Factorisation

• Jan 28th 2011, 02:18 AM
BobBali
Factorisation
Hello All, wishing all memebers a great start to the New-Year! (Happy)
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
• Jan 28th 2011, 02:28 AM
Prove It
$\displaystyle 50x^2 - 10xy - 20y^2 = 50\left(x^2 - \frac{1}{5}xy - \frac{2}{5}y^2\right)$

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

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

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

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

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

$\displaystyle = 50\left(x-\frac{\sqrt{41} + 1}{10}y\right)\left(x + \frac{\sqrt{41} - 1}{10}y\right)$
• Jan 28th 2011, 02:33 AM
FernandoRevilla
We have:

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

Solving on $x$

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

we obtain

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

So,

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

Fernando Revilla
• Jan 28th 2011, 03:58 AM
Wilmer
Quote:

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)
• Jan 28th 2011, 04:41 AM
Prove It
Quote:

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 $\displaystyle 10(x-y)(5x+4y)$...
• Jan 28th 2011, 05:01 AM
Wilmer
Quote:

Originally Posted by Prove It
I'm sure you mean $\displaystyle 10(x-y)(5x+4y)$...

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 ?
• Jan 28th 2011, 05:04 AM
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 $\displaystyle y$ terms out...
• Jan 28th 2011, 07:06 AM
BobBali
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..
• Jan 28th 2011, 07:57 AM
Unknown008
You can use FernandoRevilla's post which is the most elementary.

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

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

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

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.
• Jan 28th 2011, 08:59 AM
Wilmer
Quote:

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 $\displaystyle y$ terms out...