Here is our factoring algorithm:
Remove the Greatest Monomial Factor
Check for the Difference of Two Squares
Check for a Perfect Square Trinomial
Factor x2+bx+c
Factor ax2+bx+c
Factor by Grouping
ALWAYS FACTOR COMPLETELY!
p.282 (1-17:odd,25)
Prepare for Lesson Check 11C
12 comments:
journal entry 12/1/10
Today in Mr.Gerson's 4th period class we started off a lesson check that was just another version of yesterday's. It was very challenging, and it ended up that everyone did not do so good. After we finished the lesson check, we graded another person's. After that Mr. Gerson showed us how to do the 6th step in Factoring Algorithm: factor by grouping.
ex.
8x^3+2x^2+12x+3
=2x^2(4x+1)+3(4x+1)
=(2x^2+3)(4x+1)
note:since you have 4x+1 twice, you put what's on the outside of the parenthesis(x^2 and 3)in one parenthesis and 4x+1 in another parenthesis.
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per.4: Analissa T.
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per.6:
Analissa can u change ur display name so ur per. # and last initial are on there ty :]
December 1, 2010 Journal Entry.
Today in math we took the lesson check 11B. It was similar to the one before and I believe everyone was ready. Although a couple problems were challenging, it was easier than yesterday. Today we took short notes on the 6th and the last step we needed to learn of the Factoring Algorithm: Factor by Grouping. Then we learned this by an example problem but it was different than the other problems we saw before. It had 4 terms:
8x^3+2x^2+12x+3
=2x^2(4x+1)+3(4x+1)
=(2x^2+3)(4x+1)
The first line was the give problem.
The second line after you grouped, you would use the first step, Find the Greatest Monomial Factor.
Since you had both (4x+1), you would add 2x^2 and 3 and multiply that to (4x+1). You can also use the AMOM box method.
- Gloria Kim, Period 4.
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per.4: Analissa T. Gloria K.
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per.6:
OMG!!!! I DIDN'T UNDERSTAND HOW TO DO THE PROBS. BUT...
NOW I DOOOOO!!!!!
it's actually pretty easy!!
Journal Entry 12/1/10
In Mr.Gerson's class today, we had a lesson check. It was almost the same from yesterdays, except with different problems. It was a challenge, but some people did better than they did yesterday because they reviewed it. With the short amount of time left, due to people that were still finishing up their lesson checks, Mr. Gerson taught us how to factor by grouping.
Ex:8x^3+2x^2+12x+3
=2x^2(4x+1)+3(4x+1)
=(2x^2+3)(4x+1)
This problem had four terms
journal entry 12/1/10
Today in Mr.Gerson's 5th period class,he passed out the lesson check that was very similar to yesturday's. It was easier the 2nd time but many people didnt do well. After everyone was done we graded eachothers paper. Then Mr.Gerson showed us the 6th step in factoring.
Ex:
8x^3+2x^2+12x+3
=2x^2(4x+1)+3(4+1)
=(2x^2+3)(4x+1)
or use the box method.
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per.3: Sara L.
per.4: Analissa T. Gloria K.
per.5: Tessa P.
per.6:
i am kinda confused on #25
anyone have suggestions?
Factor each as a difference of two squares:
25) (x-y)² -z²
@ willy
for # 25...
the - of 2 ^2...
16-25=(4+5)(4-5)
#25) (x+y)²-z²
(x+4)² is a sqaure so it is part of the ( )( ) part
so the square root of (x+y)² is x+y and the square root of z² is z so u just plug it in
8)
oh. i was taking it a bit too complicated. thanks moet
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