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Factoring

Page history last edited by PBworks 16 years, 4 months ago

What is Factoring?

Factoring can be a very intimidating thing, but once you get the hang

of it you will be a pro.

 

First, let me make sure you understand what factoring is. To factor

anything, you get it into its smaller component parts, finding out

what smaller parts make up the bigger part. For instance, when you

factor a number, say 12, you break it into its smallest parts. You

could start with 6 x 2. Then 6 can be factored into 2 x 3, so your

factors of 12 could be 2 x 2 x 3. So to factor x^2 + 5x + 4 is to break

it down into simpler parts.

 

You may have already done some problems like this:

 

 

   (x + 3)(x - 4)

 

Multiplying it out gives you:

 

    x^2 - 4x + 3x - 12

 

Simplifying:

 

    x^2 - x - 12

 

There are some very important things to notice here: you get the first

term of the trinomial, x ^2 (they call it a trinomial because three

terms are being added) by multiplying the FIRST terms in each

parenthesis (x * x).

 

The last number in the trinomial, the constant (meaning it has no x's

in it)  (-12) comes from multiplying the LAST  terms in each

parenthesis (3  *  -4).

 

To get the middle term (the x term, which in this case is - x), first

multiply the first term in the first parenthesis (x) by the last term in

the second parenthesis (-4) (we can call these terms the OUTSIDE terms

because they are the outside terms of the two parentheses). Then multiply

the last term in the first parenthesis (3) by the first term in the

second parenthesis (x) (we can call the terms in the second parenthesis

the INSIDE terms because they are the inside terms of the two parentheses).

Then add the "inside" and the "outside" together. Thus the middle term

becomes 3x - 4x = -x.

 

You can also remember this by remembering the word  F O I L.  Foil

stands for First, Outside, Inside, Last.

 

To review:  in (x + 3)(x - 4)  the First is x* x or x^2,  the Outside

is x * (-4)  (or -4x)  the Inside is3 * x  (or 3x)  and the Last is

(3) * (-4) (or -12)

 

(x + 3) and (x - 4) are factors of x^2 - x - 12

 

Factoring a trinomial is just doing the reverse of what I just did in

the problem above. You want to get from the trinomial back to the

factors. First make sure your problem is written in the right way,

that is, the term with x^2 comes first (the first term), the term with

x (also called the middle term) comes next, and the constant - the term

with no x in it - comes last. In your problem x^2 is the first term,

5x is the middle term, and 4 is the last term.

 

Step 1 is to write the problem on your paper:

 

      x^2 + 5x + 4

 

Step 2 is to write two empty parentheses like this:

 

    (          )(         )    

 

We do this because we know the factors of a trinomial look this way.

 

Step 3 is to look at the first term, which is x^2. We know from the

problem I did above that the x^2 term comes from multiplying the first

term in the first parenthesis by the first term in the second

parenthesis. What are the only two things that can be multiplied together

to give you x^2?  That's right!  x * x will give you x^2. You write it

like this:

 

       (x         )(x       )

 

Step 4 is to look at signs. Remember from the problem above that the

constant term (the one with no x in it) is found by multiplying

the last terms of each parenthesis. Because the sign of the constant

term in your problem (4) is plus, the signs of both the numbers that

multiply together to give you 4 have to be either positive or negative

(because a + * + or a - * - gives you a positive number, and a + * -

will give you a negative number). So your parentheses could now look

like this:

 

       (x -  )(x -  )  or this:  (x +    )(x +  )

 

Look back at my previous problem to see what gave you a middle term. 

Remember you are adding the "outside" to the  "inside." If both signs

are minus, this will give you a minus number; if both signs are plus

it will give you a plus number. Because +5x is positive, you know that

the signs have to be positive. Thus you get:

 

         (x +    )(x +    )

 

Step 5 is the tricky part. You know that the last term is found by

multiplying the last two numbers in each parenthesis together. This

means the numbers could either be 4 and 1, or 2 and 2 (because

4 x 1 = 4 or 2 x 2 = 4). But you also know that the inside terms

multiplied together plus the outside terms multiplied together will

give you the middle term, so you have to try them out. Let's try

2 and 2:

 

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

 

Whoops! The middle term is not right. We need 5x not 4x. Now let's

try 4 and 1:

 

      (x + 4)(x + 1) = x^2 + 4x + 1x + 4  = x^2 + 5x +4

 

We got it. The factors of x^2 + 5x + 4 are (x + 4) and (x +1).

 

Let's try another one: 

 

   Factor  x^2 + x - 2

 

Here are our parentheses:

 

    (        ) (         )

 

Remember F O I L   (first, outside, inside, last)

x^2 is the first terms multiplied together and the only thing they can

be are x and x:

 

     (x        )(x         )

 

The sign in front of the constant is a minus  (-2) so the signs have

to be different (this is the only way we can get a -2, found by

multiplying the last terms).

 

       (x +  ) (x -   )

 

Now, because the term in front of the x term (+x) is a plus we know

that when we multiply the outside and inside terms together and add

them to each other, we need to end up with a plus. Because the signs

are different, the bigger term will have to be a plus. Now the only

factors of 2 are 2 and 1 so they will be the only numbers we will 

have to worry about.

 

If we put 2 and 1 in like this:

 

     (x +2) (x -1) = x^2 -  x + 2x - 2 = x^2+ x - 2

 

If we had reversed the 2 and the 1 we would have ended up with -x

instead of +x.

 

That was a long explanation. The main things to remember are FOIL

(first, outside, inside, last) and that sometimes it may take a

little trial and error to get the right facts. Always multiply the

factors back out to make sure you end up with what you started with.

 

 

 

 

 

 

 

 

 

 

Comments (20)

Anonymous said

at 9:16 pm on Oct 19, 2007

Good Job, looks like you worked hard enough to get to this part of the project. Good luck.

Anonymous said

at 10:40 pm on Oct 20, 2007

thanks

Anonymous said

at 9:29 pm on Nov 3, 2007

nice... looks like something from last yera :P

Anonymous said

at 9:48 am on Nov 5, 2007

dang brandon! great start looks awsome you should add more color though

Anonymous said

at 9:55 am on Nov 5, 2007

guud job brandon

Anonymous said

at 1:04 pm on Nov 5, 2007

wow! nice brandon you did a very nice job

Anonymous said

at 1:04 pm on Nov 5, 2007

dood this is good! add more colors to make it look more funn! :)

Anonymous said

at 1:31 pm on Nov 5, 2007

oh nice comic. long explianitions cool

Anonymous said

at 2:35 pm on Nov 5, 2007

you really need some colors, and you need to show the examples in a colorful way.

Anonymous said

at 2:43 pm on Nov 5, 2007

pretty nice brandons

Anonymous said

at 2:46 pm on Nov 5, 2007

hey brandon make your page more colorful and it will be perfect. more pictures too!

Anonymous said

at 2:54 pm on Nov 5, 2007

needs a bigger font

Anonymous said

at 2:54 pm on Nov 5, 2007

needs a bigger font

Anonymous said

at 2:58 pm on Nov 5, 2007

explain why you break down the numbers in your examples.

Anonymous said

at 4:55 pm on Nov 5, 2007

great job, maybe a bigger font, and more colors!!!

Anonymous said

at 5:57 pm on Nov 5, 2007

awsome page

Anonymous said

at 9:29 pm on Nov 5, 2007

Brandon; your page looks really good.
i like the cartoony thing at the end. it`s pretty sick.
&& there`s a lot of information that makes it very clear. :]]

Anonymous said

at 9:41 pm on Nov 5, 2007

brandon put more coloring but very nice after all

Anonymous said

at 4:48 pm on Nov 7, 2007

i love the colors you used

Anonymous said

at 5:21 pm on Nov 21, 2007

Great explanation, what about links? What about factoring things that are not trinomials, like numbers?

3 of 4 points
10.5 of 12 total project points

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