It's been driving me up the wall and this is the fastest math forum I've found on google.

Here's the problem.

8x^2-12X-6XY+9

How to factor???

Thank you in advance.

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- Mar 9th 2010, 09:28 PMnikzbordPlease help me figure this out!
It's been driving me up the wall and this is the fastest math forum I've found on google.

Here's the problem.

8x^2-12X-6XY+9

How to factor???

Thank you in advance. - Mar 9th 2010, 10:38 PMmr fantastic
- Mar 9th 2010, 10:40 PMProve It
- Mar 9th 2010, 10:46 PMmr fantastic
The given expression can't be factorised: factorise 8x^2-12x-6xy+9 - Wolfram|Alpha

- Mar 9th 2010, 11:33 PMProve It
$\displaystyle 8x^2 - (12 + 6y)x + 9$

$\displaystyle = 8\left(x^2 - \frac{6 + 3y}{4}x + \frac{9}{8}\right)$

$\displaystyle = 8\left[x^2 - \frac{6 + 3y}{4}x + \left(-\frac{6 + 3y}{8}\right)^2 - \left(-\frac{6 + 3y}{8}\right)^2 + \frac{9}{8}\right]$

$\displaystyle = 8\left[\left(x - \frac{6 + 3y}{8}\right)^2 - \frac{36 + 36y + 9y^2}{64} + \frac{72}{64}\right]$

$\displaystyle = 8\left[\left(x - \frac{6 + 3y}{8}\right)^2 - \frac{9y^2 + 36y -36}{64}\right]$

$\displaystyle = 8\left[\left(x - \frac{6 + 3y}{8}\right)^2 - \left(\frac{3\sqrt{y^2 + 4y - 4}}{8}\right)^2\right]$

$\displaystyle = 8\left(x - \frac{6 + 3y}{8} + \frac{3\sqrt{y^2 + 4y - 4}}{8}\right)\left(x - \frac{6 + 3y}{8} - \frac{3\sqrt{y^2 + 4y - 4}}{8}\right)$

$\displaystyle = 8\left(\frac{8x + 3\sqrt{y^2 + 4y - 4} - 6 - 3y}{8}\right)\left(\frac{8x - 3\sqrt{y^2 + 4y - 4} - 6 - 3y}{8}\right)$.

I believe this is factorised... - Mar 10th 2010, 01:53 AMHallsofIvy
**Any**expression can be factored if you allow any numbers (irrational and/or complex) as coefficients. Mr Fantastic is correct that this cannot be factored with**integer**coefficients which is what is usually meant by "factoring". Prove It is correct that what he gives is a factoring using irrational terms. - Mar 10th 2010, 07:09 AMnikzbord
What happened to 72/64??

- Mar 10th 2010, 08:16 AMHallsofIvy
I confess that I find this a very puzzling post. Is it missing?

Okay, I went back an looked at Prove It's post.**That**72/64!

Prove it has:

$\displaystyle = -\frac{36+ 36y+ 9y^2}{64}+ \frac{72}{64}$

$\displaystyle = -\frac{36+ 36y+ 9y^2}{64}- \frac{-72}{64}$

$\displaystyle = -\frac{9y^2+ 36y+ 36- 72}{64}= -\frac{9y^2+ 36- 36}{64}$