2x^2+9x-5
trial and error
(2x+/- )(x+/-)
since 2 and 5 are prime
experiment
(2x+5)(x-1)=2x^2+5x-2x-5=2x^2+3x-5 nope
(2x-1)(x+5)=2x^2-x+10x-5=2x^2+9x-5, yup
complete the square
first group x terms
(x^2+8x)-5
take 1/2 of 8 and squaer it
8/4=4, 4^2=16
add negative and positive inside
(x^2+8x+16-16)-5
factor perfect square
((x+4)^2-16)-5
distribute
(x+4)^2-16-5
(x+4)^2-21