Question 442441
Recognize in your problem 81s^2-144s+64 that 81 and 64 are perfect squares.
{{{(9s)^2 = 81s^2}}} and {{{(8)^2 = 64}}}

Also recognize that the middle term is negative, while the last term is positive. In order to achieve a negative sum and positive product, both constants have to be negative... for instance -3 * -3 = 9, and -3 + -3 = -6.

Given this information, we have all we need to solve this.

Remember 81s^2 = (9s)^2 , so our first term in our factor will be 9s.
Similarly with 64 = 8^2.

So our factorization will be {{{(9s-8)^2}}}

Let's FOIL to make sure.

{{{(9s-8)^2 = (9s-8)(9s-8) = 81s^2 -72s -72s + 64 = 81s^2 -144s +164}}}
Check.

Hope this helped!