Answer
$16x^4-1$
Work Step by Step
First we apply the Sum and Difference of Two Terms for the second and the third factor:
$(4x^2+1)[(2x+1)(2x-1)]$
$=(4x^2+1)[(2x)^2-1^2]$
$=(4x^2+1)(4x^2-1).$
Then we apply the Sum and Difference of Two Terms once again:
$(4x^2+1)(4x^2-1)$
$=(4x^2)^2-1^2$
$=16x^4-1.$
