How do you factor #32x^3 - 4#?

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Answer 1 · 2016-02-07T13:32:30

#4(2x-1)(4x^2+2x+1)#

the given : #32x^3-4#

common monomial factor #=4#
factor out #4# first

#4(8x^3-1)#

use the "difference of two cubes " form

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

for #8x^3-1#

so that

#8x^3-1=(2x-1)(4x^2+2x+1)#

final answer

#32x^3-4=4*(2x-1)(4x^2+2x+1)#