How to simplify this fraction $\frac{x^2+7x+1}{x^2-4}$? I'm familiar with polynomial long division but I keep getting stuck when I try to answer this question. Online math calculations also say there's no way to simplify this (but it's a homework question so there must be?) If anyone can understand this, please explain how they reach their answer!
$$\frac{x^2+7x+1}{x^2-4}$$
 A: As mentioned in the comment, the polynomial $P(x) \in \mathbb{Z}[x]$ is irreducible, where $$ P(x) = {x^2 + 7x + 1} $$
We need to show this at first.
$\bullet~\textbf{Proof:}~$ On the contrary, let's assume $P(x)$ is reducible over $\mathbb{Z}$. Then $\exists~a_{1}, a_{2} \in \mathbb{Z}$ such that
$$ x^2 + 7x + 1= (x - a_1)(x- a_2) = x^2 -(a_1 + a_2)x + a_1 a_2 $$
Hence, on the coefficient comparison, we have that $$a_{1}a_2 = 1 \quad \text{and} \quad a_{1} + a_{2} = -7$$
As $a_{1}, a_2 \in \mathbb{Z}, $ then $a_{1} = a_2 = 1$ or $a_{1} = a_2 = - 1$. But as $a_{1} + a_{2} = -7$, it's not possible!
Hence, we arrive at a contradiction!
Therefore, $P(x)$ is irreducible over $\mathbb{Z}.$
As $P(x) \not\equiv 0 \pmod{Q(x)}$ where $Q(x) = (x - 2)(x+ 2) \in \mathbb{Z}[x]$. Then the fraction $$ \frac{x^2 + 7x + 1}{x^2 - 4}  $$
is not simplifiable.

$\blacksquare~$Edit: I just saw in the comment section, that the polynomial fraction is
$$ \frac{x^2 + 7x + 10}{x^2 - 4} = \frac{(x + 2)(x + 5)}{(x + 2)(x - 2)} = \frac{(x-5)}{(x-2)} $$
