Factoring and Solving Equations with Factors(ing)

Princezz3286

Junior Member
Joined
Nov 12, 2005
Messages
66
Ok, I have used this site before and it was awesome, but I am in college.... taking a refresher course and I have never been good at this stuff even in my highschool days. I have a few questions and I would like to make sure that my overall processes are correct. Here are a few problems that I am working with....

directions are Factor Completely:

9x^2 + 12x + 4
On this I would just like to confirm that it already is factored because we can't take anything out of this and it does not break down into two factors.


16a^2 - 40a + 25
On this problem, I was able to factor it down to (4a - 5) and (4a - 5) as the factors, on this I would like to know if writing it like this (4a - 5)^2 would make my answer incorrect or if this is ok


(a+b)c^2 - 5(a+b)c - 24 (a+b)
My question on this is can I eliminate all of the (a+b)'s and work with the remaining c^2 -5c - 24? If I can do that my answer is (c - 8)(c - 3)


(c +4d)^2 - (c - 4d)^2
On this problem, I broke it down into exactally what it is:
(c +4d)(c +4d) - (c - 4d)(c - 4d)
from here I put it into trinomial form:
c^2 + 8dc + 16d^2
- c^2 - 8dc + 16d^2
----------------------------------
I cancelled out the c^2 and the 8dc, well they pretty much cancelled themselves.... and I was left with 32d^2 are my procedures here correct?

I have a few more questions but I will ask those a little later, I don't want to take up too much time here as I am aware, like myself, there are many other confused students!

Thanks in advance,
Heather
 
Princezz3286 said:
Ok, I have used this site before and it was awesome, but I am in college.... taking a refresher course and I have never been good at this stuff even in my highschool days. I have a few questions and I would like to make sure that my overall processes are correct. Here are a few problems that I am working with....

directions are Factor Completely:

9x^2 + 12x + 4 = (3x + 2)(3x + 2) = (3x + 2)[sup:14z6weq1]2[/sup:14z6weq1]

On this I would just like to confirm that it already is factored because we can't take anything out of this and it does not break down into two factors.


16a^2 - 40a + 25
On this problem, I was able to factor it down to (4a - 5) and (4a - 5) as the factors, on this I would like to know if writing it like this (4a - 5)^2 would make my answer incorrect or if this is ok

it's ok ... see the previous response


(a+b)c^2 - 5(a+b)c - 24 (a+b) = (a + b)(c[sup:14z6weq1]2[/sup:14z6weq1] - 5c - 24) = (a + b)(c - 8)(c + 3)
My question on this is can I eliminate all of the (a+b)'s and work with the remaining c^2 -5c - 24? If I can do that my answer is (c - 8)(c - 3)


(c +4d)^2 - (c - 4d)^2 = [(c + 4d) + (c - 4d)][(c + 4d) - (c - 4d)] = (2c)(8d) = 16cd
On this problem, I broke it down into exactally what it is:
(c +4d)(c +4d) - (c - 4d)(c - 4d)
from here I put it into trinomial form:
c^2 + 8dc + 16d^2
- c^2 - 8dc + 16d^2
----------------------------------
I cancelled out the c^2 and the 8dc, well they pretty much cancelled themselves.... and I was left with 32d^2 are my procedures here correct?

I have a few more questions but I will ask those a little later, I don't want to take up too much time here as I am aware, like myself, there are many other confused students!

Thanks in advance,
Heather
 
ok, so i see where I messed up on the first 3, but can you maybe explain what happens with the last one, or how to work it?

thanks!
 
look at the pattern for factoring the difference of two squares ...

a[sup:2kfnkbkh]2[/sup:2kfnkbkh] - b[sup:2kfnkbkh]2[/sup:2kfnkbkh] = (a + b)(a - b)
 
Ok, now that I see where you got [(c + 4d) + (c - 4d)][(c + 4d) - (c - 4d)] what did you do to eliminate to get (2c)(8d)?
 
combine like terms in the [ ... ]

e.g.

(c + 4d) + (c - 4d) = c + c + 4d - 4d = 2c
 
and for the other bracket c - c + 4d + 4d = 8d

(2c)(8d)= 16cd!

Oh my gosh! I get it, I just don't know if I will recognize these things on a test but..... I can at least find more practice problems..... : ) THANK YOU VERY MUCH!

Heather
 
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