• Mar 9th 2010, 09:28 PM
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???

• Mar 9th 2010, 10:38 PM
mr fantastic
Quote:

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.
• Mar 9th 2010, 10:40 PM
Prove It
Quote:

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 $(-12 - 6y)x$.
• Mar 9th 2010, 10:46 PM
mr fantastic
Quote:

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
• Mar 9th 2010, 11:33 PM
Prove It
$8x^2 - (12 + 6y)x + 9$

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

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

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

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

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

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

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

I believe this is factorised...
• Mar 10th 2010, 01:53 AM
HallsofIvy
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 AM
nikzbord
What happened to 72/64??
• Mar 10th 2010, 08:16 AM
HallsofIvy
Quote:

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

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

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