Question

factorise x3-2x2-x+2​

Answer 1

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

The constant term in f(x) is are �1 and� �2

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

f(1) = (1)3 - 2(1)2 -1 + 2

= 1 - 2 - 1 + 2 = 0

According to remainder theorem f(1) = 0 so that� (x - 1) is a factor of x3 - 2x2 - x + 2

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

f(-1) = (-1)3 - 2(-1)2 �(-1) + 2

= -1 - 2 + 1 + 2 = 0

According to remainder theorem f(-1) = 0 so that� (x + 1) is a factor of x3 - 2x2 - x + 2

Putting x =� 2 in f(x), we have

f(2) = (2)3 - 2(2)2 �(2) + 2

= 8 -82 �- 2 + 2 = 0

According to remainder theorem f(2) = 0 so that� (x � 2 ) is a factor of x3 - 2x2 - x + 2

Here maximum power of x is 3 so that its can have maximum 3 factors

So our answer is (x-1)(x+1)(x-2)

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Answer 2

Answer:

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

Step-by-step explanation:

=> x^3 - 2x^2 -x +2

=> (x^3 - 2x^2) - (x-2)

=>  x^2 (x-2) - 1 (x-2)

Therefore (x^2-1) (x-2)     (In x^2-1  use a^2-b^2)

(x-1) (x+1) (x-2)    [ After factorizing (x^2-1^2) by (a^2-b^2) = (a-b) (a+b) ]