This is the problem:

$\displaystyle (x + 3)(3x - 4)$

My math teacher taught us this:

F- first

O- outside

I- inside

L- last

So would I go something like this?

$\displaystyle (x * 3x) + (3 * 3x) + (x * - 4) + (3 * 4)$

I'm really confused.

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- Feb 5th 2008, 02:11 PMZach[SOLVED] Multiplying Polynomials?
This is the problem:

$\displaystyle (x + 3)(3x - 4)$

My math teacher taught us this:

**F**- first

**O**- outside

**I**- inside

**L**- last

So would I go something like this?

$\displaystyle (x * 3x) + (3 * 3x) + (x * - 4) + (3 * 4)$

I'm really confused. - Feb 5th 2008, 02:15 PMJhevon
no. the red 4 should be -4. you got it (except you did FIOL instead of FOIL :D)

just multiply out and simplify

you do know that "first" means "the first terms in each bracket" and analogous meanings for the others?

anyway, what FOIL basically boils down to, is taking each term in one set of brackets and multiplying every term in the other set - Feb 5th 2008, 02:17 PMZach
So:

**F**- first terms in each bracket

**O**- ??

**I**- ??

**L**- ?? - Feb 5th 2008, 02:20 PMZach
I simplified and got this:

$\displaystyle (x+3)(3x-4)$

$\displaystyle (3x^2)+(9x)+(-4x)+(12)$

$\displaystyle 3x^2+5x+12$ - Feb 5th 2008, 02:22 PMJhevon
- Feb 5th 2008, 02:23 PMJhevon
- Feb 5th 2008, 02:24 PMZach
- Feb 5th 2008, 02:29 PMZach
Another problem:

$\displaystyle (2x+7)(2x-7)$

My solution:

$\displaystyle (2x+7)(2x-7)$

$\displaystyle (4x^2)+(-14x)+(14x)+(-49)$

$\displaystyle 4x^2-49$ - Feb 5th 2008, 02:37 PMJhevon
you are correct (Clapping)

you could have also realized that this was the form of the difference of two squares and get the answer in one step (if you don't know what "the difference of two squares" is referring to, forget about it, you will soon enough. it seems that practicing FOIL is your objective, and it will always work for multiplying two binomials) - Feb 5th 2008, 02:38 PMZach
- Feb 5th 2008, 02:41 PMJhevon