Easy to think, Harder to solve

To specify a monic quartic polynomial, there are four numbers to determine: the coefficients of $1$ , $X$ , $X^2$ and $X^3$ . Since the question only gives three conditions, we have a problem. All of the polynomials $\begin{array}{rcl} f_a(X) & = & (X-1)(X-2)(X-3)(X+a) + 24(X-2)(3X-1) \\ & = & (X-2)\big[X^3 + (a-4)X^2 - (4a - 75)X + 3(a-8)\big] \\ & = & X^4 + (a-6)X^3 - (6a - 83)X^2 + (11a - 174)X - 6(a-8) \end{array}$ satisfy the conditions $f_a(1)=-48$ , $f_a(2) = 0$ and $f_a(3) = 192$ for any real $a$ . Many of them have four real roots as well. The cubic discriminant of $X^3 + (a-4)X^2 - (4a - 75)X + 3(a-8)$ is $4 (-372399 + 42288 a + 238 a^2 - 112 a^3 + a^4)$ and hence this discriminant is positive, and hence $f_a(X)$ has four real roots, provided that one of the inequalities $a < -20.4845 \qquad \qquad 12.9143 < a < 13.2389 \qquad \qquad a > 106.331$ holds.

For any value of $a$ satisfying one of the above inequalities, the four roots $\alpha,\beta,\gamma,\delta$ of $f_a(X) = 0$ are real. Using Newton's Formulae, we can show that $\alpha^3 + \beta^3 + \gamma^3 + \delta^3 \; =\; -756 + 216 a - a^3 \qquad \qquad \alpha^4 + \beta^4 + \gamma^4 + \delta^4 \; = \; 7106 + 1056 a - 288 a^2 + a^4$ and the sum of these two is $P(a) \; = \; 6350 + 1272 a - 288 a^2 - a^3 + a^4$

Since the polynomial has integer coefficients, we deduce that $a$ has to be an integer. We now need to use the fact that the four roots of the polynomial are also integers.

If $a < 8$ then $f_a(0) = 6(8-a) > 0 > -48 = f_a(1)$ , and hence one of the roots of $f_a$ lies between $0$ and $1$ , and so is not an integer. Thus there is no polynomial $f_a(X)$ with integer roots for $a < -20.4845$ . Since $f_a(1.6) = -35.9424 + 0.336 a$ , we can show that $f_a(1.6) > 0 > f_a(1)$ for all $a \ge 107$ , and so $f_a(X)$ has a root between $1$ and $1.6$ for all $a \ge 107$ . Thus there is no polynomial $f_a(X)$ with integer roots for $a \ge 107$ . Thus we deduce that the only possible polynomial $f_a(X)$ with integer coefficients and four integer roots is $f_{13}(X)$ .

The roots $\alpha,\beta,\gamma,\delta$ of $f_{13}(X)$ are equal to $-5,-3,-1,2$ , and $P(13) = 578$ , giving the answer $\boxed{6}$ .

Let $P(x) -be -the -polynomial = (x-2)(x^{3} + ax^{2} +bx+c$ So that

$a+b+ c= 47 and 3a-c=12$

Also let fourth root be 2 .

Therefore by Viete's Theoram

$\alpha + \beta+ \gamma = -a$

$\alpha . \beta+ \gamma. \alpha + \beta. \gamma= b$

$\alpha . \beta .\gamma= -c$

Adding these three equations abd one on both sides

$(1-\alpha).(1-\beta).(1- \gamma)=48$

Now its pretty easy to factorise 48 and guess the value of (1-root)

We get $\alpha = -1, \beta=-3 , \gamma=-5$

Thus answer follows.