How do you use the rational roots theorem to find all possible zeros of 3x5−7x2+x+6?

1 Answer
George C.
May 24, 2016

This quintic has no rational zeros.

Explanation:

f(x)=3x5−7x2+x+6

By the rational root 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 6 and q a divisor of the coefficient 3 of the leading term.

That means that the only possible rational zeros are:

±13, ±23, ±1, ±2, ±3, ±6

Evaluating f(x) for each of these, we find none work.

So this quintic has no rational zeros.

In fact it has one Real zero at x≈−0.873 and two pairs of Complex zeros.

These zeros are not expressible in terms of nth roots, so about the best you can do is find numerical approximations using Newton's method or the Durand-Kerner method.

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