How do you use the rational roots theorem to find all possible zeros of #f(x) = x^4 + 3x^3 - 4x^2 + 5x -12#?

1 Answer
George C.
Jun 13, 2016

#f(x)# has no rational zeros.

Explanation:

Given:

#f(x) = x^4+3x^3-4x^2+5x-12#

By the rational root 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#

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#

#color(white)()#
Footnote

Interestingly, the similar looking quartic #x^4+3x^3-4x^2+6x-12# is much easier to solve, since it factors as:

#x^4+3x^3-4x^2+6x-12 = (x^2+2)(x^2+3x-6)#

Still no rational factors, but much, much easier.

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