How do you use the rational roots theorem to find all possible zeros of f(x)=3X5−2X2+16X−2?

1 Answer
George C.
Jun 22, 2016

Use a numerical method to find:

x≈0.12701

x≈1.03966±0.998855i

x≈−1.10317±1.14378i

Explanation:

f(x)=3x5−2x2+16x−2

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 −2 and q a divisor of the coefficient 3 of the leading term.

That means that the only possible rational zeros of f(x) are:

±13, ±23, ±1, ±2

You can substitute each of these into the formula for f(x) to find that none of them is a zero.

So this quintic has no rational zeros.


By the fundamental theorem of algebra it does have exactly 5 Complex, possibly Real, zeros, counting multiplicity.

In addition, since its coefficients are Real, any non-Real Complex zeros must occur in Complex conjugate pairs.

Typically for a quintic, the zeros of this particular one are not expressible in terms of nth roots.

About the best we can do is find numeric approximations using a method like Durand-Kerner. See https://socratic.org/s/avyDEG5X for a fuller explanation, but the Durand-Kerner algorithm finds approximations for all 5 zeros at once.

The results I found were:

x≈0.12701

x≈1.03966±0.998855i

x≈−1.10317±1.14378i

Here's the C++ program I used to find them:

enter image source here

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