Assuming -b^3 should be -bx^2:
Use synthetic division for dividing x+1:
-1 | 4 -b 1 -c
4 -4 4+b -(5+b)
4 -(4+b) 5+b | 0
So -c-5-b=0 and c=-5-b.
Suppose f(x) is a function of x (polynomial with degree 2) such that:
Let f(x)=px^2+qx+r then
(2x-3)(px^2+qx+r)+30=4x^3-bx^2+x-c, because of the remainder 30.
We can equate coefficients of like terms:
x^3: 2p=4 so p=2
x^2: -3p+2q=-b, but p=2 so q=(6-b)/2
x: -3q+2r=1, but q=(6-b)/2 so 2r=1+3(6-b)/2=(2+18-3b)/2=(20-3b)/2 and r=(20-3b)/4.
Constant: -3r=-c-30, -3(20-3b)/4=-c-30, -60+9b=-4c-120, 9b+4c=-60.
But c=-5-b, so 9b-20-4b=-60, 5b=-40, b=-8 and c=-5+8=3.
The expression is 4x^3+8x^2+x-3.
The two numbers are therefore -8 and 3.
The result of dividing by x+1 is therefore 4x^2+4x-3=(2x-1)(2x+3).
The complete factorisation is (x+1)(2x-1)(2x+3).
Let's see if this is correct. 4x^3+4x^2-3x+4x^2+4x-3=4x^3+8x^2+x-3. This appears to be correct for x+1.
Now 2x-3: if we subtract 30 from the expression, 2x-3 should exactly divide into it:
Substituting in f(x) we have f(x)=2x^2+7x+11 (putting in values for p, q and r).