f is divisible by g . g = ( x - 1 )^3 f = x^4 + ( m - 1 )*x^3 + ( n + 2 )*x^2 + x + p - 3

f(x)
=x^4+(m-1)x^3+(n+2)x^2+x+p^3.

 g(x) =
(x-1)^3.

f(x) is is divisible by
(x-1)^3.

Solution :

Since
f(x)  the 4th degree is divisible by  (x-1)^3, w

 we can
write, f(x) as product of (x-1)^3 and ax+b

f(x)  = g(x) {
ax+b) Or

 x^4 + ( m - 1 )*x^3 + ( n + 2 )*x^2 + x + p - 3 =
(x-1)^3 (ax+b).

This is an  identtity,  so we can equate
coefficients of  equal powers on both sides.

x^4 = ax^4. Or
a = 1. 

 constant term p-3 = -b, or -b
=3-p

Therefore ,

 x^4 + ( m -
1 )*x^3 + ( n + 2 )*x^2 + x + p - 3 = (x-1)^3*(x+3-p)

Now
diffrentiate both sides:

4x^3+3(m-1)x^3 +2(n+2)x+1 = 
3(x-1)^2 (x+3-p) +(x-1)^3 ..(1)

If x = 0, 1 =  3
(-1)^2(3-p)+(-1)^3 = 3(3-p) -3 = 9-3p-3 = 6-p.

1=
6-p.

p = 6-1 = 5.

We rewrite
(1) with p = 5.

 4x^3+3(m-1)x^2+2(n+2)x+1 =
3(x-1)^2(x-2)+(x-1)^3...........(2).

Put x = 1, then
4+3(m-1)+2(n+2)+1 = 0

4+3m-3 +2n+4+1 =
0

3m+2n =
-6.................(3)

Differentiate  both sides of
(2):

 {4x^3+3(m-1)x^2+2(n+2)x+1}' =
{3(x-1)^2(x-2)+(x-1)^3}'

12x^2 + 6(m-1)x +2(n+2) = 
{(x-1)^2 (x-1+3x-6)}' = {(x-1)^2*(4x-7)}' = 2(x-1)(4x-7)
+(x-1)^2*(4)

12x^2+6(m-1)x +2(n+2) = (x-1){ 2(4x-7)+4(x-1)}
= (x-1)(12x-18)

12x^2+6(m-1)x +2(n+2) = 12x^2 -30x
+18

Therefore 6(m-1) = -30 and  2(n+2) =
18

m-1 = -30/6 = -5

m= -5+1 =
-4.

2(n+2) = 18

n+2= 18/2 =
9

n = 9-2 = 7.