ac method
for y=ax^2+bx+c
the ac method is normally used for when a≠1
it is used as an aid to help factor
so
0=1x^2-8x+15
multiply a and c (1 times 15=15)
what 2 numbers mulitply to get 15 and add to get -8?
-3 and -5
split the middle up to that
0=1x^2-3x-5x+15
group
0=(x^2-3x)+(-5x+15)
undistribute
0=x(x-3)-5(x-3)
undistribute (x-3)
0=(x-5)(x-3)
set to zero
x-5=0
x=5
x-3=0
x=3
solutions are x= 3 and 5
