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:

(2x-3)f(x)=4x^3-bx^2+x-c.

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.

2px^3-3px^2+2qx^2-3qx+2rx-3r=4x^3-bx^2+x-c-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:

4x^3+8x^2+x-33.

Substituting in f(x) we have f(x)=2x^2+7x+11 (putting in values for p, q and r).

(2x-3)(2x^2+7x+11)=4x^3-6x^2+14x^2-21x+22x-33=4x^3+8x^2+x-33.

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