Question

Find the possible value of p , if px-p is divided by x-4/p and remainder is p^2-p

Answer 1

Let,

P( x ) = px - p , g ( x ) = x - 4/ p and r ( x ) = p^2 - p.

Given that when p ( x ) is divided by g ( x ) then remainder is r ( x ).

Now, 
  x - 4 / p ) px - p ( p

                 px - 4
               -      +
          --------------------
                        4 - p

So, we got that that when p ( x ) is divided by g(x ) then remainder is ( 4 - p) but according to question remainder is p^2 - p.So, we came on a conclusion that ( 4 - p ) is equal to p^2 - p.

4 - p = p ^ 2 - p

4 = p^2 - p + p

4 = p^2

p = √4

p = +2 , -2. 

Proof : when p = 2,

px - p = 2 * x - 2 = 2x - 2

x - 4 / p = x - 4 / 2 = x - 2

p^2 - p = 2^2 - 2 = 4 - 2 = 2.

So, when ( 2x - 2 ) is divided by ( x -2 ) then remainder should be 2.Let's check it.

                  x - 2 ) 2x - 2 ( 2

                            2x - 4
                          -     +
                       -----------------
                                  2.

So, it is right.

Now, when p = -2.

px - p = -2x - ( -2 ) = -2x + 2

x - 4/ p = x - ( 4 / -2 ) = x - ( -2 ) = x + 2.

p^2 - p = ( -2 ) ^2 - ( -2 ) = 4 + 2 = 6.

So, when ( - 2x + 2 ) is divided by ( x + 2 ) then remainder should be 6.Let's check it.
 
                   x + 2 ) - 2x + 2 ( -2

                              - 2x - 4
                                
                              +     +
                         ------------------
                                      6.

So, it is also right.

Hence possible values for p = +2 and -2.