Answer :

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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