How do you factor the expression #6x^4+24x^2-72x#?

1 Answer
Apr 24, 2016

Answer:

#6x^4+24x^2-72x#

#=6x(x^3+4x-12)#

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

where #x_1#, #x_2# and #x_3# are the zeros of the cubic as follows...

Explanation:

Firstly, all of the terms are divisible by #6x#, so we can separate that out as a factor:

#6x^4+24x^2-72x=6x(x^3+4x-12)#

We can factor the remaining cubic factor using Cardano's method:

Let #x = u + v# and solve #x^3+4x-12 = 0#

#u^3+v^3+(3uv+4)(u+v)-12 = 0#

Add the constraint #v = -4/(3u)# to eliminate the term in #(u+v)# and get:

#u^3-(4/(3u))^3-12 = 0#

Multiply through by #27u^3# to get:

#27(u^3)^2-324(u^3)-64 = 0#

Use the quadratic formula to find:

#u^3=(324+-sqrt(324^2+4*27*64))/54#

#=(324+-sqrt(111888))/54#

#=(324+-12sqrt(777))/54#

#=(162+-6sqrt(777))/27#

Since the square root is Real, this gives us a Real value for #u^3#.so since the derivation was symmetric in #u# and #v# we can use one of these roots for #u^3# and the other for #v^3# to find the Real root:

#x_1 = 1/3(root(3)(162+6sqrt(777)) + root(3)(162-6sqrt(777)))#

and Complex roots:

#x_2 = 1/3(omega root(3)(162+6sqrt(777)) + omega^2 root(3)(162-6sqrt(777)))#

#x_3 = 1/3(omega^2 root(3)(162+6sqrt(777)) + omega root(3)(162-6sqrt(777)))#

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

Then:

#x^3+4x-12 = (x-x_1)(x-x_2)(x-x_3)#