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.
The given expression can't be factorised: factorise 8x^2-12x-6xy+9 - Wolfram|Alpha
$\displaystyle 8x^2 - (12 + 6y)x + 9$
$\displaystyle = 8\left(x^2 - \frac{6 + 3y}{4}x + \frac{9}{8}\right)$
$\displaystyle = 8\left[x^2 - \frac{6 + 3y}{4}x + \left(-\frac{6 + 3y}{8}\right)^2 - \left(-\frac{6 + 3y}{8}\right)^2 + \frac{9}{8}\right]$
$\displaystyle = 8\left[\left(x - \frac{6 + 3y}{8}\right)^2 - \frac{36 + 36y + 9y^2}{64} + \frac{72}{64}\right]$
$\displaystyle = 8\left[\left(x - \frac{6 + 3y}{8}\right)^2 - \frac{9y^2 + 36y -36}{64}\right]$
$\displaystyle = 8\left[\left(x - \frac{6 + 3y}{8}\right)^2 - \left(\frac{3\sqrt{y^2 + 4y - 4}}{8}\right)^2\right]$
$\displaystyle = 8\left(x - \frac{6 + 3y}{8} + \frac{3\sqrt{y^2 + 4y - 4}}{8}\right)\left(x - \frac{6 + 3y}{8} - \frac{3\sqrt{y^2 + 4y - 4}}{8}\right)$
$\displaystyle = 8\left(\frac{8x + 3\sqrt{y^2 + 4y - 4} - 6 - 3y}{8}\right)\left(\frac{8x - 3\sqrt{y^2 + 4y - 4} - 6 - 3y}{8}\right)$.
I believe this is factorised...
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.
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:
$\displaystyle = -\frac{36+ 36y+ 9y^2}{64}+ \frac{72}{64}$
$\displaystyle = -\frac{36+ 36y+ 9y^2}{64}- \frac{-72}{64}$
$\displaystyle = -\frac{9y^2+ 36y+ 36- 72}{64}= -\frac{9y^2+ 36- 36}{64}$