How do you factor #7c²-28c+7 #?

1 Answer
Aug 24, 2016

Answer:

#7c^2-28c+7=7(c-2-sqrt(3))(c-2+sqrt(3))#

Explanation:

The difference of squares identity can be written:

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

We use this with #a=(c-2)# and #b=sqrt(3)# as follows:

#7c^2-28c+7#

#=7(c^2-4c+1)#

#=7(c^2-4c+4-3)#

#=7((c-2)^2-(sqrt(3))^2)#

#=7((c-2)-sqrt(3))((c-2)+sqrt(3))#

#=7(c-2-sqrt(3))(c-2+sqrt(3))#