How do you factor #a^3 - 2a^2 - 4a = -8#?

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Answer 1 · 2016-02-11T04:00:17

You must mean to ask how to solve the equation.

Put everything to one side of the equation:

#a^3 - 2a^2 - 4a + 8 = 0#

Factor out a common factor from two groups: the first 2 terms and last 2 terms.

#a^2(a - 2) - 4(a - 2) = 0#

#(a^2 - 4)(a - 2) = 0#

#(a + 2)(a - 2)(a - 2) = 0#

#a = -2 and 2#

The polynomial is completely factored and the solution set is #a = +-2#.

Hopefully this helps.