It's Easier Than It Looks

The given polynomial is a reciprocal polynomial, i.e. it is of the form

$Ax^4 + Bx^3 + Cx^2 + Bx + A$

Hence to proceed we take $x^2$ common to get

$(x^2)(Ax^2 + Bx + C + \frac{B}{x} + \frac{A}{x^2}) = 0 . . . . . . (1)$

Now we make a substitution,

$t = x + \frac{1}{x}$

and hence $(1)$ is modified as:

$(x^2)(A(t^2 - 2) + B(t) + C) = 0$

This also guarantees that $x \neq 0$ , so

$A(t^2 - 2) + B(t) + C = 0$

Since $t \geq 2 \text{ or } \leq -2$ , we have to consider the solutions of $t$ fitting the conditions mentioned.

$t = \dfrac{-B \pm \sqrt{B^2 - 4AC}}{2A}$

We will proceed by substituting the values of $A,B,C$ from the question.

$t = \dfrac{3 \pm \sqrt{9 + 24}}{2}$

The solution $t = \dfrac{3 - \sqrt{33}}{2}$ is not valid as it is greater than $-2$ and less than $2$

Hence

$t = \dfrac{3 + \sqrt{33}}{2}$

$\Rightarrow x + \frac{1}{x} = \dfrac{3 + \sqrt{33}}{2}$

$\Rightarrow 2x^2 + 2 - (3 + \sqrt{33})x = 0$

Sum of the roots of the equation is $\frac{3 + \sqrt{33}}{2}$

$a = 3 , b = 33 , c =2 \Rightarrow a^{-1} + b^{-1} + c^{-1} = \frac{19}{22}$

$p = 19 , q = 22 \Rightarrow p + q = 41$

Sum of digits of $p + q = \large\color{#D61F06}{\boxed{5}}$

Since the given polynomial is a reciprocal polynomial, we can rewrite it to a square of a trinomial plus an offset

$\,$ ${ x }^{ 4 }\, -\, { 3x }^{ 3 }\, -\, { 4 }x^{ 2 }\, -\, 3x\, +\, 1\, =\, { \left( { x }^{ 2 }\, -\, \frac { 3 }{ 2 } x\, +\, 1 \right) }^{ 2 }\, -\, \frac { 33 }{ 4 } { x }^{ 2 }$

This new form can be factored. Let:

• $A\, =\, \frac { -3\, +\, \sqrt { 33 } }{ 2 }$
• $B\, =\, \frac { -3\, -\, \sqrt { 33 } }{ 2 }$

${ \left( { x }^{ 2 }\, -\, \frac { 3 }{ 2 } x\, +\, 1 \right) }^{ 2 }\, -\, \frac { 33 }{ 4 } { x }^{ 2 }\, =\, \left( { x }^{ 2 }\, +\, Ax\, +\, 1 \right) \left( { x }^{ 2 }\, +\, Bx\, +\, 1 \right)$

Now, we must check whether the two factors have real roots by finding their discriminant. We find that ${ A }^{ 2 }\, -\, 4$ is < 0, so $\left( { x }^{ 2 }\, +\, Bx\, +\, 1 \right)$ must have the real roots. The sum of the real roots must then be:

$-B\, =\, \frac { 3\, +\, \sqrt { 33 } }{ 2 } \\ \Rightarrow \, a\, =\, 3,\, b\, =\, 33,\, c\, =\, 2\\ \Rightarrow \, \frac { 1 }{ 3 } \, +\, \frac { 1 }{ 33 } \, +\, \frac { 1 }{ 2 } \, =\, \frac { 19 }{ 22 } \\ \Rightarrow \, p\, +\, q\, =\, 41,\, 4\, +\, 1\, =\, \boxed { 5 }$

As there are no evident roots or factorizations, let's assume that $f(x)$ can be expressed as a product of two quadratic polynomials, that is: $f(x) = \left( {x^2 + Ax + C} \right)\left( {x^2 + Bx + D} \right)$ We may further assume that $C = D = 1$ , as the independent term of $f(x)$ is $1$ and all the other coefficients are integral. Therefore: $f(x) = \left( {x^2 + Ax + 1} \right)\left( {x^2 + Bx + 1} \right) = x^4 + \left( {A + B} \right)x^3 + \left( {AB + 2} \right)x^2 + \left( {A + B} \right)x + 1$

The coeffcients of $f(x)$ must remain equal, so a system of equations arises:

\left\{ {\begin{array}{*{20}c} {A + B = - 3} \\ {AB = - 6} \\ \end{array}} \right.

Solving the system yields the following solution:

$A = \frac{{ - 3 + \sqrt {33} }}{2}, \qquad B = \frac{{ - 3 - \sqrt {33} }}{2}$

Notice that the discriminant of $\left( {x^2 + Ax + 1} \right)$ is negative, and thus $f(x)$ has two complex conjugate roots. On the other hand, the discriminant of $\left( {x^2 + Bx + 1} \right)$ is positive, and the sum of its roots equals $-B$ . In conclusion, the sum of the real roots of $f(x)$ is: $- B = \frac{{3 + \sqrt {33} }}{2} = \frac{{a + \sqrt b }}{c}$

Therefore:

$a^{ - 1} + b^{ - 1} + c^{ - 1} = \frac{1}{3} + \frac{1}{{33}} + \frac{1}{2} = \frac{{19}}{{22}} = \frac{p}{q}$ $p + q = 41 \Rightarrow 4 + 1 = \boxed{5}$