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
Jan 11, 2017

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#

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Fundamental Theorem of Algebra

The Fundamental Theorem of Algebra (FTOA) tells us that a polynomial in one variable of degree #n > 0# has a Complex (possibly Real) zero.

A straightforward corollary of the FTOA, often stated with it, is that a polynomial of degree #n > 0# has exactly #n# Complex (possibly Real) zeros, counting multiplicity.

In our example, #f(x)# is of degree #5#, so has exactly #5# Complex (possibly Real) zeros.

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Descartes' Rule of Signs

The signs of #f(x)# are in the pattern: #+ - + - + -#. With #5# changes of sign, Descartes' Rule of Signs tells us that #f(x)# has #5#, #3# or #1# positive Real zero.

The signs of #f(-x)# are in the pattern #- - - - - -#. With #0# changes of sign, we can deduce that #f(x)# has no negative Real zeros.

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Rational Roots Theorem

By the rational roots theorem, any rational zeros of #f(x)# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-6# and #q# a divisor of the coefficient #4# of the leading term.

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 #f(x)# has no rational zeros. It has an irrational Real zero somewhere in #(1, 3/2)#

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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 #f(x)# has one Real and four non-Real Complex zeros.

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

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