I'm really stuck with this on my assignment. I found out that -1, 1 and 2 work just fine. But I need to prove that. Can anyone help?

Hi again Erlandas;

x

^{3}-2x^{2}-x+2=0Let's factor this equation.

For the FOIL...

FIRST must be (x

^{2})(x)=x^{3}OUTER and INNER must add-up to -2x

^{2}-xLAST...(2)(1) or (1)(2), and both numbers must be either positive or negative.

(x

^{2}-1)(x-2)=0Let's FOIL...

FIRST...(x

^{2})(x)=x^{3}OUTER...(x

^{2})(-2)=-2x^{2}INNER...(-1)(x)=-1x=-x

LAST...(-1)(-2)=2

x

^{3}-2x^{2}-x+2=0So...

(x

^{2}-1)(x-2)=0Either of these parenthetical equations must =0...

x

^{2}-1=0x

^{2}=1x=1 or -1

or...

x-2=0

x=2

Let's check our results with the original equation...

x

^{3}-2x^{2}-x+2=0x=1

1

^{3}-[2(1^{2})]-1+2=01-[(2)(1)]-1+2

1-2-1+2

0=0

x=2

x

^{3}-2x^{2}-x+2=02

^{3}-[2(2^{2})]-2+28-[(2)(4)]-2+2

8-8-2+2=0

0=0

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