x^2-3x+2/x-1

Question

How would you work out this problem

Answer 1

 X^2 - 3X + 2/ ( X -1) =

 

  X^2 ( X -1 ) - 3X ( X -1 ) + 2 = 0

 

  X^3 - X^2 - 3X^2 + 3X + 2 = 0

 

  X^3 - 4 X^2 + 3X + 2 = 0 

 

   The rational roots are  ± 1, ± 2

   

     Try X =2

 

     ( 2) ^3 - 4 ( 2^2) + 3 ( 2) + 2 =

         8  - 16 + 6 + 2 = 0

 

    X^3 -4X^2 +3X + 2 = ( X -2 ) ( X^2 -2X -1 ) 

 

        X- 2 = 0    X = 2

 

          X^2 - 2X -1 = 0 

 

         X^2 - 2x +1 -1-1  =0

 

          ( X - 1 ) ^2 = 2

 

         X = 1 ±√2  

Answer 2

Hello, Samantha!  

 

I assume the problem is the whole of (x^2-3x+2) being divided by x-1, correct?

 

If so, here's what we'll do:

 

First, factor the numerator.  This way we can see if there is a factor that is the same as the denominator, to cancel those out.

 

x^2-3x+2 factors into (x-1)(x-2).  Now we have ((x-1)(x-2))/x-1.  Notice that x-1 exists in both the numerator and denominator, and because x-1 in the numerator is being multiplied, we can 'cancel out' the x-1 terms in both the numerator and the denominator.

 

This leaves us with the answer of x-2.

 

Hope this helped!

 

*Edit*  Terribly sorry, looking at Parviz's answer, it appears I made a mistake and thought you were only being asked to reduce.