Question

How to factor 4n^2+4n-8

Answer 1

$4n^2+4n-8=4\cdot n^2+4\cdot n-4\cdot2=4(n^2+n-2)\\\\=4(n^2+2n-n-2)=4(n\cdot n+2\cdot n-1\cdot n-1\cdot 2)\\\\=4[n(n+2)-1(n+2)]=\huge\boxed{4(n+2)(n-1)}$