How do you factor #c^2 + 6c + 9#?

Because the #x^2# coefficient is #1# we know the coefficient for the #c# terms in the factor will also be #1#:

#(c )(c )#

Because the constant is a positive and the coefficient for the #x# term is a positive we know the sign for the constants in the factors will both be positive because a positive plus a positive is a negative and a positive times a positive is a positive:

#(c + )(c + )#

Now we need to determine the factors which multiply to 9 and also add to 6:

#1 xx 9 = 9#; #1 + 9 = 10 # <- this is not the factor

#3 xx 3 = 9#; #3 + 3 = 6 # <- this IS the factor

#(c + 3)(c + 3)#

Or

#(c + 3)^2#

First.... look at the first term #c^2#
Now.... #c^2# can either be

• #(cxx x)(c xx y)#
• #(c^2 xx x)(1 xx y)# Now... notice that if it was the second possibility... there would be another #c^2# as: #(c^2 xx x)(1 xx y)=c^2+c^2y+x+xy# So.. possibility eliminated Which leaves us with only one possibility: #(c xx x)(c xx y)# Now.... what is #x# and #y#? Notice that both #6c# and #9# are multiples of #3# and also notice that #3+3=6# and #3xx3=9# Problem solved!! #(c+3)(c+3)# #(c+3)^2#