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` .