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

##### 1 Answer

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

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

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

where

#### Explanation:

Firstly, all of the terms are divisible by

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

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

Let

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

Add the constraint

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

Multiply through by

#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

#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

Then:

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