How do you factor completely #p(x)=(x^3)-(6x^2)-4x-10#?

1 Answer
Nov 16, 2015

Answer:

#p(x) = (x-x_1)(x-x_2)(x-x_3)#

where #x_1#, #x_2# and #x_3# are as below:

Explanation:

#p(x) = x^3-6x^2-4x-10#

Attempt to solve #p(x) = 0#

Let #t = x-2#

Then:

#t^3 - 16t -34 = (x-2)^3 -16(x-2)-34#

#= x^3-6x^2+12x-8-16x+32-34#

#= x^3-6x^2-4x-10 = p(x) = 0#

Let #t = u+v#

#0 = (u+v)^3-16(u+v)-34#

#= u^3+v^3+(3uv-16)(u+v)-34#

Add the constraint #v = 16/(3u)# to make #(3uv-16) = 0#

Then:

#0 = u^3+(16/(3u))^3-34 = u^3+4096/(27u^3)-34#

Multiply both ends by #27u^3# to get:

#0 = 27(u^3)^2 - 918(u^3) + 4096#

Use the quadratic formula to get:

#u^3 = (918+-sqrt(918^2-(4*27*4096)))/(2*27)#

#= (918+-sqrt(400396))/54#

#= (918+-6sqrt(11121))/54#

#= (3(153+-sqrt(11121)))/27#

The derivation is symmetric in #u# and #v#, hence:

#t = root(3)((3(153+sqrt(11121)))/27) + root(3)((3(153-sqrt(11121)))/27)#

#= 1/3(root(3)(3(153+sqrt(11121))) + root(3)(3(153-sqrt(11121))))#

So the Real root of #p(x) = 0# is:

#x_1 = 2+t#

#= 2+1/3(root(3)(3(153+sqrt(11121))) + root(3)(3(153-sqrt(11121))))#

The Complex roots are:

#x_2 = 2+1/3(omega root(3)(3(153+sqrt(11121))) + omega^2 root(3)(3(153-sqrt(11121))))#

#x_3 = 2+1/3(omega^2 root(3)(3(153+sqrt(11121))) + omega root(3)(3(153-sqrt(11121))))#

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