How do you find all the real and complex roots of #2x^5-4x^4-4x^2+5=0#?

1 Answer
Jul 6, 2016

Answer:

Use a numerical method to find approximations:

#x ~~ 2.2904#

#x ~~ 0.925274#

#x ~~ -0.80773#

#x ~~ -0.203974+-1.19116i#

Explanation:

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

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Fundamental theorem of algebra

The FTOA tells us that any non-zero polynomial of degree #> 0# has a Complex (possibly Real) zero. A corollary of this, often quoted as part of the FTOA, 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# zeros counting multiplicity.

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Rational root theorem

SInce #f(x)# has integer coefficients, any rational zeros must be expressible in the for #p/q# for integers #p, q# with #p# a divisor of the constant term #5# and #q# a divisor of the coefficient #2# of the leading term.

That means that the only possible rational zeros are:

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

None of these satisfy #f(x) = 0#, so #f(x)# has no rational zeros.

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Descartes rule of signs

The signs of the coefficients of #f(x)# follow the pattern #+ - - +#. With two changes of signs, that means that #f(x)# has #0# or #2# positive Real zeros.

The signs of the coefficients of #f(-x)# follow the pattern #- - - +#. With one change of sign, that means that #f(x)# has exactly #1# negative Real zero.

That leaves #2# or #4# non-Real, Complex zeros, which will occur in Complex conjugate pairs.

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Quintic

In common with most quintics (and higher degree polynomials), the zeros of this one cannot be expressed using elementary functions like #n#th roots, or even trigonometric functions. About the best we can do is find approximations using a numerical method such as Durand-Kerner to find zeros:

#x ~~ 2.2904#

#x ~~ 0.925274#

#x ~~ -0.80773#

#x ~~ -0.203974+-1.19116i#

See https://socratic.org/s/avVFw8eC for more details of the method and another example quintic.