Just need to do FOIL which is first outter inner last first one (x+4)(x-4) first: x*x=x² inner: 4*x=4x outer:x*-4=-4x last:4*-4=-16 add x²+4x-4x-16=x²-16=x²-4²
basically a²-b²=(a+b)(a-b) so
(5x+1)(5x-1)=(5x)²-1²=25x²-1
(2a+3b)(2a-3b)=(2a)²-(3b)²=4a²-9b²
you can check the foils with very little brainwork basically notice that (a+b)(a-b) ends up with a²+ab-ab+b², the ab's in middle cancle out all the time
14. 81n⁴=(9n²)² (9n²)²-5²=(9n²+5)(9n²-5) notice we can further factor the last factor (9n²-5) with (3n)²-(√5)² to obtain (3n+√5)(3n-√5) in all, the factored form complete is (9n²+5)(3n+√5)(3n-√5) not including imaginary numbers