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

2 Answers
Jul 26, 2015

Answer:

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

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

Explanation:

Factor by grouping:

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

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

Jul 26, 2015

Answer:

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

Explanation:

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))#