What are all the possible rational zeros for #f(x)=x^5+2x^4+14x^3+26x^2+40x+80# and how do you find all zeros?

1 Answer
Aug 26, 2016

Answer:

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

Explanation:

#f(x) = x^5+2x^4+14x^3+26x^2+40x+80#

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

In addition note that all of the coefficients of #f(x)# are positive, so it has no positive zeros.

So the only possible rational zeros of #f(x)# are:

#-1, -2, -4, -5, -8, -10, -16, -20, -40, -80#

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

Typically for a quintic or polynomial of higher degree, the zeros of #f(x)# are not expressible in terms of #n#th roots or elementary functions (including trigonometric functions).

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

#x_1 ~~ -1.92974#

#x_(2,3) ~~ -0.135287+-3.0955i#

#x_(4,5) ~~ 0.100156+-2.07561i#

For more details on using Durand-Kerner to find such zeros of a quintic, see https://socratic.org/s/axney67c

Here's an example C++ program that implements the Durand-Kerner algorithm for our current example...

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