Shashank Seeram
Mrs. Guzinski/Mr. Branagh
Math/Period 3
4/1/09
FACTORING AFTERMATH
2. Number 16 in the test is the problem that I could have been able to do but I "blew off" due to less time. Anyway, here is the answer.
16th question: An expression for the area of the rectangle below is xy-2x+3y-6. Find the length and width in terms of x and y.
a. x and (y-2)
b. 2xy and -3y
c. xy-2x and 3y-6
d. (x+3) and (y-2)
16th answer: we should take out something that is common from each term: x(y-2)+3(y-2) then you make the terms like this: (x+3) (y-2)
so the answer is d. (x+3) and (y-2).
3. Number 1 in the test: 2c2d4 - 8c3d5 , I got wrong and here is a similar problem
Question: 3a3n3-9a2n5
Answer: Pick smoething you could take out in everything, so it would be: 3a2n3(a-3n2). The answer would be 3a2n3(a-3n2), but on the test I just put the term I took out so that is why I lost points there. Anyhow, here the answer would be 3a2n3(a-3n2).
4. The two strengths in this test for me are factorizing completely, and solving terms when it is equal to 0. I had only one weakness because I got a 91 which is solving
word problems. I got how to factorize terms completely but I did not get how to solve one word problem which confused me alot but now I got to know my mistake
and did the problem in number 2.
5.
#1(on the test) 2c2d4 - 8c3d5 = 2c2d4(1-4cd)
#16, did this problem in number 2
6.
a. 2m3n2 - 16mn2 + 8mn = take common terms out like this: 2mn(m2n - 8n + 4)
b. 3x2a - 7xb -6c = multiply a term with c term, then you should find numbers that mutiply to -18 and add to -7. Then it is 3x2 - 9x + 2x - 6, after you take common term
out. 3x(x-3)+2(x-3), then you have to make it as (3x+2) (x-3)
c. 8ac - 2ad + 4bc - bd, you have to take common terms out so it would be 2a(4c-d)+b(4c-d). Then you make it as (2a+b) (4c-d).
