What are all the possible rational zeros for #f(x)=-6x^3+5x^2-2x+18# and how do you find all zeros?

1 Answer
Jan 20, 2018

Answer:

#x=root(3)((100345392+sqrt(100345392^2+181081080576))/68024448)+root(3)((100345392-sqrt(100345392^2+181081080576))/68024448)+5/18#

Explanation:

We can use the rational roots theorem to find that all the possible rational roots are #+-# all of the factors of #18# divided by all of the factors of #6#:
#+-18,+-9,+-6,+-9/2,+-3,+-3/2,+-1,+-2/3,+-1/2,+-1/3,+-1/6#

By using Descartes' rule of signs and showing that #f(-x)# has no sign changes, we can exclude all the negative possibilities:
#18,9,6,9/2,3,3/2,1,2/3,1/2,1/3,1/6#

After trying these, we find that none of them are zeroes. This means that the polynomial has no rational zeroes.

We can also compute the cubic discriminant and get a value of #−304580#. This means that the polynomial has one real solution and two complex ones.

To find the real solution, I will first transform the equation into a depressed cubic. This is done by first dividing so the leading coefficient is #1#:
#x^3-5/6x^2+1/3x-3=0#

Then we substitute #x=t+5/18#:
#(t+5/18)^3-5/6(t+5/18)^2+1/3(t+5/18)-3=0#

#t^3+11/108t-8603/2916=0#

Now that we have a depressed cubic (ie no #t^2#'s), we can solve by substituting #t=u+v#:
#(u+v)^3+11/108(u+v)-8603/2916=0#

#u^3+v^3+3u^2v+3uv^2+11/108u+11/108v-8603/2916=0#

Now we can factor:
#u^3+v^3+3uv(u+v)+11/108(u+v)-8603/2916=0#

#u^3+v^3+(3uv+11/108)(u+v)-8603/2916=0#

Since #u# and #v# are any arbitrary numbers, I can add a condition that #3uv+11/108=0#. This allows us to get rid of the middle term and rewrite #v# in terms of #u#:
#u^3+(-11/(324u))^3-8603/2916=0#

#u^3-1331/(34012224u^3)-8603/2916=0#

Multiply through by #u^3#:
#u^6-8603/2916u^3-1331/34012224=0#

This is a quadratic in #u^3#, so we can solve using the quadratic formula:
#u^3=(100345392+-sqrt(100345392^2+181081080576))/68024448#

#u=root(3)((100345392+-sqrt(100345392^2+181081080576))/68024448)#

We can interpret one of the roots as the value for #u#, and the other for #v#. This gives that #t# is equal to:
#t=u+v=root(3)((100345392+sqrt(100345392^2+181081080576))/68024448)+root(3)((100345392-sqrt(100345392^2+181081080576))/68024448)#

Now, we just put back the solution for #t# into #x=t+5/18# to get the solution for #x#:
#x=root(3)((100345392+sqrt(100345392^2+181081080576))/68024448)+root(3)((100345392-sqrt(100345392^2+181081080576))/68024448)+5/18#