How do you list all possible roots and find all factors of x5+7x3−3x−12?

1 Answer
George C.
Dec 23, 2016

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)=x5+7x3−3x−12


Rational roots theorem

By the rational roots theorem, any rational zeros of f(x) are expressible in the form pq 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 |x|≥3 we have:

∣∣7x3∣∣+|3x|+|12|≤∣∣7x3∣∣+|3x|+|4x|=∣∣7x3∣∣+|7x|<∣∣7x3∣∣+∣∣x3∣∣=∣∣8x3∣∣<∣∣x5∣∣

So no x with |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)


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:

x1≈1.22622

x2,3≈0.101096±2.734i

x4,5≈−0.714207±0.892944i

Here's the C++ program I used:

enter image source here

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