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???

2. Originally Posted by nikzbord
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???

You can't. Clearly the given expression is incorrect.

3. Originally Posted by mr fantastic
You can't. Clearly the given expression is incorrect.
Actually you can complete the square if you notice that you can write the middle term as $(-12 - 6y)x$.

4. Originally Posted by Prove It
Actually you can complete the square if you notice that you can write the middle term as $(-12 - 6y)x$.
The given expression can't be factorised: factorise 8x&#x5e;2-12x-6xy&#x2b;9 - Wolfram|Alpha

5. $8x^2 - (12 + 6y)x + 9$

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

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

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

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

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

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

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

I believe this is factorised...

6. 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.

7. What happened to 72/64??

8. Originally Posted by nikzbord
What happened to 72/64??
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:
$= -\frac{36+ 36y+ 9y^2}{64}+ \frac{72}{64}$

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

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