Factorize expression $(4a+3)^2-a^4$ The question is to factorize $(4a+3)^2-a^4.$

The answer I ended up with was $(4a+3-a^2)(4a+3-a^2).$ 

I got this by difference of two squares $\cdots.$

I would like to know if this is correct.
You wrote $(4a+3-a^2)(4a+3-a^2)$, it should have been $(4a+3-a^2)(4a+3+a^2)$.

Furthermore, notice that: $(4a+3+a^2)=(a+3)(a+1)$.
SIMPLE

The question was to factorize $(4a+3)^2-a^4.$

IF you use difference of two squares we will get

LET $4a+3$  be $X$ and $a^2$ be $Y$ 

the $X^2-Y^2$

WE GET

$(4a+3-a^2)(4a+3+a^2).$ 

UNDERSTOOD 

Further each of the quadratic equation can be factorised

$(x-1)(x-3)(x-/7+2)(x+/7+2)$

SO THE FINAL AWSER IS HERE