One To Eight

$|LHS|$ is the product of the distances from points $-1,-2,-3,-4$ , and likewise for $|RHS|$ . By symmetry, $LHS=RHS$ for $x=\boxed{-4.5}$ . For $x>-4.5$ we have $|LHS|<|RHS|$ and vice versa for $x<-4.5$ .

$\begin{aligned}\implies(x+1)(x+2)(x+3)(x+4)+1=&(x^2+5x+4)(x^2+5x+4+2)+1\\=&(x^2+5x+4)^2+2(x^2+5x+4)+1\\=&(x^2+5x+5)^2\\\implies(x+5)(x+6)(x+7)(x+8)+1=&(x^2+13x+40)(x^2+13x+40+2)+1\\=&(x^2+13x+40)^2+2(x^2+13x+40)+1\\=&(x^2+13x+41)^2\end{aligned}$

From Equation

$\begin{aligned}(x+1)(x+2)(x+3)(x+4)=&(x+5)(x+6)(x+7)(x+8)\\(x^2+5x+5)^2-1=&(x^2+13x+41)^2-1\\(x^2+5x+5)^2-(x^2+13x+41)^2=&0\\(x^2+5x+5-x^2-13x-41)(x^2+5x+5+x^2+13x+41)=&0\\(-8x-36)(2x^2+18x+46)=&0\\(2x+9)(x^2+9x+23)=&0\end{aligned}$

Now,

$\begin{aligned}&\implies2x+9=0\implies x=-\dfrac92\\&\implies x^2+9x+23=0\implies\Delta=9^2-4(1)(23)=-11<0\quad\text{[Complex Solution]}\end{aligned}$

Thus, the sum of all real solutions is $-\dfrac92=\boxed{-4.5}$

$\begin{aligned} (x+1)(x+2)(x+3)(x+4) & = (x+5)(x+6)(x+7)(x+8) \quad \quad \small \color{#3D99F6}{\text{Let }y=x+\frac{9}{2}} \\ \left(y-\frac 72\right) \left(y-\frac 52\right) \left(y-\frac 32\right) \left(y-\frac 12\right) & = \left(y+\frac 12\right) \left(y+\frac 32\right) \left(y+\frac 52\right) \left(y+\frac 72\right) \\ y^4 - 8 y^3 + \frac{43}{2}y^2 - 22y + \frac{105}{16} & = y^4 + 8 y^3 + \frac{43}{2}y^2 + 22y + \frac{105}{16} \\ \implies 16y^3 + 44y & = 0 \\ 4y^3 + 11y & = 0 \\ y(4y^2+11) & = 0 \\ \implies y & = 0 \quad \quad \small \color{#3D99F6}{\text{The other root is not real.}} \\ \implies x + \frac 92 & = 0 \\ x & = - \frac 92 \end{aligned}$

The sum of all real roots is therefore $\boxed{-4.5}$ .

My solution was similar to that of Chew-Seong Cheong, but I cut down on some calculations.

Define $y = x - 4\tfrac12$ ; that symmetrizes the equation to $(y-\tfrac7 2)(y-\tfrac 52)(y-\tfrac 32)(y-\tfrac12) = (y+\tfrac12)(y+\tfrac32)(y+\tfrac52)(y+\tfrac72).$ Due to the symmetry, we know that the terms with even exponents on each side will be the same. Therefore the equation reduces to $-ay^3 - by = ay^3 + by\ \ \ \therefore\ \ \ ay^2 + b = 0\ \vee\ y = 0.$ Now $a$ and $b$ are both positive, so that $ay^2 + b = 0$ has no real solutions. That leaves $y = 0$ and thus $x = \boxed{4\tfrac12}$ .

From $16x^3+216x^2+1016x+1656 = 0$
We factor the left side of the equation and set the factors equal to zero. The factorazation: 8(2x+9) $(x^2+9x+23)$ . And 8 = 0, 2x+9 = 0 and x^2+9x+23=0. 8=0 is not true, and the answer of $x^2+9x+23=0$ is a complex solution.
Where the only real answer is from $2x+9=0$ , the answer is x = -(9÷2). => (-4.5).
if I was wrong in something, please tell me.
Glad I can help.

$(X+1)(X+2)(X+3)(X+4)=(X+5)(X+6)(X+7)(X+8) \\ \therefore\ (X^2+5X+4)(X^2+5X+6)=(X^2+13X+40)(X^2+13X+42)\\ Let M=X^2+9X,\ \ \ \ and\ \ \ \ N=M+23.........Note:-\ 9=\frac 1 2 *(5+13),\ \ \ 23=\frac 1 4 (4+6+40+42)\\ \therefore\ (M-4)(M-6)=(M+4)(M+6) \ \ \ \therefore\ -24M=24M,\ \ \ \therefore\ M=0, \implies \ X=0\ \ Or\ \ X^2+9=0,\\ But\ X\neq\ 0, so\ \color{#D61F06}{X=\ -\ 4.5.}\\ Also\ (N-4X-19) (N-4X-17)= (N+4X+19) (N+4X+17) \ \ \therefore\ -N(8X+34+34X)=N(8X+34+34X)\ \implies\ N=0.\\ That\ is \ X^2\ +9X+23=0.\ \ But\ these \ roots\ are\ not\ real.$