SOLUTION: When the polynomials f(x)=(a-1)x^3+ax^2+bx+c where a,b,and c are constants is divided by (x+2) and (x-1),the remainder are -5 and 4 respectively.if (x+1) is a factor of f(x),find t

If f(x) = (a-1)x³+ax²+bx+c is divided by (x+2), the remainder is -5.

That tells us that f(-2) = -5

f(-2) = (a-1)(-2)³+a(-2)²+b(-2)+c = -5
f(-2) = (a-1)(-8)+a(4)-2b+c = -5
f(-2) = -8(a-1)+4a-2b+c = -5
f(-2) = -8a+8+4a-2b+c = -5
f(-2) = -4a+8-2b+c = -5
or
        -4a-2b+c = -13

f(x)=(a-1)x³+ax²+bx+c where a,b,and c are constants is divided by (x-1), the remainder is 4.

That tells us that f(1) = 4

f(1) = (a-1)(1)³+a(1)²+b(1)+c = 4
f(1) = (a-1)+a+b+c = 4
f(1) = a-1+a+b+c = 4 
f(1) = 2a+b+c-1 = 4
or
       2a+b+c = 4  

(x+1) is a factor of f(x)

That tells us that f(-1) = 0

f(-1) = (a-1)(-1)³+a(-1)²+b(-1)+c = 0
f(-1) = -1(a-1)+a-b+c = 0
f(-1) = -a+1+a-b+c = 0
f(-1) = 1-b+c = 0
or
            -b+c = -1

 So we have the system of equations:

        -4a-2b+c = -13
         2a +b+c =   5 
            -b+c =  -1

Solve that and get a=3,b=0, c=-1

So

f(x) = (a-1)x³+ax²+bx+c becomes:

f(x) = (3-1)x³+(3)x²+(0)x+(-1)
f(x) = 2x³ + 3x² - 1

Since we are told that (x+1) is a factor, -1 is a zero (or root)
and we may divide by synthetic division:

-1|2  3  0 -1
  |  =2 -1  1
   2  1 -1  0

Thus we have factored (factorised, as you say in the UK), f(x) as

f(x) = (x+1)(2x²+x-1)

and the complete factorization (you spell it factorisation) is

f(x) = (x+1)(2x-1)(x+1)

or since (x+1) is a factor twice,

f(x) = (x+1)²(2x-1)

Edwin