How do you factor #2x^3 -128x#?

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Answer 1 · 2015-10-18T11:42:12

#2x(x-8)(x+8)#

Start by breaking down your initial expression in search of possible common factors

#2x^3 = color(blue)(2 * x) * x^2#

#128x = 64 * color(blue)(2 * x)#

This means that you can use #color(blue)(2x)# as a common factor for these two terms

#2x^3 - 128x = 2x * (x^2 - 64)#

Notice that #64# is a perfect square

#64 = 8 * 8 = 8^2#

which means that you are actually dealing with the difference of two squares

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

The bracket can thus be written as

#x^2 - 64 = x^2 - 8^2 = (x-8)(x+8)#

The expression will thus be equivalent to

#2x^3 - 128x = color(green)(2x(x-8)(x+8))#