Answer :

For finding zeroes, we’ll use the hit and trial method.

For that, we’ll put a value which will make our polynomial zero.

Since, here the whole equation is a combination of positive and negative that means, zero can be achieved by putting the positive or negative values.

So, first we’ll put x = 1.

⇒ 1^{3} – 2(1)^{2} – 1 + 2 = 1 – 2 + 1 – 2

= 0

⇒ x = 1 is one of the zeroes.

Now, we’ll write the equation in terms of (x – 1) to make equation quadratic and hence to find the other zeroes.

⇒ x^{3} – 2x^{2} – x + 2 = x^{2} (x – 1) – x(x – 1) – 2(x – 1)

= (x – 1)(x^{2} – x – 2)

Now, using factorization by splitting the middle term method

= (x – 1)(x^{2} – (2 – 1)x – 2)

= (x – 1)(x^{2} – 2x + 1x – 2)

= (x – 1)(x(x – 2) + 1(x – 2))

= (x – 1)(x + 1)(x – 2)

Now, to get the zeroes we’ll put these equals to zero.

⇒ x – 1 = 0, x + 1 = 0, x – 2 = 0

⇒ x = 1, x = – 1, x = 2 are the zeroes of x^{3} – 2x^{2} – x + 2

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