Question

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

Answer 1

`(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`

Answer 2

`(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)`