What are all the zeroes of #f(x) = 2x^3 + 3x^2 + 4x - 10#?

2 Answers
Jan 11, 2016

Answer:

#1.060419 and -1.280210+-i*1.753940#

Explanation:

Rewriting the function:
#f(x)=2*(x^3+1.5x^2+2x-5)#

The easy way (trying as root #+-1, +-5 and +-1/5#) doesn't help us to find the zeros.

Then we can go the hard way as described in
https://en.wikipedia.org/wiki/Cubic_function#General_formula_for_roots

So for the general cubic equation

#a x^3 + b x^2 + c x + d = 0#

#x_k = -(1/(3a)) (b+u_k C+Delta_0/(u_k*C))#, #k=1,2, 3#

where
#u_1 = 1#, #u_2 = (-1 + i*sqrt(3))/2#, #u_3 =(-1 - i*sqrt(3))/2#

#C = root(3)((Delta_1 + sqrt(Delta_1^2 - 4*Delta_0^3))/2)# with
#Delta_0 = b^2-3 a c#
#Delta_1= 2 b^3-9 a b c+27 a^2 d#

In case:
#Delta_0=1.5^2-3*1*2=2.25-6=-3.75#
#Delta_1=2*3.375-9*1*1.5*2+27*1*(-5)=6.75-27-135=-155.25#

#C=root(3) ((-155.25+sqrt((-155.25)^2-4*(-3.75)^3))/2)#
#C=root(3) ((-155.25+sqrt(24313.5))/2)=root(3)(.338934)=.697223#

For #k=1 and u_1=1#
#x_1=(-1/3)(1.5+.697223+(-3.75)/.697223)=(-1/3)(2.197223-5.378480)# => #x_1=1.060419#

We could continue with this formula, but dividing #(x^3+1.5x^2+2x-5)# by # (x-1.060419)# we get #(x^2+2.560419x+4.715117)#, whose roots we can obtain in this way:
#(x^2+2.560419x+4.715117)=0#
#Delta=6.555745-18.860468=-12.304723#
#x=(-2.560.419+-i*3.507809)/2# => #x_2=-1.280210+i*1.753904#, #x_3=-1.280210-i*1.753904#

Jan 16, 2016

Answer:

#x_1 = 1/6(-3 + root(3)(621+6sqrt(10806)) + root(3)(621-6sqrt(10806)))#

#x_2=1/6(-3 + omega root(3)(621+6sqrt(10806)) + omega^2 root(3)(621-6sqrt(10806)))#

#x_3=1/6(-3 + omega^2 root(3)(621+6sqrt(10806)) + omega root(3)(621-6sqrt(10806)))#

Explanation:

When a cubic has one Real and #2# non-Real Complex zeros, I like to use Cardano's method...

Given #f(x) = 2x^3+3x^2+4x-10#

Let #t_1 = 2x#

Let #t_2 = t_1+1#

Then:

#4f(x) = 8x^3+12x^2+16x-40#

#=(2x)^3+3(2x)^2+8(2x)-40#

#=t_1^3+3t_1^2+8t_1-40#

#=(t_1^3+3t_1^2+3t_1+1) + (5t_1+5)-46#

#=(t_1+1)^3+5(t_1+1)-46#

#=t_2^3+5t_2-46#

Next let #t_2 = u+v#

#0 = (u+v)^3+5(u+v)-46#

#= u^3+v^3+(3uv+5)(u+v)-46#

To eliminate the term in #(u+v)# add the constraint #v = -5/(3u)#

#= u^3 - (5/(3u))^3-46#

#= u^3 - 125/(27u^3)-46#

Multiply through by #27u^3# and rearrange to obtain this quadratic in #u^3#:

#27(u^3)^2-1242(u^3)-125 = 0#

Use the quadratic formula to find:

#u^3 = (1242+-sqrt(1242^2+(4*27*125)))/54#

#=(1242+-sqrt(1556064))/54#

#=(1242+-12sqrt(10806))/54#

#=(621+-6sqrt(10806))/27#

So the Real value of #u# is given by:

#u = root(3)((621+-6sqrt(10806))/27) =1/3 root(3)(621+-6sqrt(10806))#

Then since the derivation was symmetric in #u# and #v# we can deduce that one of these values is #u# and the other #v#.

#t_2 = u+v#

#= 1/3 root(3)(621+6sqrt(10806)) + 1/3 root(3)(621-6sqrt(10806))#

Hence the Real root of the original cubic is:

#x_1 = (t_2-1)/2#

#=1/6(-3 + root(3)(621+6sqrt(10806)) + root(3)(621-6sqrt(10806)))#

The Complex roots come from replacing the Real value of #u# with #omega u# or #omega^2 u#, where #omega = -1/2+sqrt(3)/2i# is the primitive cube root of #1# ...

#x_2=1/6(-3 + omega root(3)(621+6sqrt(10806)) + omega^2 root(3)(621-6sqrt(10806)))#

#x_3=1/6(-3 + omega^2 root(3)(621+6sqrt(10806)) + omega root(3)(621-6sqrt(10806)))#