Biquadric Equation

Vieta's Formula explains it all...

Considering the coefficients of the equation, if $a = 1$ and $b = -16$ , then $-\dfrac{b}{a} = 16$ . Thus, the sum of all roots is $16$ aka $x_1 + x_2 + x_3 + x_4 = 16$ where $x_1, x_2, x_3, x_4$ are solutions to the quartic equation.

The equation is

$x^4-2x^3-14x^3+28x^2+61x^2-122x-84x+168=0$

$\Rightarrow x^3(x-2)-14x^2(x-2)+61x(x-2)-84(x-2)$

Implies that, $(x-2)(x^3-3x^2-11x^2+33x+28x-84)=0$

So, $(x-2)(x^2(x-3)-11x(x-3)+28(x-3))=0$

$(x-2)(x-3)(x^2-11x+28)=0$

$\Rightarrow (x-2)(x-3)(x-4)(x-7)=0$

So, the zeroes are: $2,3,4,7$ and the sum is $\boxed{16}$