Help with math problem

bhaskar2 · #1 $Old$ October 18th, 2009, 10:46 AM

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.

skeeter · #2 $Old$ October 18th, 2009, 11:15 AM

Quote:

Originally Posted by bhaskar2 $View Post$

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.

$f(x) = (2x+4)^4$

$f(a) = (2a+4)^4$

$f(x) - f(a) =$

$(2x+4)^4 - (2a+4)^4 =$

$[(2x+4)^2 + (2a+4)^2][(2x+4)^2 - (2a+4)^2] =$

$[(2x+4)^2 + (2a+4)^2][(2x+4) + (2a+4)][(2x+4)-(2a+4)] =$

$[(2x+4)^2 + (2a+4)^2][2(x+a+4)][2(x-a)]$

divide the last expression by $x-a$

redsoxfan325 · #3 $Old$ October 18th, 2009, 11:19 AM

Quote:

Originally Posted by bhaskar2 $View Post$

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.

$(2x+4)^4-(2a+4)^4=((2x+4)^2-(2a+4)^2)((2x+4)^2+(2a+4)^2)$ by the difference of squares identity.

$(2x+4)^2-(2a+4)^2=((2x+4-(2a+4))(2x+4+2a+4)=4(x-a)(x+a+4)$ by difference of squares again.

So the whole thing is $\frac{4(x-a)(x+a+4)((2x+4)^2+(2a+4)^2)}{x-a}$

Cancel the $x-a$ and then simplify.

EDIT: Beat to it!