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

2 Answers
Feb 25, 2017

Answer:

#x^3-2x^2-4x+8 = (x-2)^2(x+2)#

Explanation:

Note that the ratio of the first and second terms is the same as that of the third and fourth terms. So this cubic will factor by grouping:

#x^3-2x^2-4x+8 = (x^3-2x^2)-(4x-8)#

#color(white)(x^3-2x^2-4x+8) = x^2(x-2)-4(x-2)#

#color(white)(x^3-2x^2-4x+8) = (x^2-4)(x-2)#

#color(white)(x^3-2x^2-4x+8) = (x^2-2^2)(x-2)#

#color(white)(x^3-2x^2-4x+8) = (x-2)(x+2)(x-2)#

#color(white)(x^3-2x^2-4x+8) = (x-2)^2(x+2)#

Feb 26, 2017

Answer:

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

Explanation:

Although this question has already been answered, here is an alternative way of grouping the 4 terms, which results in the same factors.

#color(blue)(x^3) color(red)(-2x^2) color(blue)(-4x)color(red)(+8)#

#=color(blue)(x^3-4x) color(red)(-2x^2+8)#

#=x(x^2-4)+(-2x^2+8)#

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

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

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

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