Question 106574
Fully factor:
{{{4a^2-(3a+1)^2}}} Do you recognise this as the difference of two squares?
The difference of two squares is factorable as follows:
{{{A^2-B^2 = (A-B)(A+B)}}} In your problem, A = 2a while B = (3a+1), so...
{{{4a^2-(3a+1)^2 = (2a-(3a+1))(2a+(3a+1))}}} Now simplfy this to get:
{{{(-a-1)(5a+1)}}} 
You can check this solution by multiplying these two factors to see if you get back the original expression. Let's try it using FOIL.
{{{(-a-1)(5a+1) = -5a^2+6a+1}}} = {{{4a^2-(9a^2+6a+1) = 4a^2-(3a+1)^2}}}