what times \(n\) is \(4n^2\)?
and what times \(5\) gives \(-60\)?
ok put them together are you are done
at the back of my textbook the answer is 4(n-3)
not really factored completely that would be \[4(n-3)(n+5)\]
yeah because the terms of \(4n-12\) have a common factor of 4
so you divide (4n-12) by 4 to get 4(n-3) ?
that is not "division" that is factoring, aka the distributive property \[4\times x-4\times 3=4(x-3)\]