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
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) = 4x^4-16x^3+12x-30 = 2(2x^4-8x^3+6x-15)#
By the rational roots theorem, any rational zeros of
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
So we can discard
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.