ok so the problem is this :
Let f(x) = (2x+4)^4
Simplify ->
f(x)-f(a)
(x-a)
Stated in words: f of x minus f of a, all divided by x minus a.
so can anyone help me with this ASAP.
$\displaystyle f(x) = (2x+4)^4$
$\displaystyle f(a) = (2a+4)^4$
$\displaystyle f(x) - f(a) =$
$\displaystyle (2x+4)^4 - (2a+4)^4 = $
$\displaystyle [(2x+4)^2 + (2a+4)^2][(2x+4)^2 - (2a+4)^2] = $
$\displaystyle [(2x+4)^2 + (2a+4)^2][(2x+4) + (2a+4)][(2x+4)-(2a+4)] =
$
$\displaystyle [(2x+4)^2 + (2a+4)^2][2(x+a+4)][2(x-a)]
$
divide the last expression by $\displaystyle x-a$
$\displaystyle (2x+4)^4-(2a+4)^4=((2x+4)^2-(2a+4)^2)((2x+4)^2+(2a+4)^2)$ by the difference of squares identity.
$\displaystyle (2x+4)^2-(2a+4)^2=((2x+4-(2a+4))(2x+4+2a+4)=4(x-a)(x+a+4)$ by difference of squares again.
So the whole thing is $\displaystyle \frac{4(x-a)(x+a+4)((2x+4)^2+(2a+4)^2)}{x-a}$
Cancel the $\displaystyle x-a$ and then simplify.
EDIT: Beat to it!