Harmonic Roots

If $f(x) = 0$ has roots that form a Harmonic Progression, then $f \left ( \frac {1}{x} \right ) = 0$ has roots that form an Arithmetic Progression.

So $4x^3 + 12x^2 + 11x + 3 = 0$ has roots that form Arithmetic Progression.

Or $x^3 + 3x^2 + \frac {11}{4} x + \frac {3}{4} = 0$

Let the roots of the above equation be $a-d,a,a+d$ for $d>0$ , by Vieta's formula

$3a = -3 \Rightarrow a = -1$ and $a(a-d)(a+d) = -\frac {3}{4} \Rightarrow d = \frac {1}{2}$

Which means the roots of the given equation is $\frac {1}{-1}, \frac {1}{-1 + \frac{1}{2}}, \frac {1}{-1 - \frac{1}{2}}$ or $-2,-1, \frac {-2}{3}$

Thus $A = -2, B = -1, C = -2, D = 3 \Rightarrow A+B+C-D = \boxed{-8}$

First, note that one solution is x = -1 [A,B, are integers. So, there is atleast one integer solution. You can hunt for this integer by plugging in factors of the product of the constant term and the coefficient of x^3]

This leads to the idea that $(x+1)$ is a factor of the expression.

Divide the expression with $(x+1)$ to get $3 x^2+8 x+4$

$3 x^3 + 11 x^2 + 12 x + 4 = 0$

$\implies (x+1)(3x^2+8x +4) = 0$

$\implies (x+1)(3x^2+6x+2x+4)=0$

$\implies (x+1)(3x(x+2)+2(x+2))=0$

$\implies (x+1)(3x+2)(x+2)=0$

Any of the three factors must equal zero:

So, the three roots are $-1,-2,\frac{-2}{3}$

Its simple if you find the roots. -1,-2,-2/3