How do you factor completely #2a^2 - 4a - 2#?

1 Answer
Oct 28, 2016

Answer:

#2a^2-4a-2 = 2(a-1-sqrt(3))(a-1+sqrt(3))#

Explanation:

The difference of squares identity can be written:

#A^2-B^2 = (A-B)(A+B)#

Use this with #A=(a-1)# and #B=sqrt(3)# as follows:

#2a^2-4a-2 = 2(a^2-2a-2)#

#color(white)(2a^2-4a-2) = 2(a^2-2a+1-3)#

#color(white)(2a^2-4a-2) = 2((a-1)^2-(sqrt(3))^2)#

#color(white)(2a^2-4a-2) = 2((a-1)-sqrt(3))((a-1)+sqrt(3))#

#color(white)(2a^2-4a-2) = 2(a-1-sqrt(3))(a-1+sqrt(3))#