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

1 Answer
Apr 16, 2016

Answer:

#x=12# or #x=-4#, each with multiplicity #1#

Explanation:

Complete the square then use the difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

with #a=(x-4)# and #b=8# as follows:

#0 = x^2-8x-48#

#=(x-4)^2-16-48#

#=(x-4)^2-8^2#

#=((x-4)-8)((x-4)+8)#

#=(x-12)(x+4)#

So #x=12# or #x=-4#, each with multiplicity #1#