I am failing to understand partial fraction decomposition in cases like the following:
Provide the partial fraction decomposition of the following: $$\frac{x+2}{(x-4)^3(x^2 + 4x + 16)}$$
I see this and I think of
$$\frac{A}{x-4} + \frac{Bx+C}{(x-4)^2} + \frac{Dx^2 + Ex + F}{(x-4)^3} + \frac{Gx+H}{x^2 + 4x + 16}$$
But I am told that the correct answer is $$\frac{A}{x-4} + \frac{B}{(x-4)^2} + \frac{C}{(x-4)^3} + \frac{Dx+E}{x^2 + 4x + 16}$$
What exactly is the numerator of each fraction based on?
You may see the redundancy of the $Bx$ term as follows: \begin{align} \frac{Bx + C}{(x - 4)^2} & = \frac{Bx}{(x - 4)^2} + \frac{C}{(x - 4)^2} \\ & = \frac{B(x - 4)}{(x - 4)^2} + \frac{4B}{(x - 4)^2} + \frac{C}{(x - 4)^2}\\ & = \frac{B}{x - 4} + \frac{4B + C}{(x - 4)^2}. \end{align}
You have the correct understanding that the numerator is one degree lower than the denominator. For example: $$\frac{A}{x+1}$$ $$\frac{Ax+B}{x^2+3x+2}$$ $$\frac{Ax^2+Bx+C}{x^3+2x^2+5x+1}$$ However, repeated terms are an exception. The numerator is one degree less than the repeated polynomial in the denominator. You have as many fractions as the power on the repeated term. With each fraction's denominator with a different power such that one fraction's denominator is raised to the first power, and so on raising each denominator to a higher power, until the power on the original denominator is reached. Let's look at an example: $$\frac{x}{(x+1)^2}=\frac{A}{x+1}+\frac{B}{(x+1)^2}$$ $$\frac{x}{(x+1)^3}=\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C}{(x+1)^3}$$ $$\frac{x}{(x^2+x+1)^2}=\frac{Ax+B}{(x^2+x+1)}+\frac{Cx+D}{(x^2+x+1)^2}$$
The partial fraction expansion (over the real numbers) has denominators that are powers of either linear terms (such as $x-4$) or quadratic terms with no real roots (such as $x^2 + 4 x + 16$). In the case of linear terms, the numerators are constants ($A$, $B$, $C$), while in the case of quadratic terms, they are linear terms ($D x + E$).
The numerator is a polynomial with a degree lower than that of the base polynomial (the base of the exponentiation) in the denominator. In your example, the numerators over the powers of $x-4$ must be constant to have degree less than one.
See the last bullet point in the section Procedure here.
Hint. Observe that you have $$ \frac{Bx+C}{(x-4)^2} =\frac{B(x-4)+4B+C}{(x-4)^2}=\frac{B}{(x-4)}+\frac{4B+C}{(x-4)^2} $$ and $$ \begin{align} \frac{Dx^2 + Ex + F}{(x-4)^3} &=\frac{D(x-4)^2+(E+8D)(x-4)+C+4E+32D}{(x-4)^3}\\\\ &=\frac{D}{(x-4)}+\frac{E+8D}{(x-4)^2}+\frac{C+4E+32D}{(x-4)^3}\\\\ \end{align} $$
