Factorisation

**BobBali** · January 28th 2011, 02:18 AM

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

**Prove It** · January 28th 2011, 02:28 AM

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

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

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

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

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

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

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

**FernandoRevilla** · January 28th 2011, 02:33 AM

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

**Wilmer** · January 28th 2011, 03:58 AM

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)

**Prove It** · January 28th 2011, 04:41 AM

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 $\displaystyle 10(x-y)(5x+4y)$ ...

**Wilmer** · January 28th 2011, 05:01 AM

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 ?

**Prove It** · January 28th 2011, 05:04 AM

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...

**BobBali** · January 28th 2011, 07:06 AM

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..

**Unknown008** · January 28th 2011, 07:57 AM

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.

**Wilmer** · January 28th 2011, 08:59 AM

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.

Of course...my bad.

Originally Posted by Prove It

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

Geesh...having a real bad day...
With those, I drop the "y"s while doing "the work", then put 'em back in
once finished...IF I don't forget

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