**x^3-2x^2-x+2 = 0**

**Solution......**

x^2(x-2)-1(x-2) = 0

=>(x-2)(x^2-1) = 0

=>(x-2)(x+1)(x-1) = 0 [x^2-1 = x^2-1^2 = (x+1)(x-1)]

=> x-2=0 I => x+1=0 I => x-1=0

=>x=2 I => x=-1 I => x=1

Hence x={2,-1,1}

To solve x^3-2x^2-x+2 = 0.

Solution:

Grouping as

x^2(x-2)-1(x-2) = 0. Or

(x-2)(x^2-1) = 0. Or

(x-2)(x+1)(x-1) = 0, as x^2-1 = x^2-1^2 = (x+1)(x-1).

So x-2= 0, Or x+1 = 0, Or x-1) = 0. From these 3 equations, we get:

x=2,

x=-1 and

x= 1.

To solve the equation, we'll group the first 2 terms together and the next 2 terms together, so that we can factorize:

(x^3 - 2x^2) - (x - 2) = 0

x^2(x - 2) - (x - 2) = 0

We'll factorize again:

(x - 2)(x^2 - 1) = 0

We'll put each factor as being 0.

x-2 = 0

x = 2

x^2 - 1 = 0

The expression is a difference of squares:

x^2 - 1 = (x - 1)(x + 1) = 0

x-1 = 0

x = 1

x + 1 = 0

x = -1

**The solutions of the equation are: {-1,1,2}.**

The equation x^3 - 2x^2 - x + 2 = 0 can be solved using factorization.

x^3 - 2x^2 - x + 2 = 0

x^2(x - 2) - 1(x - 2) = 0

(x^2 - 1)(x - 2) = 0

Use the expansion x^2 - 1 = (x - 1)(x + 1)

(x + 1)(x - 1)(x - 2) = 0

This gives the roots of the equation as -1, 1 and 2

`x^3 - 2x^2 - x + 2 = 0 `

group the factors

`(x^3 - 2x^2) (- x + 2) `

take out the numbers the numbers in the parentheses have in common

`x^2(x-2) -1(x-2) `

`(x^2-1)(x-2) `

set those parenthesis equal to 0

x-2=0

`x=2`

`x^2-1=o`

`x^2=1`

`x=+-1`

`x=1 x=-1`

so x is 1,-1 and 2

`x^3 - 2x^2 - x + 2 = 0`

You can solve by factoring

`(x^3 - 2x^2) (- x + 2)`

`x^2(x-2) -1(x-2)`

`(x^2-1) (x-2)`

x-2=0 x=2

`x^2-1 = (x-1)(x+1) = 0`

`x=1 x=-1`

**x=1 x=-1 x=2**