How do you factor #x^3 - 6x^2 + 3x - 10#?

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Answer 1 · 2016-05-09T20:22:48

See explanation...

Use Cardano's method.

First a simple Tschirnhaus transformation, to eliminate the term of degree #2#...

#0 = x^3-6x^2+3x-10=(x-2)^3-9(x-2)-20#

Let #t=x-2# to get the simplified cubic equation:

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

Next substitute #t = u + v# to get:

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

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

#u^3+(3/u)^3-20=0#

Multiply through by #u^3# to get:

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

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

#=10+-sqrt(400-108)/2#

#=10+-sqrt(292)/2#

#=10+-sqrt(4*73)/2#

#=10+-sqrt(73)#

The derivation was symmetric in #u# and #v#, hence (noting #x = t+2#) we can deduce that the Real zero of the original cubic is:

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

and Complex zeros:

#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))#

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

Then:

#x^3-6x^2+3x-10 = (x-x_1)(x-x_2)(x-x_3)#