To solve for p, q and r, apply the remainder theorem. To do so, set each divisor equal to zero and solve for x.

For the first divisor,

`x+2=0`

`x=-2`

For the second divisor,

`x-1=0`

`x=1`

And for the third divisor,

`x-3=0`

`x=3`

Then, plug-in the values of x to the dividend and set it equal to their corresponding remainder.

For the first divisor, x=-2 and the remainder is -48.

`-48=(-2)^3+p(-2)^2+q(-2)+r`
`-48=-8+4p-2q+r `
`-40=4p-2q+r`                          (Let this be EQ1.)

For the second divisor, x=1 and the remainder is 0.

`0=1^3+p*1^2+q*1+r`
`0=1+p+q+r`
`-1=p+q+r`                                  (Let this be EQ2.)

And for the third divisor, x=3 and the remainder is 2.

`2=3^3+p*3^2+q*3+r`
`2=27+9p+3q+r`
`-25=9p+3q+r`                         (Let this be EQ3.)

Next, apply elimination method. So, subtract EQ2 from EQ1 to eliminate r.

      `-40=4p-2q+r`
`(-)`     `-1=p+q+r `
`--------------`
           `-39=3p-3q`
           `-13=p-q`                          (Let this be EQ4.)

Then, subtract EQ3 from EQ2 to eliminate r again.

          `-1=p+q+r`
`(-)`   `-25=9p+3q+r `
`--------------`
              `24=-8p-2q`
           `-12=4p+q `                     (Let this be EQ5.)

Then, add EQ4 and EQ5 to eliminate q.

          `-13=p-q`
(+)     `-12=4p+q`
---------------------------
          `-25=5p`
            ` -5=p`

Plug-in the value of p to EQ4, to get the value of q.

`-13=p-q`
`-13=-5-q`
`-8=-q` 
`8=q`

And, plug-in the values of p and q to EQ2 to get r.

`-1=-5+8+r`
`-1=3+r`
`-4=r`

Hence, `f(x)=x^3-5x^2+8x-4` .

To factor, consider the divisor that has a remainder zero which is  x-1. Since the remainder is zero, it indicates that the polynomial is divisible by x-1.

Dividing x^3-5x^2+8x-4 by x-1 result to:

`(x^3-5x^2+8x-4)/(x-1)=x^2-4x+4`

So,

`f(x)=x^3-5x^2+8x-4=(x-1)(x^2-4x+4)`

Applying the reverse of FOIL, factor x^2-4x+4.

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

Hence, the factor form of the function is `f(x)=(x-1)(x-2)^2` .