How do you use the rational roots theorem to find all possible zeros of f(x)=x4+3x3−4x2+5x−12?

1 Answer
George C.
Jun 13, 2016

f(x) has no rational zeros.

Explanation:

Given:

f(x)=x4+3x3−4x2+5x−12

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 −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

None of these work, so f(x) has no rational zeros, but in the process of checking, we find:

f(−6)=462>0

f(−4)=−32<0

f(1)=−7<0

f(2)=22>0

So f(x) has a couple of Real irrational zeros in (−6,−4) and (1,2)

Here's a graph of f(x)10 ...

graph{(x^4+3x^3-4x^2+5x-12)/10 [-10.705, 9.295, -6.56, 3.44]}

This particular quartic is messy to solve algebraically.

You can use a numerical method such as Durand Kerner, to find approximations to the zeros of our quartic:

−4.3358

1.4332

−0.04871±1.38880i


Footnote

Interestingly, the similar looking quartic x4+3x3−4x2+6x−12 is much easier to solve, since it factors as:

x4+3x3−4x2+6x−12=(x2+2)(x2+3x−6)

Still no rational factors, but much, much easier.

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