Fourth degree polynomial

Division by $x^2$ yields

$x^2+5x-4+5/x+1/x^2=0$

Rearranging and adding $2-2$ yields

$(x^2+1/x^2+2)-2-4+5(x+1/x)=0$

Now the expression can be written as

$(x+1/x)^2+5(x+1/x)=6$

Then substituting $x+1/x=t$

$t^2+5t-6=0$

This is just a standard quadratic equation with solutions $t=1$ and $t=-6$

This means that $x+1/x=1$ and $x+1/x=-6$

Multiplying by x on both sides yields

$x^2-x+1=0$ and $x^2+6x+1=0$

The first equation has no real solutions since $(-1)^2-4*1*1<0$

The second equation has two real solutions $-3 \pm 2 \sqrt2$

Thus $a+b+c=-3+2+2=1$

$\begin{aligned} x^4 + 5x^3 - 4x^2 + 5x + 1 & = 0 & \small \color{#3D99F6} \text{Divide throughout by }x^2 \\ x^2 + 5x - 4 + \frac 5x + \frac 1{x^2} & = 0 \\ {\color{#3D99F6}\left(x + \frac 1x\right)}^2 + 5{\color{#3D99F6}\left(x + \frac 1x\right)} - 6 & = 0 & \small \color{#3D99F6} \text{A quadratic of }\left(x + \frac 1x\right) \\ \left({\color{#3D99F6} x + \frac 1x} - 1\right)\left({\color{#3D99F6} x + \frac 1x} + 6\right) & = 0 \end{aligned}$

$\implies \begin{cases} x + \dfrac 1x - 1 = 0 & \implies x^2 - x + 1 = 0 & \small \color{#D61F06} \text{Has no real roots.} \\ x + \dfrac 1x + 6 = 0 & \implies x^2 + 6x + 1 = 0 & \implies x = \dfrac {-6 \pm \sqrt{6^2-4}}2 = - 3 \pm 2\sqrt 2 \end{cases}$

Therefore, $a+b+c = -3 + 2+2 = \boxed 1$ .

$(x^2+rx+s)(x^2+tx+u)=x^4+(r+s)x^3+(rt+s+u)x^2+(ru+st)x+su.$

$r+t=5$

$rt+s+u=-4$

$ru+st = 5$

$su=1$

The only integer solutions to the 4th equation are $s=1, u=1$ and $s=-1, u=-1$ .

First, try $s=1, u=1$ . Then $rt=-6$ . Combining that with $r+t=5$ gives $r=6, t=-1$ as a possible solution.

This gives us the factorization.

Set each part to zero:

$x^2+6x+1=0$

$x^2-x+1=0$

The second equation has no real roots. Applying the quadratic formula to the first equation yields $-3\pm 2\sqrt{2}$ for the two roots.

Divide the original equation by x^2, resulting in: x^2 + 5x -4 +5/x + 1/x^2. Let y = x + 1/x, and substituting, we have y^2 + 5y - 6 = 0, or (y + 6)(y - 1) =0. So y = -6 and y = 1. This gives rise to the 2 equations for x: (1) -6 = x + 1/x, or x^2 + 6x +1 =0. By the quadratic formula, x = [-6 +/- sqrt(36 - 4)]/2 or x = -3 +/- 2sqrt(2), which gives a + b + c = -3 +2 +2 =1. The other root for y results in complex roots for x. Ed Gray