# Help with math problem

• Oct 18th 2009, 08:46 AM
Help with math problem
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.
• Oct 18th 2009, 09:15 AM
skeeter
Quote:

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 $\displaystyle x-a$
• Oct 18th 2009, 09:19 AM
redsoxfan325
Quote:

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.

$\displaystyle (2x+4)^4-(2a+4)^4=((2x+4)^2-(2a+4)^2)((2x+4)^2+(2a+4)^2)$ 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!