Expand: (2n + 5k)2 polynomial identeties

(2n + 5k)^2 = (2n + 5k)(2n + 5k)
Here we need to use FOIL
First - 2n * 2n = 4n^2
Outside - 2n * 5k = 10nk
Inside - 5k * 2n = 10nk
Last - 5k * 5k = 25k^2

That gives us 4n^2 + 10nk + 10nk + 25k^2, we can combine 10nk + 10nk to give 20nk, and a final equaion of 4n^2 + 20nk + 25k^2

(x2 + y2)2 = (x2 – y2)2 + (2xy)2 is true for any x and y. That is an example so in your case i think it would be 

hello,

Step-by-step explanation:

a=12*x³yz⁵=4*3*x³yz⁵

b=16*x⁴y³z²=4*4*x⁴*y³*z²

hcf(a,b)=4*x³*y*z²=4x³yz²