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
1.
4=2²
a²-2²=(a+2)(a-2)
2.
64=8²
n²-8²=(n+8)(n-8)
3.
81=9²
9²-x²=(9+x)(9-x)
4.
100=10²
c²-10²=(c+10)(c-10)
5.
25=-25*-1=-5²*i²=-(5i)² (i=√-1)
k²-(5i)²=(k+5i)(k-5i)
6.
49y²=(7y)²
1²-(7y)²=(1+7y)(1-7y)
7.
9b²=(3b)²
(3b)²-10²=(3b+10)(3b-10)
8.
25x²=(5x)²
(5x)²-7²=(5x+7)(5x-7)
9.
16a²=(4a)²
121=11²
(4a)²-11²=(4a+11)(4a-11)
10.
81y²=(9y)²
x²-(9y)²=(x+9y)(x-9y)
11.
4h²=(2h)²
25g²=(5g)²
(2h)²-(5g)²=(2h+5g)(2h-5g)
12.
64u²=(8u)²
(8u)²-v²=(8u+v)(8u-v)
13.
x²y²=xxyy=(xy)(xy)=(xy)²
(xy)²-1²=(xy+1)(xy-1)
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
