#x^3+6x^2+3x+10# ?

1 Answer
George C.
May 17, 2017

#x^3+6x^2+3x+10# has real zero:

#x_1 = -2+root(3)(-10+sqrt(73))+root(3)(-10-sqrt(73))#

and related complex zeros as derived below...

Explanation:

Whether you want to find the zeros, factor or graph this cubic function, finding the zeros is probably the first step...

#f(x) = x^3+6x^2+3x+10#

Discriminant

The discriminant #Delta# of a cubic polynomial in the form #ax^3+bx^2+cx+d# is given by the formula:

#Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd#

In our example, #a=1#, #b=6#, #c=3# and #d=10#, so we find:

#Delta = 324-108-8640-2700+3240 = -7884#

Since #Delta < 0# this cubic has #1# Real zero and #2# non-Real Complex zeros, which are Complex conjugates of one another.

Tschirnhaus transformation

To make the task of solving the cubic simpler, we make the cubic simpler using a linear substitution known as a Tschirnhaus transformation.

#0=f(x)=x^3+6x^2+3x+10#

#=(x+2)^3-9(x+2)+20#

#=t^3-9t+20#

where #t=x+2#

Cardano's method

We want to solve:

#t^3-9t+20=0#

Let #t=u+v#.

Then:

#u^3+v^3+3(uv-3)(u+v)+20=0#

Add the constraint #v=3/u# to eliminate the #(u+v)# term and get:

#u^3+27/u^3+20=0#

Multiply through by #u^3# and rearrange slightly to get:

#(u^3)^2+20(u^3)+27=0#

Use the quadratic formula to find:

#u^3=(-20+-sqrt((20)^2-4(1)(27)))/(2*1)#

#=(-20+-sqrt(400-108))/2#

#=(-20+-sqrt(292))/2#

#=(-20+-2sqrt(73))/2#

#=-10+-sqrt(73)#

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

#t_1=root(3)(-10+sqrt(73))+root(3)(-10-sqrt(73))#

and related Complex roots:

#t_2=omega root(3)(-10+sqrt(73))+omega^2 root(3)(-10-sqrt(73))#

#t_3=omega^2 root(3)(-10+sqrt(73))+omega root(3)(-10-sqrt(73))#

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

Now #x=-2+t#. So the zeros of our original cubic are:

#x_1 = -2+root(3)(-10+sqrt(73))+root(3)(-10-sqrt(73))#

#x_2 = -2+omega root(3)(-10+sqrt(73))+omega^2 root(3)(-10-sqrt(73))#

#x_3 = -2+omega^2 root(3)(-10+sqrt(73))+omega root(3)(-10-sqrt(73))#

Related questions
Impact of this question
531 views around the world
You can reuse this answer
Creative Commons License