How do you list all possible roots and find all factors of #x^5+7x^3-3x-12#?

1 Answer
Dec 23, 2016

Answer:

Possible rational zeros are:

#+-1, +-2, +-3, +-4, +-6, +-12#

but none are actually zeros.

This quintic is not solvable using radicals and elementary functions.

Explanation:

#f(x) = x^5+7x^3-3x-12#

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Rational roots theorem

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 #-12# 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, +-12#

Note that when #abs(x) >= 3# we have:

#abs(7x^3)+abs(3x)+abs(12) <= abs(7x^3)+abs(3x)+abs(4x) = abs(7x^3)+abs(7x) < abs(7x^3)+abs(x^3) = abs(8x^3) < abs(x^5)#

So no #x# with #abs(x) >= 3# can be a zero.

Checking the other possible rational zeros, we find:

#f(-2) = -32-56+6-12 = -94#

#f(-1) = -1-7+3-12 = -17#

#f(1) = 1+7-3-12 = -7#

#f(2) = 32+7-6-12 = 21#

So #f(x)# has no rational zeros, but has an irrational zero somewhere in #(1, 2)#

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Quintic

In fact this is a typical quintic with #1# Real zero and #4# non-Real complex zeros, none of which are even expressible in terms of radicals and elementary functions - including trigonometric, exponential or logarithmic ones.

About the best you can do is find approximations using numerical methods such as Durand Kerner.

See https://socratic.org/s/aAGsRKkf for another example and a description of the Durand-Kerner algorithm for a quintic.

Using this algorithm, I found the following approximations:

#x_1 ~~ 1.22622#

#x_(2,3) ~~ 0.101096+-2.734i#

#x_(4,5) ~~ -0.714207+-0.892944i#

Here's the C++ program I used:

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