What are all the possible rational zeros for #f(x)=x^5-6x^4+12x^2-8x+36# and how do you find all zeros?

1 Answer
Aug 27, 2016

Answer:

Find there are no rational zeros. Use a numerical method to find approximations.

Explanation:

#f(x) = x^5-6x^4+12x^2-8x+36#

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 #36# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1, +-2, +-3, +-4, +-6, +-9, +-12, +-18, +-36#

None of these works, so #f(x)# has no rational zeros.

About the best we can do is use the Durand-Kerner or similar numerical method to find approximations:

#x_1 ~~ 5.63048#

#x_2 ~~ 2.05347#

#x_3 ~~ -1.84949#

#x_(4,5) ~~ 0.0827684+-1.29486i#

See https://socratic.org/s/axpvv6Hk for some more description of the method.

For the current example, the above approximations were found using this C++ program...

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