How do you factor #8c^3-8c#?

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Answer 1 · 2018-05-21T20:12:09

See a solution process below:

There common factor for each term is: #color(red)(8c)#:

#8c^3 = color(red)(8c) * c^2#

#8c = color(red)(8c) * 1#

We can then factor as:

#8c^3 - 8c => (color(red)(8c) * c^2) - (color(red)(8c) * 1) => color(red)(8c)(c^2 - 1)#

#c^2 - 1# is a special form of the quadratic where:

#c^2 - 1 = (c + 1)(c - 1)#

#color(red)(8c)(c^2 - 1) => color(red)(8c)(c + 1)(c - 1)#