How do you find all the zeros of #f(x) = 9x^4 – 17x^3 + 2x^2 – 3x + 33#?

1 Answer
Jun 3, 2016

Answer:

Check that it has no rational zeros, then resort to a numerical method to find approximations.

Explanation:

#f(x) = 9x^4-17x^3+2x^2-3x+33#

By the rational root theorem, any rational zeros of #f(x)# can be expressed in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #33# and #q# a divisor of the coefficient #9# of the leading term.

So the only possible rational zeros are:

#+-1/9#, #+-1/3#, #+-1#, #+-11/9#, #+-3#, #+-11/3#, #+-11#, #+-33#

In addition, notice that if you invert the signs of the coefficients on the terms of odd degree, then all of the resulting coefficients are positive. Hence #f(x)# has no negative zeros.

So the only possible rational zeros are:

#1/9, 1/3, 1, 11/9, 3, 11/3, 11, 33#

We find that #f(x) > 0# for all of these values of #x#. So #f(x)# has no rational zeros.

It is possible to solve such a quartic algebraically, but it gets very messy and is not necessarily very useful.

You can use a numerical method such as Durand-Kerner to find good approximations to the two pairs of Complex zeros that it does have.

See https://socratic.org/s/av2PtWLY for an example finding approximations for the zeros of a quintic.

In our example, we find approximations:

#x = 1.57163 +-0.656656i#

#x = -0.627188 +-0.932989i#

An example C++ program that performs the calculations would be:

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