Math Help - Even more factoring!

1. Even more factoring!

9-4x^2: I don't know where to begin with this one. There are no examples in the book and no odd numbered similar problems so that I could possibly reverse engineer the problem.

12a^2 +24a: for this one I started with 12(a+1)(a+2). From there I went on to several different variations but came up empty. Wait... is 12a(a+2) the right answer?

Any and all help is much appreciated. Thanks!

2. The easiest way to do this problem would be to first factor out the -4
$-4\left(x^2 - \frac{9}{4}\right)$
Now notice this is a difference of squares so we can factor it as
$-4\left(x-\frac{3}{2}\right)\left(x+\frac{3}{2}\right)$

3. Originally Posted by Ingersoll
9-4x^2: I don't know where to begin with this one. There are no examples in the book and no odd numbered similar problems so that I could possibly reverse engineer the problem.

12a^2 +24a: for this one I started with 12(a+1)(a+2). From there I went on to several different variations but came up empty. Wait... is 12a(a+2) the right answer?

Any and all help is much appreciated. Thanks!
Assuming you want to write these as a pair of factors...

$9-4x^2$ is the "difference of two squares"

$3^2-(2x)^2=(3-2x)(3+2x)$

Notice when you multiply this out you get $(3+2x)(3-2x)=3(3-2x)+2x(3-2x)=3^2-6x+6x+2x(-2x)$

You correctly corrected yourself on the other.

Sorry Bacterius...let's call it a dead-heat!

4. There is an easier way for the first one ... Just note that $4x^2 = (2x)^2$ and you already have a difference of squares, without fractions.

EDIT : beaten to it

5. Okay. The reason I was having difficulty with it is because it appeared to have been written backwards. As 4x^2 -9, I'm okay. So then what I should do when a difference of squares problem is written this way is to simply rewrite it as 4x^2 -9 and proceed normally?

You correctly corrected yourself on the other.

Sorry Bacterius...let's call it a dead-heat![/QUOTE]

6. So then what I should do when a difference of squares problem is written this way is to simply rewrite it as 4x^2 -9 and proceed normally?
Why would you want to do that ? Just notice that $4x^2$ is a square as well as $9$ and off you go, right ?

Sorry Bacterius...let's call it a dead-heat!
It's ok, I really need we need some sort of mutex system on MHF where people can be warned that someone has posted just before them ... so they can reconsider their post. I might bump the idea soon if no-one can be bothered to notify it.

7. [QUOTE=Ingersoll;541232]Okay. The reason I was having difficulty with it is because it appeared to have been written backwards. As 4x^2 -9, I'm okay. So then what I should do when a difference of squares problem is written this way is to simply rewrite it as 4x^2 -9 and proceed normally?
[QUOTE]

$4x^2-9=(2x)^2-(3)^2=(2x-3)(2x+3)$

$9-4x^2=(3)^2-(2x)^2=(3-2x)(3+2x)=-(2x-3)(3+2x)=-(4x^2-9)$

8. Originally Posted by Ingersoll
Okay. The reason I was having difficulty with it is because it appeared to have been written backwards. As 4x^2 -9, I'm okay. So then what I should do when a difference of squares problem is written this way is to simply rewrite it as 4x^2 -9 and proceed normally?
As others have pointed out $x^2- y^2$ and $y^2- x^2$ are both the "a difference of squares"- but not the same thing: $4x^2- 9$ and $9- 4x^2$ are different expressions and have different factorings.

You could have written $9- 4x^2= -(4x^2- 9)= -(2x-3)(2x+3)$ but $9- 4x^2= (3- 2x)(3+ 2x)$ is simpler.

You correctly corrected yourself on the other.

Sorry Bacterius...let's call it a dead-heat!