How do you completely factor #x^3+2x-4x-8#?

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Answer 1 · 2018-06-16T16:08:00

#(x+2)^2(x-2)=0#

I assume that the correct equation is #x^3+color(red)(2x^2)-4x-8=0#

Now group the first two terms and the last two terms

#(x^3+2x^2)+(-4x-80)=0#

Take out the common terms from each group

#x^2# from first group

#-4# from last group

#x^2(x+2)-4(x+2)=0#

Since #x+2# is common in both take it out

#(x+2)(x^2-4)=0#

#x^2-4# can be factorize into #(x+2)(x-2)#

Then the factorized form of the equation will be

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

#=>(x+2)^2(x-2)=0#