How do you factor completely: #12x^5 + 6x^3 + 8x^2#?

1 Answer
Jul 14, 2015

Answer:

#12x^5+6x^3+8x^2 = 2x^2(6x^3+3x+4)#
#= 12x^2(x-x_1)(x-x_2)(x-x_3)#

where #x_1, x_2, x_3# are defined below.

Explanation:

#12x^5+6x^3+8x^2 = 2x^2(6x^3+3x+4)#

Let #f(x) = 6x^3+3x+4#.

This is way too messy to solve, but for the record...

Use Cardano's method to solve #f(x) = 0#.

Let #x = u + v#

#f(x) = 6(u+v)^3 + 3(u+v) + 4#

#=6u^3+6v^3+(18uv+3)(u+v) + 4#

Let #v = -1/6u#.

Then #18uv+3 = 0# and

#f(x) = 6u^3-1/(36u^3)+4#

If #f(x) = 0# then:

#6u^3-1/(36u^3)+4 = 0#

Multiply through by #36u^3# to get:

#216(u^3)^2+144(u^3)-1 = 0#

From the quadratic formula:

#u^3 = (-144 +-sqrt(144^2+4*216))/(2*216)#

#=(-2^4 3^2+-sqrt(2^8 3^4 + 2^5 3^3))/(2^4 3^3)#

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

#=-1/3 +-sqrt(150)/36#

#=-1/3 +-(5sqrt(6))/36#

Since this derivation has been symmetric in #u# and #v#:

Let #u = root(3)(-1/3 +(5sqrt(6))/36)#

and #v = root(3)(-1/3 -(5sqrt(6))/36)#

The roots of #f(x) = 0# are:

#x_1 = u + v#
#x_2 = omega u + omega^2 v#
#x_3 = omega^2 u + omega v#

where #omega = -1/2 + i sqrt(3)/2#

#x_1# is the real root of #f(x) = 0#
#x_2# and #x_3# are a pair of complex conjugate roots.

Hence #f(x) = 6(x - x_1)(x - x_2)(x - x_3)#

and #12x^5+6x^3+8x^2 = 12x^2(x - x_1)(x - x_2)(x - x_3)#