# [SOLVED] Multiplying Polynomials?

• Feb 5th 2008, 02:11 PM
Zach
[SOLVED] Multiplying Polynomials?
This is the problem:

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

My math teacher taught us this:
F - first
O - outside
I - inside
L - last

So would I go something like this?
$(x * 3x) + (3 * 3x) + (x * - 4) + (3 * 4)$

I'm really confused.
• Feb 5th 2008, 02:15 PM
Jhevon
Quote:

Originally Posted by Zach
This is the problem:

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

My math teacher taught us this:
F - first
O - outside
I - inside
L - last

So would I go something like this?
$(x * 3x) + (3 * 3x) + (x * - 4) + (3 * {\color{red}4})$

I'm really confused.

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 PM
Zach
So:

F - first terms in each bracket
O - ??
I - ??
L - ??
• Feb 5th 2008, 02:20 PM
Zach
I simplified and got this:

$(x+3)(3x-4)$
$(3x^2)+(9x)+(-4x)+(12)$
$3x^2+5x+12$
• Feb 5th 2008, 02:22 PM
Jhevon
Quote:

Originally Posted by Zach
So:

F - first terms in each bracket
O - the outside terms, that is, the terms on the ends
I - the inside terms, the terms that are closest together
L - the last terms in each bracket, that is, the right-most terms

so, if we are talking about (a + b)(c + d) for example, we have:

F: a and c, so we do a*c

O: a and d, so we do a*d

I: b and c, so we do b*c

L: b and d, so we do b*d
• Feb 5th 2008, 02:23 PM
Jhevon
Quote:

Originally Posted by Zach
I simplified and got this:

$(x+3)(3x-4)$
$(3x^2)+(9x)+(-4x)+(12)$
$3x^2+5x+12$

you made the same mistake i told you about last time. it is a minus 4. so you must do (3 * -4) = -12
• Feb 5th 2008, 02:24 PM
Zach
Quote:

Originally Posted by Jhevon
so, if we are talking about (a + b)(c + d) for example, we have:

F: a and c, so we do a*c

O: a and d, so we do a*d

I: b and c, so we do b*c

L: b and d, so we do b*d

Thank you so much I finally understand!
• Feb 5th 2008, 02:29 PM
Zach
Another problem:

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

My solution:

$(2x+7)(2x-7)$
$(4x^2)+(-14x)+(14x)+(-49)$
$4x^2-49$
• Feb 5th 2008, 02:37 PM
Jhevon
Quote:

Originally Posted by Zach
Another problem:

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

My solution:

$(2x+7)(2x-7)$
$(4x^2)+(-14x)+(14x)+(-49)$
$4x^2-49$

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 PM
Zach
Quote:

Originally Posted by Jhevon
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)

Yeah I'm in 8th grade Saxon Algebra 1.

Thanks so much you were of great help. (Rofl)
• Feb 5th 2008, 02:41 PM
Jhevon
Quote:

Originally Posted by Zach
Yeah I'm in 8th grade Saxon Algebra 1.

Thanks so much you were of great help. (Rofl)

you're welcome :). you were a great helpee!

"666" appears in my post count again!