How do you find all the real and complex roots and use Descartes Rule of Signs to analyze the zeros of # P(x) = 2x^5 + 7x^3 + 6x^2 - 2 #?

1 Answer
Jul 31, 2016

Answer:

See explanation...

Explanation:

#P(x)=2x^5+7x^3+6x^2-2#

The pattern of signs of the coefficients is #+ + + -#. Since there is only one change of signs, this quintic has exactly #1# positive Real zero.

#P(-x) = -2x^5-7x^3+6x^2-2#

The pattern of signs of the coefficients of #P(-x)# is #- - + -#. Since there are #2# changes of signs, #P(x)# has exactly #0# or #2# negative Real zeros.

By the fundamental theorem of algebra, #P(x)# has exactly #5# zeros counting multiplicity, so that leaves #4# or #2# Complex zeros, occuring in Complex conjugate pairs.

By the rational root theorem, any rational zeros of #P(x)# must be expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-2# and #q# a divisor of the coefficient #2# of the leading term.

That means that the only possible rational zeros are:

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

None of these work, so #P(x)# has no rational zeros.

In common with most quintics and higher order polynomials, the zeros of #P(x)# are not expressible using elementary functions, including #n#th roots and trigonometric functions.

About the best you can do is use a numerical method such as Durand-Kerner to find approximations for the zeros:

#x_1 ~~ 0.460754#

#x_(2,3) ~~ -0.617273+-0.377799i#

#x_(4,5) ~~ 0.386896+-1.99853i#