How do you factor #3x^3-24#?

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Answer 1 · 2016-02-14T05:39:24

#3x^3-24=3(x-2)(x^2+2x+4)#
# = 3(x-2)(x+1+sqrt(3)i)(x+i-sqrt(3)i)#

Using the difference of cubes formula #a^3 - b^3 = (a-b)(a^2+ab+b^2)#
we have

#3x^3-24 = 3(x^3-8)#

#=3(x^3-2^3)#

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

If we only allow real numbers, we are done. If we allow complex numbers, we can use the quadratic formula to factor #x^2+2x+4# by finding its roots.

#x^2+2x+4 = 0#
#<=> x = (-2+-sqrt(2^2-4(1)(4)))/(2(1))#

#=(-2+-sqrt(-12))/2#

#=-1+-sqrt(-3)#

#=-1+-sqrt(3)i#

Thus

#3x^3-24 = 3(x-2)(x+1+sqrt(3)i)(x+i-sqrt(3)i)#