Answer
The statement is False.
$$
\frac{x^{2}+4}{x^{2}\left(x-4\right) }
$$
It can be put in the form:
$$
\frac{A}{x^{2}}+\frac{B}{x}+\frac{C}{x-4}
$$
In fact, $A = −1, B = \frac{-1}{4} $ and $C =\frac{5}{4} $.
Work Step by Step
Resolve into partial fractions
$$
\frac{x^{2}+4}{x^{2}\left(x-4\right) }
$$
We’ll first factor the denominator and then get the form of the partial fraction decomposition as follows:
$$
\begin{aligned}
\frac{x^{2}+4}{x^{2}\left(x-4\right) } & =\frac{A}{x^{2}}+\frac{B}{x}+\frac{C}{x-4}\\
&=\frac{A\left(x-4\right)+Bx\left(x-4\right)+Cx^{2} } {x^{2 }\left(x-4\right) }
\end{aligned}
$$
multiplying both sides by the denominator, we get:
$$
x^2+4=A\left(x-4\right)+Bx\left(x-4\right)+Cx^{2}
$$
Substituting $x = 0 $ into this equation gives $A = −1$, substituting $x = 4$ gives $ C= \frac{5}{4}$, and substituting $x =1 $ gives $ B = \frac{-1}{4}$ so that
$$
\frac{x^{2}+4}{x^{2}\left(x-4\right) }=\frac{-1}{x^{2}}+\frac{\frac{5}{4}}{x}-\frac{\frac{1}{4}}{x-4}
$$
Thus, the statement is False. It can be put in the form:
$$
\frac{A}{x^{2}}+\frac{B}{x}+\frac{C}{x-4}
$$
In fact, $A = −1, B = \frac{-1}{4}, C = \frac{5}{4}$
