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

2 Answers
Jun 28, 2016

Answer:

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

Explanation:

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

grouping:
#color(white)("XXX")=color(red)(""(x^3-2x^2))+color(blue)(""(-4x+8))#

#color(white)("XXX")=color(red)(x^2(x-2))+color(blue)((-4)(x-2))#

#color(white)("XXX")=(color(red)(x^2)color(blue)(-4))(x-2)#

#color(white)("XXX")=color(green)(""(x^2-4))(x-2)#

then using difference of squares:
#color(white)("XXX")=color(green)((x+2)(x-2))(x-2)#

#color(white)("XXX")=(x+2)(x-2)^2#

Jun 28, 2016

Answer:

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

Explanation:

grouping in 'pairs' gives.

#[x^3-2x^2]+[-4x+8]#

factorising each pair.

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

'Taking out' a common factor of (x - 2)

#(x-2)(x^2-4)........ (A)#

#x^2-4 color(blue)" is a difference of squares"# and factorises

#color(red)(|bar(ul(color(white)(a/a)color(black)(a^2-b^2=(a-b)(a+b))color(white)(a/a)|)))#

#x^2=(x)^2" and " 4=(2)^2rArra=x" and "b=2#

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

Substituting in (A) : #(x-2)(x-2)(x+2)#

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