Question

Can you help me with this fundamental math problem? (a^2 - 3a -9)/(a-3) How many "a" values (in natural numbers) are there that make this equation's result a positive integer number?

Answer 1

@HolsterEmission

Answer 2

You can eliminate \(a=3\), since that makes the quotient undefined.

Some polynomial division: for \(a\neq3\),
\[\frac{a^2-3a-9}{a-3}=\frac{a(a-3)-9}{a-3}=a-\frac{9}{a-3}\]This number will be positive whenever \(a>\dfrac9{a-3}\) and an integer whenever \(a-3\) divides \(9\) exactly. This only happens for three instances of \(a\in\mathbb N\), namely \(a\in\{2,6,12\}\).