Let, f (x) = x³+2x²-x-2

The constant term in f (x) is equal to -2 and factors of -2 are .

Putting x = 1 in f (x), we have

f (1) = (1)³ + 2 (1)² – 1 – 2

= 1 + 2 – 1 – 2

= 0

Therefore, (x – 1) is a factor of f (x).

Similarly, (x + 1) and (x + 2) are the factors of f (x).

Since, f (x) is a polynomial of degree 3. So, it cannot have more than three linear factors.

Therefore, f (x) = k (x – 1) (x + 1) (x + 2)

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

Putting x = 0 on both sides, we get

0 + 0 – 0 – 2 = k (0 – 1) (0 + 1) (0 + 2)

-2 = -2k

k = 1

Putting k = 1 in f (x) = k (x – 1) (x + 1) (x + 2), we get

f (x) = (x – 1) (x + 1) (x + 2)

Hence,

x³+2x²-x-2 = (x – 1) (x + 1) (x + 2)