Question 7201
Factor:

{{{16k^2 - 80k + 100}}}  Factor a 4 first.

{{{4(4k^2 - 20k + 25)}}} Now try:

{{{4(2k - 5)(2k - 5))}}}

The reason to try a -5 constant term in the factors is because you want to get a +25 as the constant in the second equation The only way is to have (-5)(-5) in the factors. Well, (-1)(-25) also = +25 but you can quickly see that this would not work.

So the complete factorisation is: 4(2k-5)(2k-5) 

Check the solution by using FOIL:

{{{4(2k-5)(2k-5) = 4(4k^2 - 10k - 10k + 25)}}} = {{{16k^2 - 80k + 100}}}