Q. 7 D5.0( 1 Vote )

Find the zeroes of the polynomials:

x3 – 2x2 – x + 2

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.


13 – 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.


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


= (x – 1)(x2 – x – 2)


Now, using factorization by splitting the middle term method


= (x – 1)(x2 – (2 – 1)x – 2)


= (x – 1)(x2 – 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 x3 – 2x2 – x + 2


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