Help with Factoring

how do i factor these:

1. 16a^2-8ab+b^2+c^2+6c-9

2. 9^3-b^3-a-b

pls dont flame me.........
thanks in advance to those who will help me

well my good son the answer is simply
3.14159265

Quote:

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Originally Posted by Saaaaaad

1. umm...are you sure thats -9 and not +9?
The first part with a and b factorises to: (4a-b)^2
if its a +9, then it would be:
(4a-b)^2+(c+3)^2

2. Not sure about this one...

EDIT: @Jerov, your solution to 2 seems a little simple to me. Also, 9^3 = 729, you're thinking of 3^3

EDIT2: I just realised Tasha's only 3 posts...are to ask a maths question. Who joins a game forum to ask a maths question?

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Thank you mr jerov and saddd
yes thats -9, if it was a positive one, it will be very simple but as it it a negative one it makes my head twirl. Uhm I have 3 posts although uhm cant say it.

:( I am so lost

****! This is just like my proofs class!

Alright, HERE'S the factorization that I got for question 1:

Now, we can factorize the a's and b's, so gonna ignore that and focus on the c's.

Now, it's c^2 + 6c - 9. We know that we're not solving for the roots, which takes a load off the brain. Notice something funny about the -9? It's -9 = -7 - 2. And we know that if it's c^2 + 6c - 7, it's easy to factorize: (c + 7)(c - 1). Ssoo what do we get?

(4a - b)^2 + (c + 7)(c - 1) - 2 = factor-x

And I can legally do this since I'm just breaking the -9 into its parts to make it easier.

If this is the wrong answer though, then I'm completely stumped

Edit: uuhh, actually probably wrong up there. All that I get is c + 3 = +/- 3*root(2), so c = (-3)(1 + root(2) and (-3)(1 - root(2)). So it's either gonna be one of these:

A: (4a - b)^2 + (c + 3(1 + root(2))(c + 3(1 - root(2))

B: (4a - b)^2 + (c + 3)^2 - 18

I personally prefer B in terms of simplification, but dunno if it's complete factorization

don't make us do your homework you lazy bastard

Quote:

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Originally Posted by Jerov

****! This is just like my proofs class!

Alright, HERE'S the factorization that I got for question 1:

Now, we can factorize the a's and b's, so gonna ignore that and focus on the c's.

Now, it's c^2 + 6c - 9. We know that we're not solving for the roots, which takes a load off the brain. Notice something funny about the -9? It's -9 = -7 - 2. And we know that if it's c^2 + 6c - 7, it's easy to factorize: (c + 7)(c - 1). Ssoo what do we get?

(4a - b)^2 + (c + 7)(c - 1) - 2 = factor-x

And I can legally do this since I'm just breaking the -9 into its parts to make it easier.

If this is the wrong answer though, then I'm completely stumped

Edit: uuhh, actually probably wrong up there. All that I get is c + 3 = +/- 3*root(2), so c = (-3)(1 + root(2) and (-3)(1 - root(2)). So it's either gonna be one of these:

A: (4a - b)^2 + (c + 3(1 + root(2))(c + 3(1 - root(2))

B: (4a - b)^2 + (c + 3)^2 - 18

I personally prefer B in terms of simplification, but dunno if it's complete factorization

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hey thanks now your B answer i think it the most logical as it broke down the -9 at the 2nd tri, I really appreciated it sir. I sincerely thank you.

@neutral
uhm sir if i am lazy, i wont really do anything about finding solution for these probs. And its not a homework =_=, its a refresher as Im switching career and already forget some of my math lessons. but its your opinion and i cant do anything about it.

You made an account on here to ask us math problems?