Question

4n^2-16 factor the quadratic

Answer 1

$4n^2-16=4(n^2-4)=4(n^2-2^2)=(*)\\\\use:a^2-b^2=(a-b)(a+b)\\\\(*)=4(n-2)(n+2)$

Answer 2

(2n+4)(2n-4)
This way when you distribute it out, you get 4n^2-8n+8n-16.
Then the -8n and the 8n cancel out and you are left with 4n^2-16.
Another way of factoring it is to start out by taking out a common factor which is 4.
This then becomes 4(n^2-4) You can then factor that by using the method above into 4(n+2)(n-2).
The second way is just a bit more factored but they're both an answer.