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

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Answer 1 · 2015-07-26T17:54:10

Factor $x^3 - 2x^2 - 4x + 8$

Ans: $(x - 2)^2(x + 2)$

Factor by grouping:

$x^2(x - 2) - 4(x - 2) = (x - 2)(x^2 - 4) =$

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

Answer 2 · 2015-07-26T17:56:00

Factor by grouping first, then use the formula for the difference of two squares.

Your expression looks like this

$x^3 - 2x^2 - 4x + 8$

You can factor this expression by grouping the first and third terms

$(x^3 - 4x) - 2x^2 + 8$

This is equivalent to

$x(x^2 - 4) - 2(x^2 - 4) = (x^2 - 4)(x-2)$

The first binomial is actually the difference of two perfect squares, which means that you can write it as

$x^2 - 4 = x^2 - 2^2 = (x-2)(x+2)$

Your expression will thus be

$x^3 - 2x^2 - 4x + 8 = color(green)((x-2)(x+2)(x-2))$

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