How do you factor completely #a^2 - 1 - 4b^2 - 4b#?

1 Answer
George C.
Dec 18, 2016

#a^2-1-4b^2-4b = (a-2b-1)(a+2b+1)#

Explanation:

The difference of squares identity can be written:

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

Use this with #A=a# and #B=2b+1# as follows:

#a^2-1-4b^2-4b = a^2-(4b^2+4b+1)#

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

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

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

Related questions
See all questions in Factoring Completely
Impact of this question
1853 views around the world
You can reuse this answer
Creative Commons License