Setting up the particular solution to $y''+4y'=e^{-4x}+3$ $y''+4y'=e^{-4x}+3$
The characteristic polynomial is $r^2+4r=0 \to r = 0,-4$
the complementary solution is: $y_c = C_1+C_2e^{-4x}$
The part that is throwing me off is the addition of 3. I can't even set that up as a polynomial?

In this particular case, and similar cases, there is a second derivative with constant coefficients equaling a constant. This suggests a linear polynomial of the order of the highest derivative. This leads to the form (the exponential term being equal to a known solution so other rules apply):
$$f(x) = a_{0} + a_{1} \, x + a_{2} \, x^2 + b \, x \, e^{-4 x}.$$ 
This solution is to satisfy
$$f'' + 4 \, f' = e^{-4 x} + 3.$$
Now:
\begin{align}
3 + e^{-4 x} &= \frac{d}{dx}(a_{1} + 2 a_{2} x + b (-4 x + 1) \, e^{-4 x}) +  4 \, (a_{1} + 2 a_{2} x + b (-4 x + 1) \, e^{-4 x}) \\
&= 2 a_{2} + b (16 x - 8) \, e^{-4 x} + 4 a_{1} + 8 a_{2} x + b (-16 x + 4) \, e^{-4 x} \\
&= 4 a_{1} + 2 a_{2} + 8 a_{2} \, x - 4 b e^{-4 x}
\end{align}
leading to $4 a_{1} = 3$, $a_{2} = 0$, $-4 b = 1$, and 
$$f(x) = a_{0} + \frac{3 \, x}{4} - \frac{x}{4} \, e^{-4 x}.$$
Since the general solution also contains a constant the $a_{0}$ constant in the particular solution can be removed to present the total solution of
$$y(x) = c_{0} + \frac{3 \, x}{4} + \left( c_{1} - \frac{x}{4} \right) \, e^{-4 x}.$$
A second method
By integration one can show that:
$$y' + 4 y = - \frac{1}{4} \, e^{-4 x} + 3 \, x + b_{1}.$$
Now this has equation has the form
\begin{align}
e^{- 4 x} \, \frac{d}{dx} (e^{4 x} \, y) &= - \frac{1}{4} \, e^{-4 x} + 3 \, x + b_{1} \\
\frac{d}{dx} (e^{4 x} \, y) &= - \frac{1}{4} + 3 \, x \, e^{4 x} + b_{1} \, e^{4 x} \\
e^{4 x} \, y &= - \frac{x}{4} + 3 \, \left( \frac{x}{4} - \frac{1}{16} \right) \, e^{4 x} + \frac{b_{1}}{4} \, e^{4 x} + b_{0} \\
y(x) &= \frac{4 b_{1} - 3}{16} + \frac{3 x}{4} + \left(b_{0} - \frac{x}{4}\right) \, e^{- 4 x}.
\end{align} 
This solution is in agreement with the first solution.
