How do you factor #2x^3 - 4x#?

1 Answer
Apr 21, 2016

Answer:

#2x^3-4x = 2x(x-sqrt(2))(x+sqrt(2))#

Explanation:

The difference of squares identity can be written:

#a^2-b^2 = (a-b)(a+b)#

We use this with #a=x# and #b=sqrt(2)# later.

#color(white)()#
First note that both terms are divisible by #2# and by #x#, so they are both divisible by #2x#, so separate that out as a factor first...

#2x^3-4x = 2x(x^2-2) = 2x(x^2-(sqrt(2))^2) = 2x(x-sqrt(2))(x+sqrt(2))#