How do you factor #16x^3 - 250#?

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Answer 1 · 2015-09-27T10:01:41

Separate out the common scalar factor #2#, then use the difference of cubes identity to find:

#16x^3-250 = 2(2x-5)(4x^2+10x+25)#

Both #16# and #250# are divisible by #2# so separate that out first.

#16x^3-250 = 2(8x^3-125)#

Now #8 = 2^3# and #125 = 5^3#

So we find:

#2(8x^3-125)#

#=2((2x)^3-5^3)#

#=2(2x-5)((2x)^2+(2x)*5+5^2)#

#=2(2x-5)(4x^2+10x+25)#

...using the difference of cubes identity:

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