Topic ID: 25898
|
View previous topic :: View next topic |
| Author |
Message |
1dering at stars
Senior Writer
Gender: 
Age: 15
Joined: 07 Dec 2007
Posts: 194
Reviews: 99
Country: East of the sun and West of the moon
300 Points
|
 Posted: Thu Feb 14, 2008 12:58 am Post subject: Algebra 1 |
 |
|
OK, so its algebra one and were factoring expressions. The problem is:
2x squared + 9x + 9
The way we have learned to do this is to make a diamond problem with the 9 from 9x on the bottom (sum) and the 9 on the top (product). So you would have to find two numbers that have a sum and a product of 9, right? Then you add x to both of them. So it would be: (x + #) (x + #) OK. So I'm wondering what to do with the 2x squared after you've got this far? |
_________________
.... And I wish I could do something great no one's ever done;
Catch a cloud, or steal a star, or sail 'round the sun.... |
|
| Back to top |
|
|
|
Flemzo
Now With 50% More Flem!!
Novelist
Gender: 
Age: 18
Joined: 31 Dec 2006
Posts: 412
Reviews: 134
Country: United States
336 Points
|
| Posted: Thu Feb 14, 2008 1:18 am Post subject: |
|
|
We learned kind of differently in my Algebra I class. We were told to, when factoring, to find the "Head" (which, in this case, is 2x^2), the "Tail" (which is 9), and the "Guts" (which is 9x).
When checking your factoring [(x + #) (x + #)], the "Head" is found by multiplying the x's, the "Tail" by multiplying the numbers, and the "Guts" by multiplying the outsides and the insides, and adding them together. It's commonly called the FOIL method (Firsts, Outsides, Insides, Lasts), which is basically taking the x's, the first x and the last number, the first number and the second x, and the two numbers.
For example, if you have the problem x^2 + 8x + 16, and you want to factor it, you need to take the "Head" and "Tail", then the "Guts" will just come.
Let's start with the x^2. Obviously, you get x^2 when you multiply two x's. So you can just put them in your factoring problem:
(x + #) (x + #)
Now, go to 16. It has to be something times something, so write out your factors:
(1.16) (2,8) (4,4) (8,2) (16,1)
Now, out of those factors, which two, when added together, equal your "Guts", 8? Obviously, four. So plug those in:
(x + 4) (x + 4)
Now check, using FOIL:
FIRSTS: x * x = x^2
OUTSIDES: x * 4 = 4x
INSIDES: 4 * x = 4x
LASTS: 4 * 4 = 16
Add your ousides and insides:
4x + 4x = 8x
And see if it fits:
x^2 + 8x + 16
Hooray! You did it!
Now, as for your 2x^2, you just need to figure out what times what makes 2x^2. Obviously it'll be x * x, but where does the two come in?
HINT: your factoring might be (#x + #) (x + #) |
_________________
"How strange life is. How fragile. You never know what stunning development lies around the next corner."
-- From The Corner Of His Eye, by Dean Koontz |
|
| Back to top |
|
1dering at stars
Senior Writer
Gender:
Age: 15
Joined: 07 Dec 2007
Posts: 194
Reviews: 99
Country: East of the sun and West of the moon
300 Points
|
| Posted: Thu Feb 14, 2008 1:26 am Post subject: |
|
|
| Wow this really helps a lot. I'm starting to remember learning this. Thank you so much. You just completely saved my grade on this 60 problem worksheet. |
_________________
.... And I wish I could do something great no one's ever done;
Catch a cloud, or steal a star, or sail 'round the sun.... |
|
| Back to top |
|
Sleeping Valor
^_^ Back for summer!
Speaker of the Forum
Gender:
Age: 17
Joined: 12 Jan 2006
Posts: 940
Reviews: 207
Country: I'm on the other side of the reflection you see, living in a world of fantasy.
300 Points
|
| Posted: Thu Feb 14, 2008 1:34 am Post subject: |
|
|
My notes are all in french, so here's hoping my translations make sense.
When I write "#^2" I mean the number is squared.
Factorizing y=ax^2+bx+c
Case 1 : a=0 (y=x^2+bx+c)
1) Add and substract the square of half x's coefficient (half of b quared).
2) Groupe the termes of the perfect trinome together.
3) Put the perfect trinome into the form of a square of a binome.
Example: x^2+6x+8
1) y = x^2 + 6x + 9 - 9 + 8 (9=half of 6 squared)
2) y = [x^2 + 6x + 9) - 9 + 8
3) y = (x + 3)^2 - 1 (You do the sum product thing)
In this case, it's when you can't do sum/product right away. The technique is called completing the square(diamond in english, maybe).
Case 2 : a = Anything other than 0 (y=ax^2+bx+c)
1) Group de terms with the variable x.
2) Factorize by the coefficient of x^2 (with braquets)
3) Complete the diamond (square) inside the braquets.
4) Inside the braquets, find the perfect trinome and turn it into the square of a binome.
5) Developp to get rid of the braquets.
Example : y=2x^2+9x+9
1) y = [2x^2 + 9x] + 9
2) y = 2[x^2 + (9/2)x] + 9
3) y = 2[x^2 + (9/2)x + (9/4)^2 - (9/4)^2] + 9 (o.O I hate this technique, it gets super complicated with uneven numbers >.< I'll try and remember if there's another way we learned it)
4a) y = 2[(x^2 + (9/2)x + (9/4)^2) - (9/4)^2] + 9
4b) y = 2[(x+[9/4])^2 - (9/4)^2] + 9 (sum/product)
5) y = 2[x+(9/4)]^2 - 2(9/4)^2 + 9 (Sorry, i'd give you actual number but I have no calculator)
You'll notice, that you end up with an annoying amoung of decimal numbers. Not much fun, easier with a good calculator. I'll hunt around my notes and see if there was another way.
Now, did any of that make sense?
Edit: lol. Someone beat me to it. Their way is simpler, so hopefully that works for you. =P |
_________________
Here's a free coupon! Good for:
1) A new friend
2) A free review
3) Advice on problems and general YWS assistance
^_^ PM me anytime to use! |
|
| Back to top |
|
|
