# Using factor theo

Let, f (x) = x3-2x2-x+2

The factors of the constant term +2 are

Putting x = 1, we have

f (1) = (1)3 – 2 (1)2 – (1) + 2

= 1 – 2 – 1 + 2

= 0

So, (x - 1) is a factor of f (x)

Let us now divide

f (x) = x3-2x2-x+2 by (x - 1) to get the other factors of f (x)

Using long division method, we get

x3-2x2-x+2 = (x - 1) (x2 - x - 2)

x2 - x - 2 = x2 - 2x + x - 2

= x (x - 2) + 1 (x - 2)

= (x + 1) (x - 2)

Hence, x3-2x2-x+2 = (x - 1) (x + 1) (x - 2)

= (x - 1) (x + 1) (x – 2)

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