1. ## Factorisation

Hello All, wishing all memebers a great start to the New-Year!
I'm not sure if my working is entirely ok below... Tried using latex..no idea why it didnt work..?

Factorise: [tex] 50x^2-10xy-20y^2 [\math]
10(5x^2-xy+2y^2)
10[(5x^2 - 2y^2) - xy]
10[( \sqrt5x + \sqrt2y)^2 -xy )]

10[( \sqrt5x + \sqrt2y) ( \sqrt5x + \sqrt2y) - 7.32xy ] * completing the square

2. $\displaystyle \displaystyle 50x^2 - 10xy - 20y^2 = 50\left(x^2 - \frac{1}{5}xy - \frac{2}{5}y^2\right)$

$\displaystyle \displaystyle = 50 \left[x^2 - \frac{1}{5}xy + \left(-\frac{1}{10}y\right)^2 - \left(-\frac{1}{10}y\right)^2 - \frac{2}{5}y^2\right]$

$\displaystyle \displaystyle = 50\left[\left(x - \frac{1}{10}y\right)^2 - \frac{1}{100}y^2 - \frac{40}{100}y^2\right]$

$\displaystyle \displaystyle = 50\left[\left(x - \frac{1}{10}y\right)^2 - \frac{41}{100}y^2\right]$

$\displaystyle \displaystyle = 50\left[\left(x - \frac{1}{10}y\right)^2 - \left(\frac{\sqrt{41}}{10}y\right)^2\right]$

$\displaystyle \displaystyle = 50\left(x - \frac{1}{10}y - \frac{\sqrt{41}}{10}y\right)\left(x - \frac{1}{10}y + \frac{\sqrt{41}}{10}\right)$

$\displaystyle \displaystyle = 50\left(x-\frac{\sqrt{41} + 1}{10}y\right)\left(x + \frac{\sqrt{41} - 1}{10}y\right)$

3. We have:

$\displaystyle p(x,y)=10(5x^2-xy-2y^2)$

Solving on $\displaystyle x$

$\displaystyle 5x^2-xy-2y^2=0$

we obtain

$\displaystyle x=\dfrac{y\pm \sqrt{41}y}{10}=\{x_1(y),x_2(y)\}$

So,

$\displaystyle p(x,y)=10(x-x_1(y))(x-x_2(y))=\ldots$

Fernando Revilla

4. Originally Posted by BobBali
Factorise: 50x^2-10xy-20y^2
Are you sure that's not 50x^2 - 10xy - 40y^2 ?
Factors "nicely" to: (x - 1)(5x + 4)

5. Originally Posted by Wilmer
Are you sure that's not 50x^2 - 10xy - 40y^2 ?
Factors "nicely" to: (x - 1)(5x + 4)
I'm sure you mean $\displaystyle \displaystyle 10(x-y)(5x+4y)$...

6. Originally Posted by Prove It
I'm sure you mean $\displaystyle \displaystyle 10(x-y)(5x+4y)$...
Hmmm....
I was taught (50 years ago!) to go this way:
50x^2 - 10xy - 40y^2
Divide by 10:
5x^2 - xy - 4y^2
Now factor...

HOW serious is it to "drop" that 10 ?

7. When you're asked to factorise you shouldn't "drop" terms... You can only "drop" terms when you're solving an equation.

Besides, the bigger mistake is the fact that you left the $\displaystyle \displaystyle y$ terms out...

8. No Sir, unfortunately its does not end with -40y^2; it is -20y^2 and so far the replies i have gotten from some serious "super members", i mean the math they show to factorise this is just beyond me..

9. You can use FernandoRevilla's post which is the most elementary.

$\displaystyle 5x^2-xy-2y^2=0$

Here, use the quadratic formula $\displaystyle x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ to get:

$\displaystyle x=\dfrac{y\pm \sqrt{41}y}{10}$

You only need to substitute those now in the previous expression.

Oh and the LaTeX works with the forward slash '/' and not backslash '\' in the math tabs.

10. Originally Posted by Prove It
When you're asked to factorise you shouldn't "drop" terms... You can only "drop" terms when you're solving an equation.
Besides, the bigger mistake is the fact that you left the $\displaystyle \displaystyle y$ terms out...