Can [math]y^2-2xy-3x[/math] be factored?

I tried to do it:

[math]y^2-(2y-3)x[/math]

[math]\rightarrow[/math][math]y(y-3)x-1(y-3)x[/math] and that finally gave me

[math](y-3)x(1+y)[/math] Is this correct?

Edited April 18, 20169 yr by Chikis

No it cannot, since if you multiply [latex](y-3)x(1+y)[/latex] it would be: [latex]y^2x-2xy-3x[/latex]
If you wanted to factor

it would be
[latex]y^2-x(2y+3)[/latex]

OR

[latex]y(y-2x)-3x[/latex]

but that would be pretty much the end of it. In my eyes at least, it may be wrong

Edited April 18, 20169 yr by bluescience

So nothing can be done again?

So nothing can be done again?

What do you want to do?

 

 

You can write y as a function of x for example. You can construct the general solution for the equation you have given.

Just take look at this expression, am required to reduce it to its lowest term.

[math]\frac{3(x^2-y^2)}{y^2-2xy-3x}[/math]

Factoring the numerator, we get

[math]\frac{3(x+y)(x+y)}{y^2-2xy-3x)}[/math], but doing a perfect factoring at the denominator to see how terms could cancel out becomes a major problem. So how does one go about this?

Edited April 19, 20169 yr by Chikis

You have a minus sign wrong. You need the difference of two squares formula.

 

 

I am not sure what the simplest way of writing your expression is. Do you have a target expression given to you?

You have a minus sign wrong.

Which minus sign are you talking about here?

In the difference of two squares...

 

Expand out (x+y)(x+y) and then (x+y)(x-y) you see that the minus sign makes a big difference.

In the difference of two squares...Expand out (x+y)(x+y) and then (x+y)(x-y) you see that the minus sign makes a big difference.

Oh! Sorry! I made a mistake there. I intended writing [math]\frac{3(x+y)(x-y)}{y^2-2x-3x}[/math] before I typed that trash. Now that the mistake has been corrected, can the denominator be factored out completely to see the terms that could cancel out?

Edited April 21, 20169 yr by Chikis

[math]\frac{3(x+y)(x-y)}{y^2-2xy-3x}[/math]

Well the solution to the problem above is [math]\frac{(y-x)}{(x-\frac{y}{3})}[/math]. The issue now is that I cannot give any justification to the solution. Any help here?

Edited April 22, 20169 yr by Chikis