Expert: Paul Klarreich - 5/27/2006

Question
Hi Paul:
Just starting back to school after 5 years and I have been stuck on a problem for a while, if you could please refresh my memory i wouod greatly appreciate it.
I have to show my work for G(x)=x³ when x=(2+h)
I have forgotten how to foil when it is G(x)=x³
Thanks
Lisa

Answer
Hi, Lisa,

You wrote:
Subject:  garphing a function
Question:  Hi Paul:
Just starting back to school after 5 years and I have been stuck on a problem for a while, if you could please refresh my memory i wouod greatly appreciate it.
I have to show my work for G(x)=x³ when x=(2+h)
I have forgotten how to foil when it is G(x)=x³
Thanks
Lisa
-----------------------

The proper vocabulary is 'removing parentheses' or 'simplifying'.  Foil is good for leftovers, but not for either baked potatoes or math.

Note:  Exponents don't come through this crude interface very well.  I saw yours -- you wrote 'x to the third' -- but I don't know how to do that, and so I'll write:

G(x) = x^3.

Now you want G(2 + h).  That means substitute (2 + h) for x into the expression for the function, and ALWAYS USE PARENTHESES.

G(2 + h) = (2 + h)^3.

You might remember something called the binomial expansion and can write this as an example of:

(a + b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3

with  a = 2  and  b = h.  Then it comes out to:

(2 + h)^3 = 2^3 + 3(2^2) h + 3(2)h^2 + h^3
= 8 + 12h + 6h^2 + h^3

If not, then you write:

(2 + h)^3 = (2 + h)^2 (2 + h) =

(4 + 4h + h^2)(2 + h) =

at this point you just have to multiply out and there are SIX PARTIAL PRODUCTS:

Three terms times 2 and
three terms times h.

8 + 8h + 2h^2   +   4h + 4h^2 + h^3

Now you combine like terms:
   --   ====       --   ====
8 + 8h + 2h^2   +   4h + 4h^2 + h^3

8 + 12h + 6h^2 + h^3