How do you factor #2x^3+162# ?

1 Answer
Aug 3, 2015

Answer:

#2x^3+162 = 2(x^3+3^4)#

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

Explanation:

First separate out the common scalar factor #2# to get:

#2x^3+162 = 2(x^3+81) = 2(x^3+3^4)#

To factor this further, we want to treat #(x^3+3^4)# as a sum of cubes, then use the identity #a^3+b^3 = (a+b)(a^2-ab+b^2)#.

Unfortunately, #root(3)(3^4)# is not rational. It can be expressed as #root(3)(81)# or #3root(3)(3)# or #3^(4/3)#. I'll use the last of these.

Let #a = x# and #b = 3^(4/3)# to find:

#x^3+3^4 = a^3+b^3 = (a+b)(a^2-ab+b^2)#

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

If we go beyond real numbers to complex numbers we can factor this further as:

#x^3+3^4 = (x+3^(4/3))(x+3^(4/3)omega)(x+ 3^(4/3)omega^2)#

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