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

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

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

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