What are all the possible rational zeros for #f(x)=4x^4-16x^3+12x-30# and how do you find all zeros?

1 Answer
Jan 8, 2017

Answer:

The "possible" rational zeros are:

#+-1/2, +-1, +-3/2, +-5/2, +-3, +-5, +-15#

but this quartic has two irrational and two complex zeros.

Explanation:

Note that all of the coefficients of #f(x)# are divisible by #2#, so separate that off as a factor before applying the rational roots theorem:

#f(x) = 4x^4-16x^3+12x-30 = 2(2x^4-8x^3+6x-15)#

By the rational roots theorem, any rational zeros of #2x^4-8x^3+6x-15# are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-15# and #q# a divisor of the coefficient #2# of the leading term.

That means that the only possible rational zeros are:

#+-1/2, +-1, +-3/2, +-5/2, +-3, +-5, +-15#

Note that:

#2(color(blue)(5)^4) = 1250 > 1045 = 8(color(blue)(5)^3)+6(color(blue)(5))+15#

Hence any zeros lie strictly inside the circle #abs(x) = 5# in the complex plane.

So we can discard #+-5# and #+-15# as possibilities.

To cut a long story short, none of the "possible" rational zeros are zeros of this quartic.

It is a typical "nasty" quartic with complicated irrational zeros.

It is possible to solve algebraically, but much simpler to find numerical approximations to the zeros using a method such as Durand-Kerner.

#x_1 ~~ -1.2905#

#x_2 ~~ 3.9293#

#x_(3,4) ~~ 0.68061+-1.00785i#

See https://socratic.org/s/aB9Ee9wQ for another example.