How do you find all the zeros of #x^2 - 8x - 48 = 0#?

1 Answer
Apr 29, 2016

Answer:

The important thing is to be able to identify factors of #48# that add / subtract to give #-8#.

Explanation:

The important thing is to be able to identify factors of #48# that add / subtract to give #-8#.

#48# could be #6*8# which doesn't work, or it could be #4*12# which does work - #4-12 = -8#

Then #x^2 - 8x - 48 = (x-12)(x+4)#