How do you factor completely #x^2 + 16#?

1 Answer
Jan 15, 2017

Answer:

No factor in #RR#.
#x = 4i# multiplicity 2 or #x = -4i# multiplicity 2 in #CC#

Explanation:

In #RR#,

#x^2 + 16 = 0#

#=> x^2 = -16#

No solution in #RR# since #x^2 >= 0# in #RR#.


In #CC#,

#x^2 = -16#

#=> x = sqrt(-16)#

#=> x = sqrt16*sqrt(-1)#

#=> x = +-4i#

x = 4i multiplicity 2 or x = -4i multiplicity 2