How do you find the zeros of #f(x) = 4x^5 + x^4 + x^3 + x^2 - 2x - 2#?

1 Answer
Jul 25, 2018

Answer:

Numerically...

Explanation:

Given:

#f(x) = 4x^5+x^4+x^3+x^2-2x-2#

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 #-2# and #q# a divisor of the coefficient #4# of the leading term.

That means that the only possible rational zeros are:

#+-1/4, +-1/2, +-1, +-2#

None of these is a zero and hence we can tell that #f(x)# has no rational zeros.

Note that the signs of the coefficients are in the pattern #+ + + + - -#. With one change of sign, Descartes' Rule of Signs tells us that #f(x)# has exactly one positive real zero.

The signs of the coefficients of #f(-x)# are in the pattern #- + - + + -#. With #4# changes of signs, that means that #f(x)# has #4#, #2# or #0# negative real zeros.

In fact it has two complex conjugate pairs of non-real complex zeros.

Typically for a quintic polynomial none of these zeros is expressible in terms of #n#th roots or elementary functions. In theory they can be found using modular functions, but in practice it's probably better to stick with numerical approximations.

Using a Durand-Kerner method, I found approximate zeros:

#x_1 ~~ 0.854463#

#x_(2,3) ~~ 0.144877+-1.00403i#

#x_(4,5) ~~ -0.697109+-0.287521i#

Here's the C++ program I used:

enter image source here

See https://socratic.org/s/aSVeQs9C for another example and a fuller description of this method.