How do you find all the zeros of f(x)=x^5+3x^3-x+6?

1 Answer
George C.
Jun 22, 2016

Find there are no rational zeros.

Use Durand-Kerner or similar to find approximations.

Explanation:

f(x) = x^5+3x^3-x+6

By the rational root 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 1 of the leading term.

That means that the only possible rational zeros are:

+-1, +-2, +-3, +-6

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

In common with most quintic polynomials, the zeros are not expressible in terms of nth roots, so you are left with numerical methods such as Durand-Kerner to help find approximations:

x ~~ -1.17826

x ~~ -0.202554+-1.89313i

x ~~ 0.791684+-0.882051i

See https://socratic.org/s/avxUUEiJ for another example.

Here's a sample C++ program that implements the Durand-Kerner algorithm for this example:

enter image source here

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