i#

See https://socratic.org/s/aBf43swQ for another example and more information on this method.

So we find that

Here's the C++ program I used to find the zeros: