How do you find all the zeros of #x^3 + 2x^2 + 5x +1#?

1 Answer
Mar 12, 2016

Answer:

Use Cardano's method...

Explanation:

#f(x) = x^3+2x^2+5x+1#

To cut down on the number of fractions we need to work with, multiply by #3^3 = 27# first...

#0 = 27f(x)#

#=27x^3+54x^2+135x+27#

#=(3x+2)^3+33(3x+2)-47#

Substitute #t = 3x + 2# ...

#=t^3+33t-47#

Using Cardano's method, substitute #t = u+v# ...

#=u^3+v^3+3(uv+11)(u+v)-47#

Add the constraint #v = -11/u# to eliminate the #(u+v)# term ...

#=u^3-(11^3/u^3)-47#

#=u^3-1331/u^3-47#

Multiply through by #u^3# to get this quadratic in #u^3# ...

#(u^3)^2-47(u^3)-1331 = 0#

Solve using the quadratic formula to get:

#u^3 = (47+-sqrt(47^2+4*1331))/2#

#= (47+-sqrt(7533))/2#

#= (47+-9sqrt(93))/2#

Since the derivation was symmetric in #u# and #v#, we can use one of these roots for #u^3# and the other for #v^3# to derive the Real root:

#t_1 = root(3)((47+9sqrt(93))/2) + root(3)((47-9sqrt(93))/2)#

and the Complex roots:

#t_2 = omega root(3)((47+9sqrt(93))/2) + omega^2 root(3)((47-9sqrt(93))/2)#

#t_3 = omega^2 root(3)((47+9sqrt(93))/2) + omega root(3)((47-9sqrt(93))/2)#

where #omega = -1/2+sqrt(3)/2i# is the primitive Complex cube root of #1#

Then #x = (t-2) / 3# hence zeros of the original function:

#x_1 = 1/3(-2+root(3)((47+9sqrt(93))/2) + root(3)((47-9sqrt(93))/2) )#

#x_2 = 1/3(-2 + omega root(3)((47+9sqrt(93))/2) + omega^2 root(3)((47-9sqrt(93))/2))#

#x_3 = 1/3(-2 + omega^2 root(3)((47+9sqrt(93))/2) + omega root(3)((47-9sqrt(93))/2))#