How do you find all the zeros of #f(x) = x^2 + 6x + 18#?

1 Answer
Mar 10, 2016

Answer:

Complete the square and use the difference of squares identity to find zeros:

#x = -3+-3i#

Explanation:

Complete the square then use the difference of squares identity:

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

with #a=(x+3)# and #b=3i# as follows:

#f(x) = x^2+6x+18#

#= x^2+6x+9+9#

#= (x+3)^2+3^2#

#=(x+3)^2-(3i)^2#

#= ((x+3)-3i)((x+3)+3i)#

#= (x+3-3i)(x+3+3i)#

Hence #f(x) = 0# when #x=-3+-3i#